CBSE Class 9 Science Chapter 9 Question Answer
CBSE Class 9 Science Chapter 9 Question Answer – Force and Laws of Motion
Science studies various multidisciplinary phenomenons of the universe and its constituents. One can learn about its various functions and principles through multiple theories and laws put forward by different scientists. Science as a subject lays the foundation for numerous sciences and thus is an integral part of a student’s academic studies.
The chapter ‘Force and Law of Motion’ studies different forces we use daily. Also, one learns about the various laws of motion that exist in the universe for working consistently. The vital topics included in Chapter 9 Class 9 Science important questions are:
- Balanced and unbalanced focus
- First law of motion
- Inertia and the mass
- Second law of motion
- Third law of motion
- Conservation of momentum
One can find all the questions based on the key concepts and the vital points covered in the important questions in Class 9 Science Chapter 9 in a detailed manner. That gives students a clear understanding of every concept they have learnt and helps them retain it for a longer period of time. As a result, they get good command over the subject and score well in their examinations.
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Get Access to CBSE Class 9 Science Important Questions 2022-23 with Solutions
Also, get access to CBSE Class 9 Science Important Questions for other chapters too:
CBSE Class 9 Science Important Questions | ||
Sr No | Chapters | Chapter Name |
1 | Chapter 1 | Matter in Our Surroundings |
2 | Chapter 2 | Is Matter Around Us Pure |
3 | Chapter 3 | Atoms and Molecules |
4 | Chapter 4 | Structure of Atom |
5 | Chapter 5 | The Fundamental Unit of Life |
6 | Chapter 6 | Tissues |
7 | Chapter 7 | Diversity in Living Organisms |
8 | Chapter 8 | Motion |
9 | Chapter 9 | Force and Laws of Motion |
10 | Chapter 10 | Gravitation |
11 | Chapter 11 | Work and Energy |
12 | Chapter 12 | Sound |
13 | Chapter 13 | Why Do We Fall ill |
14 | Chapter 14 | Natural Resources |
15 | Chapter 15 | Improvement in Food Resources |
Force and Laws of Motion CBSE Class 9 Science Chapter 9 Question Answer
The following important some of the important questions and their solutions have been picked from the Science Class 9 Chapter 9 important questions:
Question 1. Which of the following statements is not correct for an object moving along a straight path in an accelerated motion?
(a) A force is always acting on it
(b) Its velocity constantly changes
(c) It always moves away from the Earth’s surface
(d) Its speed keeps changing
Answer 1.
The answer is ( c ): It always goes moves from the Earth’s surface
Explanation:
To move away from Earth, an object’s acceleration must be more than the acceleration due to gravity. To escape from gravity, only moving along a straight path is not only required; thus, option (c) is the wrong statement.
As per the third law of motion, action and reaction (a) always act on the same given body, (b) always act on the different bodies in the opposite directions, and (c) have the same magnitude as well as directions
Question 2. According to the third law of motion, action and reaction
(a) act on either body at average to each other
(b) always act on different bodies in the opposite direction
(c) have the same magnitude and directions
(d) always work on the same body
Answer 2.
The answer is (b) always act on different bodies in the opposite directions
Explanation:
Newton’s Third Law states: “For every action, there is an equal and opposite reaction.” So, the answer is b)
Question 3. A goalkeeper in the game of football pulls his hands backwards after holding the ball shot at the goal. That enables the goalkeeper to
(a) decrease the rate of change in momentum
(b) reduce the force exerted by the ball on the hands
(c) increase the rate of change in momentum
(d) exert more significant force on the ball
Answer 3.
The answer is (b) to reduce the force exerted by the ball on the hands.
Explanation:
Pulling hands backwards would help the goalkeeper reduce the ball’s momentum, which will, in turn, reduce the force exerted on the Goalkeeper’s hands.
Question 4. The inertia of an object tends to cause the object
(a) to decelerate due to friction
(b) to decrease its speed
(c) to resist any change in its state for motion
(d) to increase its speed
Answer 4.
The answer is (c) to resist any change in its state of motion
Explanation:
Inertia is a property that resists the state of motion of the given object. The object remains in the existing condition of rest or uniform motion in the straight line unless any external force changes that state.
Question 5. A passenger in a moving train tosses the coin which falls behind him. It means that the motion of the train is
(a) accelerated
(b) along circular tracks
(c) retarded
(d) uniform
Answer 5.
The answer is (a) accelerated
Explanation:
If the motion of the particular train is uniform, the coin would have fallen entirely in his hand. When the action is retarded, the currency would have fallen ahead of him. As the coin falls behind the person, the motion of that train is accelerated.
Question 6. An object of mass 2 kg is sliding with a constant velocity of 4 m/s on a frictionless horizontal table. The force required to keep the thing moving at the same speed is
(a) 2 N
(b) 0 N
(c) 32 N
(d) 8 N
Answer 6.
The answer is (b) 0 N
Explanation:
In this case, the friction opposes the force; thus, no force is required to keep the object in a uniform motion. So, the answer is 0 N.
Question 7. Rocket works on the principle of the conservation of
(a) mass
(b) energy
(c) momentum
(d) velocity
Answer 7.
The answer is (c) momentum
Explanation:
The velocity of the hot gases coming out of the rocket provides strong momentum in the opposite direction, making the missile move upwards. Here the conservation of the required momentum takes place.
Question 8. A water tanker filled up to 2/3 of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would
(a) move backwards
(b) move forward
(c) be unaffected
(d) rise upwards
Answer 8.
The answer is (b) move forward
Explanation:
On applying brakes, the given water tanker comes to rest however water will be in motion. That makes the water come forward.
Question 9. A batsman hits a cricket ball which then rolls on level ground. After covering a short distance, the ball comes to rest. The ball comes s to a stop because –
- Velocity is proportional to the force exerted on the ball.
- The batsman didn’t hit the ball hard enough.
- There is a force on the ball opposing the motion.
- There is no unbalanced force on the ball. Thus the ball would come to rest.
Answer 9. (c) There is a force on the ball opposing the motion.
Question 10. What is the momentum of an object of mass m, moving with the velocity v?
- (mv)2
- 1/2mv2
- mv2
- Mv
Answer 10. (d) mv
Question 11. What is the S.I. unit of momentum?
- Kgms
- Mskg
- kgms−1
- kgms
Answer 11. (c) kgms−1
Question 12. What is the numerical formula for force?
- F=ma
- F=ma
- F=ma2
- F=m2a
Answer 12. (a) F=ma
Question 13. If the initial velocity is zero, the force acting is
- Retarding
- Acceleration
- Both
- None
Answer 13. (a) Acceleration
Question 14. What is the S.I. unit of the force?
- kgms2
- kgms
- kgm2s2
- kgm2s2
Answer 14. (a) kgms2
Question 15. Newton’s first law for motion is also called
- Law of Inertia
- Law of Momentum
- Law of Action & Reaction
- None of these
Answer 15. (a) Law of Inertia
Question 16. Which law explains swimming?
- Newton’s first law
- Newton’s second law
- Newton’s third law
- All of these
Answer 16. (c) Newton’s third law
Question 17. The S.I. unit of weight is:
- N
- Nm
- Ns
- Nms
Answer 17. (a) N
Question 18. Which equation defines Newton’s Second law of motion?
- F=ma=dpdt
- F=mdadt=P
- dFdt=ma=P
- F=ma=P
Answer 18. (d) F=ma=P
Question 19. The people on the bus are pushed backwards if the bus starts suddenly due to
- Inertia due to rest
- Inertia due to motion
- Inertia due to direction
- Inertia
Answer 19. (a) Inertia due to rest
Question 20. If the force acting on the body is zero. Its momentum is
- zero
- constant
- Both
- None
Answer 20. (b) constant
Question 21. The inability of the body to change its state of rest or motion is
- Momentum
- Force
- Inertia
- Acceleration.
Answer 21. (c) Inertia
Question 22. We want to move a wooden cabinet across the floor with a horizontal force of 200N200N at a constant speed. What is the frictional force exerted on the case?
Answer 22. To move the cabinet across the floor at a constant speed, the net force it experiences could be zero. Thus, the frictional force of 200N must be applied to the cabinet to move it across the floor at a constant speed against the horizontal force of 200N.
Question 23. There are three bodies made of aluminium, steel and wood that have the same shape and volume. Which of them would have greater indolence?
Answer 23. The inertia depends entirely on the mass of the object. For solids of the same size and volume, the inertia is determined by the mass. So steel has the highest inertia.
Question 24. Two balls of the same size but different materials, rubber and iron, are held on the smooth floor of a moving train. The brakes are suddenly applied to stop the train. Are the balls rolling? If yes, in which direction? Will they move at the same speed? Justify your answer.
Answer 24. Yes, the balls will start rolling in the given direction. When the brakes are applied, the train stops and then the balls try to stop while the balls’ inertia keeps moving and starts rolling. Because the mass between two balls is not equal, the inertia of the iron ball is greater than the inertia of the rubber ball. So the rubber ball rolls faster than the iron ball.
Question 25. Two identical bullets are fired, one from a light rifle and another one from a heavy rifle of the same power. Which rifle will hurt the shoulder more, and why?
Answer 25. The amount of movement of the bullet depends entirely on the amount of reverse movement of the rifle. According to Newton’s third law for the motion: The recoil speed is equal to the speed of the bullet, making it lighter. Hence the momentum of the lighter rifle is greater than that of the heavier rifle. The bullet fired from the lighter rifle will do more damage to the shoulder.
Question 26. A horse continues to exert force to move a cart at a constant speed. Explain why.
Answer 26. When a cart moves on the given path, it must encounter friction. To keep the wagon moving, the horse must overcome friction which is acting against the motion of the cart. So the horse has to keep applying force to overcome this friction.
Question 27. Suppose a ball of the mass m is thrown vertically upwards with an initial velocity v, where its velocity decreases continuously until it becomes zero. After that, the ball starts falling and reaches the speed v before hitting the ground. That implies that the magnitude of the ball’s initial and final moments are equal. However, it is not an example of conservation of momentum. Explain why.
Answer 27. When the external force is not acting on the given system, the momentum of the given system remains constant. In the above case, the velocity change results from the Earth’s gravitational pull and is therefore not an example of conservation of momentum.
Question 28. Two friends on roller skates face each other 5 m apart. One throws a 2kg ball to the other, who catches it. How does this activity affect the position of the two? Explain your answer
Answer 28. The distance between them would increase. Before throwing the ball, the momentum of both would be zero. To maintain momentum, the person throwing the ball moves backwards. The person catching the ball experiences the net force as it catches and then moves backwards.
Question 29. The lawn sprinkler starts rotating as soon as the water is added. Explain the principle on which it works.
Answer 29. The water sprinkler works on the principle of the third law of motion. As water exits the sprinkler head, an equal and opposite force is exerted on it. That would cause the sprinkler to rotate.
Question 30. Which has more inertia:
(a) a rubber ball and a rock of the same size?
(b) a bicycle and a train?
(c) a five rupee coin and a one rupee coin?
Answer 30. Because inertia depends on the object’s mass, the object with the greater mass has greater inertia. The following things have greater inertia because of their mass:
- A rock of the same size
- train
- five rupee coin
Question 31. Try to determine how often the speed of the ball changes in the following example: “A soccer player kicks a soccer ball at another player of his team who kicks the ball into the goal. The opposing team’s goalkeeper picks up the ball and kicks it towards a player on his team. Also, identify the agent providing the force in each case.
Answer 31. It can be seen that the speed of the soccer ball changes four times.
First, when the soccer player kicks the ball to another player. Second, when that particular player kicks the ball to the Goalkeeper. Third when the Goalkeeper stops that football. Fourth, when the Goalkeeper kicks that ball toward his teammate. Agent providing the force:
- a) The First case is for the First player
- b) The Second case is for the Second player
- c) The Third case is for the Goalkeeper
- d) The Fourth case is for the Goalkeeper
Question 32. Explain why some leaves may fall from a tree when its branch is violently shaken.
Answer 32. When the branch of this tree is shaken, the unit moves in the reciprocating motion. But the inertia of the leaves attached to the branch resists the movement of that branch. Thus, the leaves loosely attached to the branch fall by inertia, while the leaves firmly attached to the branch remain attached.
Question 33. Why do you fall forward when a moving bus slows down and fall back when it accelerates from a standstill?
Answer 33. When the bus accelerates forward from a standstill, the passengers first experience the force that is exerted on them in the backward direction. This is due to their inertia opposing the forward motion. Once the bus starts moving ahead, passengers are in a forward motion state. When the brake is applied, the bus drives towards the parking position. Now the forward force is exerted on the passengers as their inertia resists the change in the motion of the bus. This causes bus occupants to fall forward when sudden brakes are applied.
Question 34. Explain how a horse can pull a cart if action is always equal to the reaction.
Answer 34. When the horse moves forward with the cart attached, it exerts force in the backward direction on the Earth. In the given case, the Earth exerts an equal force in the horse’s opposite direction (forward direction). This force moves the horse and carriage forward. Hence the car moves forward.
Question 35 Explain why it is difficult for a firefighter to hold onto a hose discharging large volumes of water at high speed.
Answer 35. When the firefighter holds the hose that is ejecting large amounts of water at high speed, the reaction force is exerted on him with the water that he is ejecting backwards. That is due to Newton’s third law of motion. Due to the backward force, the firefighter’s stability decreases. Therefore, it is difficult for him to remain stable while holding the hose.
Question 36. A bullet with a mass of 50 g is fired from a gun with a mass of 4 kg with an initial velocity of 35 m s–1. Calculate the rifle’s initial recoil velocity.
Answer 36. Given the mass of the bullet (m1) = 50 g
The mass of the rifle (m2) = 4 kg = 4000 g
The initial velocity of the given bullet fired (v1) = 35 m/s
Suppose the Reverse speed is v2. Since the rifle was initially at rest, the initial momentum for the rifle = 0
The total momentum for the rifle and bullet after the shot = m1v1 + m2v2
According to the law for the conservation of momentum, the total momentum of the given rifle and bullet after the Ignition = 0 (same as initial momentum)
Hence m1v1 + m2v2 = 0
That implies that
v2 = – m1v1/ m2
= – (50 g x 35 m/s)/4000 g
= -0.4375 m/s
The negative sign indicates that the required recoil velocity is opposite to the bullet’s movement.
Question 37. Two objects of mass 100 g and 200 g, are moving along the same line and direction with velocities of 2 ms–1 and 1 ms–1, respectively. They collide after the collision; the first object moves with a speed of 1.67ms-1. Find the speed of the second object.
Answer 37. Let the first object be objected A and the second object B; it is given as:
mass of A (m1) = 100g
mass of B (m2) = 200g
initial velocity of A (u1) = 2m /s
initial velocity of B (u2) = 1 m/s
final velocity of A (v1) = 1.67 m/s
the final velocity of B (v2) =?The total initial impulse is given as follows;
= the initial momentum of A + the initial momentum of B
= m1u1 + m2u2
= (100g) × (2m/s) + (200g) × (1m/s) = 400 g.m.sec-1
By conserving momentum, the total momentum before the collision must equal the total momentum after the collision.
Therefore m1u1 + m2u2 = m1u1 + m2u2 = 400 g.m.sec-1
Solving for v2, we get
v2 = (100 g) × (2 m/s) + (200 g) × (1 m/s) = 400 g.m.sec-1
Therefore v2 = (400 -167) / 200 m.sec-1
v2 = 1.165 m/s
Therefore, the speed of object B after the collision is 1.165 metres per second.
Question 38. An object experiences a net external unbalanced force of zero. Can the object travel at a speed other than zero? If so, indicate the conditions that must be placed on the magnitude and direction of the velocity. If not, please provide a reason.
Answer 38. Yes, the object may be moving at a speed other than zero. An object moving in one direction at a certain constant speed would continue in its state of motion as long as no external and unbalanced forces act on it. To change the object’s motion, an unbalanced external force must act on it.
Question 39. If you hit a carpet with a stick, dust comes out. Explain.
Answer 39. When a certain rug is struck with a stick, the stick exerts a force on the rug that sets it in motion. The inertia of the dust particles in the mat counteracts the change in the movement of the mat. Therefore, the forward movement of the mat exerts a backward force on the dust particles, setting them in motion in the opposite direction. That will help the dust come out of the mat when hit.
Question 40. Why is it recommended to tie down luggage carried on the roof of a bus with a rope?
Answer 40. If the luggage is placed on the roof of the initially stationary bus, the forward acceleration of the bus will exert a reverse force on the luggage. Similarly, when the initially running bus comes to an abrupt stop due to the application of the brakes, the force in the forward direction is applied to the luggage. Depending on the mass of the luggage and the force applied, the luggage can fall off the bus due to inertia. Latching the luggage would secure its position and prevent it from falling off the bus.
Question 41. A batsman hits a cricket ball which then rolls on level ground. After a short distance, the ball comes to rest. The ball stops because
(a) the batsman did not hit the ball hard enough.
(b) Velocity is proportional to the force exerted on the ball.
(c) there is a force acting on the ball resisting movement.
(d) there is no unbalanced force on the ball, so the ball would come to rest. .
Answer 41. When the ball rolls on the flat surface of the floor, its motion is reversed due to the frictional force created between the floor and the ball. This frictional force eventually causes the ball to stop. Therefore the correct answer is (c).
If the flat ground surface is lubricated with oil or other given lubricant, the friction generated between the ball and the ground will be reduced, which allows the ball to roll longer.
Question 42. A truck starts from a standing start and rolls downhill with constant acceleration. It covers a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it when its mass is 7 tons (hint: 1 ton = 1000 kg).
Answer 42. We have,
Distance travelled by truck (s) = 400 metres
The time it takes to travel the distance (t) = 20 seconds
The initial speed of the truck (u) = 0 (since the start of rest position)
From the equation of motion, we already know
s = ut + 1/2at2
So 400 = 0(20s) + ½(a)(400s2) = 2 ms-2
For the given condition, the acceleration of the truck is 2 ms-2
According to the second law of motion,
force = mass x acceleration
mass of the truck = 7 tons = 7000 kg
The force acting on the truck is
= 7000 kg x 2 ms -2
= 14000 kg.ms-2
= 14000 N
So 14000 N force acts on the truck
Question 43. A 1 kg stone is thrown at a speed of 20 ms-1 over the surface frozen in the sea and stops after a distance of 50 m. How big is the frictional force between stone and ice?
Answer 43. Assume
Mass of the stone (m) = 1 kg
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 m/s (the stone reaches a state of rest)
Of the stone(s)( en) distance travelled = 50 m
According to the third equation of motion
v² = u² + 2as
Substituting the values into the previous equation gives
0² = (20)² + 2(a )(50)
-400 = 100a
a = -400/100 = -4m/s² (deceleration)
We already know that
F = m×a
Let’s add the value obtained from a = -4 in F = m x a
We get
F = 1 × (-4) = -4N
Where the negative sign indicates the opposing force, which is friction.
Question 44. An 8000 kg locomotive pulls a train of 5 cars on a horizontal track, each weighing 2000 kg. If the locomotive exerts a force of 40,000 N and the track offers a frictional force of 5,000 N, then calculate: (a) the net acceleration force and (b) the acceleration of the train
Answer 44. (a)
Suppose the force exerted by the train (F) = 40,000 N
frictional force = -5000 N (the negative sign implies that the force is exerted in the opposite direction)
Then the net acceleration force = sum of all forces = 40,000 N + (-5,000 N) = 35,000 N
(b) The total mass of train = mass of the locomotive + mass of each car = 8,000 kg + 5 × 2,000 kg
The total mass of the train is 18,000 kg.
According to the second law of motion, F = ma (or: a = F/m)
Therefore, the acceleration of train = (net acceleration force) / (total mass of train)
= 35,000/ 18,000 = 1.94 ms-2,
Hence, the acceleration of the train is 1.94 m.s-2.
Question 45. A motor vehicle has a mass of 1500 kg. How large must the force between the car and the road be for the car to stop with a negative acceleration of 1.7 ms-2?
Answer 45. Suppose
vehicle mass (m) = 1500 kg
acceleration (a) = -1.7 ms-2
By the second law of motion, F = ma
F = 1500 kg × (-1.7 ms -2) = -2550 N
So the force acting between the car and the road is -2550 N, acting in the opposite direction from the direction of the car’s motion..
Question 46. What is the momentum of a body of mass m moving with velocity v?
(a) (mv)2 (b) mv2 (c) ½ mv2 (d) mv
Answer 46.
The momentum of an object is given by the product of its mass m and its velocity v
Momentum = Mass x Velocity
As a result, the correct answer is mv, i.e. option (d)
Question 47. We want to move a wooden cabinet with a horizontal force of 200 N at a constant speed over the floor. What is the frictional force exerted on the case?
Answer 47. Since the cabinet’s velocity is constant, its acceleration could be zero. Therefore, the effective force acting on it is also zero. That shows that the magnitude of the opposing frictional force is the same as the force exerted on the case, i.e. 200N. So the total frictional force is -200N.
Question 48. Two bodies with a mass of 1.5 kg each move in the same straight line but in opposite directions. The speed of each object is 2.5ms-1 before the collision, during which they stick to each other. What speed will the combined object have after the collision?
Answer 48. Given
mass of the first object, m1 = 1.5 kg
mass of the second object, m2 = 1.5 kg
velocity of the first object before impact, v1 = 2.5 m/s
The speed of the second object moving in the opposite direction, v2 = -2.5 m/s
We already know that
total momentum before collision = total momentum after the collision
m1v1 + m2v2 = (m1 + m2) v
1.5 (2.5) + 1.5 (-2.5) = (1.5 + 1.5)v
3.75 – 3.75 = 3v
v = 0
Therefore, the speed of the combined object after the collision is 0 m/s
Question 49. According to the third law of motion, when we press on an object, the object presses on us with an equal and opposite force. If the object is a huge truck parked on the side of the road, it probably won’t move. One student justifies this by saying that the two equal and opposite forces cancel each other out. Comment on this logic and explain why the truck is not moving.
Answer 49. The truck has a very high mass, and the static friction between the roadway and the truck is high. By pushing the truck with a small force, the applied frictional force cancels the applied force, and the truck does not move. That shows that the two forces are equal in magnitude, Therefore, the student’s logic is correct.
Question 50. A hockey ball with a mass of 200 g and moving at ten ms-1 is hit by a hockey stick so that it returns to its original path with a speed of 5 ms-1. Calculate the magnitude of the change in momentum in the motion of the hockey ball due to the force exerted by the hockey stick.
Answer 50. Given
mass of bullet (m) = 200 g
initial velocity of bullet (u) = 10 m/s
final velocity of bullet (v) = -5 m/s
initial momentum of bullet = mu = 200 g × 10ms-1 = 2000g.m.s-1
Final momentum of the bullet = mv = 200g × –5 ms-1 = –1000 g.m.s-1
So, the change of momentum (mv – mu) = –1000 g.m.s-1 – 2000 g.m.s-1 = –3000 g.m.s-1
That shows that the momentum of the bullet is reduced to 1000 g.m.s-1 when the hockey stick hits.
Question 51. A projectile with a mass of 10 g moving horizontally at a speed of 150 m s–1 hits a stationary block of wood and stops in 0.03 s.Calculate the penetration distance of the bullet into the block. Also, calculate the magnitude of the force the block of wood exerts on the ball.
Answer 51. Assume
bullet mass (m) = 10 g (or 0.01 kg)
initial bullet velocity (u) = 150 m/s
final bullet velocity (v) = 0 m/s
duration (t) = 0.03 s
To find the penetration distance, one could calculate the acceleration of the bullet
Assuming the penetration distance to be s
By the first law of motion,
v = u + at
0 = 150 + a ( 0.03 )
a = – 5000 ms-2
v2 = u2 + 2as
0 = 1502 + 2 x (-5000)s
s = 2.25 m
According to the second law of motion,
F = ma
F = 0.01 kg × (-5000 ms -2)
F = -50N
Question 52. An object of mass 1 kg moving in a straight line at a speed of 10 ms–1 collides with a stationary block of wood of mass 5 kg and sticks to it. Then both walk away together in the same straight line. Calculate the total momentum just before impact. . Also, calculate the speed of the combined object.
Answer 52. Suppose
The mass of the object (m1) = 1 kg
The mass of the block (m2) = 5 kg
The initial velocity of the object (u1) = 10 m/s
The initial velocity of the block ( u2 ) = 0
mass of resulting object = m1 + m2 = 6kg
the velocity of the resulting object (v) =?
Total momentum before impact
= m1u1 + m2u2
= (1kg) × (10m/s) + 0
= 10 kg.m.s-1
According to the law for the conservation of momentum, the total momentum before impact is equal to the total momentum after impact. Hence the total momentum after the collision is also 10 kg.m.s-1
Also (m1 + m2) × v = 10kg.m.s-1`
Therefore v = (10kg.m.s-1 )/ 6kg = 1.66 m.s-1
The resulting object moves at a specified speed of 1.66 metres per second.
Question 53. A body with a mass of 100 kg is accelerated uniformly from a speed of 5 ms–1 to 8 ms–1 in 6 s. Calculate the starting and ending moments of the object. Also, find the magnitude of the force being exerted on the object.
Answer 53. Assume
mass of object (m) = 100 kg
initial velocity (u) = 5 m/s
final velocity (v) = 8 m/s
duration (t) = 6 s
also amount of initial motion (m × u) = 100 kg × 5 m/s = 500 kg.m.s-1
final momentum (m × v) = 100 kg × 8 m/s = 800 kg.m.s-1
acceleration of the object (a) = (v-u)/t = (8-5)/6m.s-2
hence the object accelerates at 0.5 ms-2. This shows that the force (F = ma) acting on the object will be:
F = (100 kg) × (0.5 ms-2) = 50 N
Therefore the force of 50 N on the object will be 100 kg applied, accelerating in 0.5 ms-2.
Question 54. Akhtar, Kiran and Rahul were driving in a car on a highway at high speed when an insect hit the windshield and got stuck on the windshield. Akhtar and Kiran began to think about the situation. Kiran suggested that the bug experienced a greater change in momentum as compared to the change in the momentum of the car (because the bug’s velocity change was much larger than the car’s). Akhtar said that as the car moved faster, it exerted a greater force on the insect. And as a result, the insect died. Rahul, who gave an entirely new explanation, said that both the car and the bug experienced the same force and a change of momentum. Comment on their suggestions.
Answer 54. According to the law for the conservation of momentum, the total momentum before the collision between the bow and the car is equal to the total momentum after the collision. Therefore, the change in momentum of the bug is much larger than the change in momentum of the car because the force is proportional to the mass.
Thus, Akhtar’s assumption is partially correct. Since the car’s mass is quite high, the force exerted on the insect during the collision is also quite high. Kiran’s statement is completely wrong. The momentum change of the bug and car’s momentum change equals the conservation of momentum. The bug’s speed will change accordingly due to its mass since it is quite small compared to the car. Likewise, the speed of a car is quite insignificant since its mass is very large compared to that of an insect.
Rahul’s statement is correct. According to the third law of motion, the force the bug exerts on the car is equal and opposite to the force, the car exerts on the bug. But Rahul’s suggestion that the change in momentum is equal contradicts the law of conservation of momentum.
Question 55. How much momentum does a dumbbell with a mass of 10 kg transfer to the ground when it falls from a height of 80 cm? Assume its downward acceleration to be 10 ms–2.
Answer 55. Given that,
The mass of the dumbbell (m) = 10kg
Distance covered (s) = 80cm = 0.8m
Initial velocity (u) = 0 (it dropped from the state of rest)
Acceleration (a) = 10ms-2
Terminal velocity (v) =?
The momentum of the dumbbell if it hits the ground = mv
According to the third law of motion
v2 – u2 = 2as
Thus, v2 – 0 = 2 (10 ms-2) (0.8m) = 16 m2s-2
v = 4 m/s
The momentum that is transferred by the dumb-bell to the floor = (10kg) × (4 m/s) = 40kgms−1
Question 56. The following is a distance-time table of an object in motion:
Time (seconds) | Distance (metres) |
0 | 0 |
1 | 1 |
2 | 8 |
3 | 27 |
4 | 64 |
5 | 125 |
6 | 216 |
7 | 343 |
(a) What conclusion could you draw about acceleration? Is it constant, increasing, decreasing or zero? (b) What do you conclude about the forces acting on the object?
Answer 56. (a) The distance travelled by an object in a given time interval is greater than any distance travelled in the previously given time intervals. Therefore, the acceleration of the object increases.
(b) By the second law of motion, force = mass × acceleration. Since the object’s mass remains constant, the increase in acceleration shows that the force acting on the object is also increasing.
Question 57. Two people manage to push a car with a mass of 1200 kg at a constant speed on a flat road. Three people can push the same car to produce an acceleration of 0.2 ms-2. With what force does each person push the car? (Assume that all people push the car with the same physical force))
Answer 57. Given
The mass of the car (m) = 1200 kg
If a third person starts to push the car, the acceleration (a ). is 0 .2ms-2. Hence, the force applied by the third person is (F = ma):
F = 1200 kg × 0.2 ms-2 = 240 N
The force applied by the third person on the car is 240 N. Since the three people apply the same physical energy, the force exerted by each person on the car is 240 N.
Question 58. A hammer with a mass of 500 g moving at 50 m s-1 strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
Answer 58. Suppose
The mass of the hammer (m) = 500 g = 0.5 kg
Initial speed of the hammer (u) = 50 m/s
The final speed of the hammer (v) = 0 (the hammer stops and reaches a resting state).
Duration (t) = 0.01 s
So the acceleration of the hammer is:
a = (v-u)/t = (0-50 m/s)/0.01 s
a = -5000 ms-2
So, the force exerted by the hammer on the nail (F = ma) could be calculated as:
F = (0.5kg) * (-5000 ms-2) = -2500 N
According to the third law for motion, the nail exerts an equal and opposite force on the hammer. Since the force that the hammer exerts on the nail is equal to -2500 N, the force that the nail exerts on the hammer is equal to +2500 N.
Question 59. A car with a mass of 1200 kg moves in a straight line at a constant speed of 90 km/h. An unbalanced external force reduces its speed to 18 km/h in 4 s. Calculate the acceleration and the change in momentum. Also, calculate the magnitude of the force required.
Answer 59. Assuming
The mass of the car (m) = 1200 kg
Initial speed (u) = 90 km/h = 25 metres /sec.
Terminal velocity (v) = 18 km/h = 5 metres /sec
Duration (t) = 4 seconds
The acceleration of the car could be calculated with the help of the formula:
a = (v-u)/t,
a = ( 5-25)/4ms-2
= – 5ms– 2
Therefore the acceleration of the car is -5ms-2.
Car initial momentum = m × u = (1200 kg) × (25 m/s) = 30,000 kg.m.s-1
Car final momentum = m × v = (1200 kg) × (5 m/s ) = 6000kg.m.s-1
Therefore the momentum change is
= end momentum – start the momentum
= (6,000 – 30,000) kg.m.s-1
= -24,000 kg.m.s-1
External force applied = mass of car × acceleration = (1200 kg) × (-5 ms-2) = -6000 N
Hence the magnitude of the force required to bring the car’s speed to 18 km/h to reduce is 6000 N
Question 60. An object experiences a net external unbalanced force of zero. Is it possible that the object is travelling at a speed other than zero? If so, indicate the conditions that must be placed on the magnitude and direction of the velocity. If not, please justify.
Answer 60.
- Yes, any object can move with nonzero velocity if it experiences a zero, unbalanced net external force. That is based on Newton’s first law for motion, which states that an object would change its state of motion only if the net force exerted on it is nonzero.
- Therefore, to change the motion or to bring about the motion, an external and unbalanced force is required.
- In this case, the object experiences a net zero imbalance force, which will cause it to move at a nonzero velocity, which previously caused the object to move at a constant velocity.
Question 61. If you hit a carpet with a stick, dust comes out. To explain.
Answer 61. If you hit a carpet with a stick, dust comes out due to Newton’s first law for motion, the law of inertia. Initially, dust particles and the carpet are in a dormant state. When the mat is hit with a stick, the mat moves while the dust particles resist the change in motion due to the inertia of rest.. Therefore, the forward movement of the mat exerts a backward force on the dust particles, causing them to move in the opposite direction.
Hence the dust comes off the mat.
Question 62. Why is it recommended to tie luggage carried on the roof of a bus with a rope?
Answer 62. Due to Newton’s first law of motion, the law of inertia, it is recommended that any luggage kept on the roof of the bus must be tied with a rope.
When the bus is moving, the baggage will move in inertia, and when the bus stops suddenly, the baggage tends to resist this change in motion, causing it to move forward and fall off if not bound by the rope. .
In the same way, if the bus decelerates while turning or changes its direction of motion, the inertia of motion of the baggage would try to counteract this change of action, which would cause the baggage to move in the opposite direction and fall if it is not tied with a rope
Question 63. A 1 kg stone is thrown over the frozen surface of a lake at a speed of 20 ms−1 and stops after a distance of 50 m. What is the frictional force between the stone and the ice?
Answer 63. Given:
Mass of the stone: m=1kg
The initial speed of the stone: u=20ms−1
The final speed of the stone: v=0ms−1 (at a standstill)
Distance travelled on ice: s=50m
Found: Frictional force between stone and ice. First, we need to calculate the delay:
We already know that
v2 = u2+2as
So
02= (20)2+2a(50)
0 = 400+100a
−400 =100a
a = − 4ms– 2
The negative sign indicates delay.
Next,
We need to find the frictional force:
F = ma
F = (1)(−4)
F = −4N
So the frictional force between the rock and the ice is −4N.
Question 64. A motor vehicle has a mass of 1500 kg. How large must the force between the vehicle and the road be if the vehicle is to stop with a negative acceleration of 1.7ms−2
Answer 64. Given:
mass of the vehicle: m = 1500kg
negative acceleration: a =− 1 .7ms−2
Found: Frictional force acting between road surface and vehicle.
We already know that – F=ma
F=(1500)(−1.7)
F=−2550N
Therefore, when the vehicle has to stop with negative acceleration, the force between the vehicle and the road is 1 .7 ms−2 equals −2550 N.
Question 65. An object with a mass of 100 kg accelerates smoothly from a speed of 5ms−1 to 8ms−1 in 6s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force being exerted on the object.
Answer 65. Given:
Object mass: m=100kg
Initial object velocity: u= 5ms-1
Final object velocity: v= 8ms-1
Acceleration duration: t=6s
To find:
Initial impulse
Final impulse
Force exerted on the given object
We already know that
momentum = mass × velocity
initial momentum = mass × initial velocity
initial momentum = 100 × 5
initial momentum = 500 kgms-1
final momentum = mass × final velocity
final momentum=100×8
final momentum=800kgms-1
Well, the force – F=ma
F=m(v−ut)
F=100(8−56)
F=100(36)
F=50N
So
initial impulse: 500 kgs-1
final impulse: 800 kgs -1
Force exerted on the object: 50N
Question 66. What is the momentum of a frame of mass 200g shifting with a speed of 15ms−1?
Answer 66. Given that:
Mass of the frame:m=200g=0.2kg
The velocity of the frame: v=15ms-1
To find the momentum of the frame.
We already recognize that
momentum=mass×speed
momentum=0.2×15
momentum=3kgms-1
Hence, the momentum of the frame is 3kgms-1
Question 67. Define pressure and what are the diverse kinds of forces.
Answer 67. Force approaches the frenzy or pulls at the item, which produces extra withinside the kingdom or form of an object. It may also purpose the extra withinside the pace and the route of the item’s movement.
The diverse kinds of forces performing are:
- Mechanical pressure
- Gravitational pressure
- Frictional pressure
- Electrostatic pressure
- Electromagnetic pressure
- Nuclear pressure
Question 68. A pressure of 25N25N acts on a mass of 500g resting on a frictionless surface. What is the acceleration produced?
Answer 68. Given that:
Mass:m=500g=0.5kg
Force exerted: F=25N
To find acceleration.
We already recognize that
F=ma
a=F/m
a=25/0.5
a=50ms-2
So, the acceleration produced is 50ms-3
Question 69. Give Newton’s first law of motion.
Answer 69. Newton’s first law for motion is also known as the law of inertia. It states that an object would remain in a state of rest or uniform motion in a straight line unless compelled to change state by an applied force. An object at rest would continue to be still, just as a moving object would continue to move unless some external force acts on it.
Question 70. A body with a mass of 5 kg takes off and rolls down an inclined plane 32 m in 4 s. Find the force acting on the body.
Answer 70. Given:
mass of the body: m=5kg
the initial velocity of the body: u=0ms−1 (while it starts rolling)
distance travelled on the incline: s=32m s=32m
duration of rolling: time: t =4s
Found: Force acting on the body.
First
We need to find the acceleration:
We already know that
s=ut+1/2at2
So 32=(0×4)+1/2(a×42)
32=12(a×16)
32 =( a ×8)
a=4ms-2
Then
Find the force:
F=maF=ma
F=(5×4)
F=20N
Therefore the force acting on the body is 20N
Question 71 On a given planet, a small rock thrown vertically at 15 ms−1 takes 7.5 s to return to the ground. What is the gravitational acceleration on the planet?
Answer 71. Given:
the initial speed of the stone: u=15ms−1
the final speed of the stone: v=0ms−1 (since it becomes zero at the highest point)
the total duration of flight when thrown up and falling to the ground: t= 7.5st=7.5s
Found: The gravitational acceleration of the planet.
We already know that
v=u+at
So 0=15+(at)
t=−15a
That denotes the time needed for half the whole flight. The total duration of the flight is therefore twice as long.
7.5 s=2t
7.5=2(–15a)
So the gravitational acceleration is equal to
a=2×(–15)7.5
a=–4ms–2
Therefore the gravitational acceleration of the planet is −4ms−2
Question 72. Why is the object’s weight greater at the poles than at the equator?
Answer 72. The object’s weight is greater towards the poles than at the equator since the gravitational acceleration is slightly stronger at the poles than at the equator. Because g=GMr2, the acceleration due to gravity is inversely proportional to the square of the given radius. Since the Earth’s radius is larger at the equator than at the poles, the acceleration due to gravity is slightly lower at the equator than at the poles. In addition, we also know that weight is directly proportional to the acceleration of gravity (∵w=m×g⇒wαg)
Using these implications, we can conclude that at the equator, where the radius is larger, the acceleration due to gravity is relatively less, and weight is less. At the poles, where the radius is smaller, the gravitational acceleration is relatively greater, hence, the weight of the object is greater.
Question 73. Why is the passenger pushed forward in a moving bus when the bus suddenly stops?
Answer 73. Passengers seated in the moving bus will be pushed forward when the bus suddenly stops due to inertia because the passenger’s upper body keeps moving while the lower part of the body is that is in contact with the seat remains at rest. This pushes the passenger’s upper body forward in the direction the bus was moving before it stopped.
Question 74. Why does the ship move backwards when the sailor jumps forward?
Answer 74. The ship moves backwards when the sailor jumps forward due to Newton’s third law of motion, which states that there is an equal and opposite reaction for every action. So when the sailor jumps forward, he creates an action force that causes the ship to move backwards. In response, the boat exerts an equal and opposite force (the reaction force) on the sailor, causing her to be pushed forward.
Question 75. Derive the law of conservation of momentum from Newton’s third law.
Answer 75. Newton’s third law of motion shows that for every action, there is an equal and opposite reaction acting on different given bodies.
For example, if we have two bodies, A and B, with masses mA and mB, they move in the same direction along the straight line with different speeds, uA and uB, respectively.
. Believes that other external and unbalanced forces do not act on them.
Suppose uA>uB and any two objects collide with each other.
For example, during the collision, which lasts t, A exerts a force FAB on B, and B exerts a force FBA on A.
For vA and vB, the velocities of the two objects A and B after the collision. We already know that
momentum=mass×velocity
momentum from A before the collision:
mA×uA
Momentum from A after the collision:
mA×vA
Momentum from B before the collision:
mB×uB
Momentum from B after the collision:
mB×vB
We also know that force can also be defined as the rate of change of momentum,
F = ma = m(v−ut) = mv−mut
Also, the rate of change of momentum of A during collision is mAvA – mAuAt, which is the FAB force.
And the rate of change of momentum of B during the collision is mBvB−mBuBt, and the force is FBA.
According to Newton’s third law of motion, the force FAB exerted by A on B and the force FBA exerted by B on A is equal but opposite in direction.
FAB=−FBA
Using the formulas, we get
FAB=−FBA
mAvA−mAuAt=−(mBvB−mBuBt)
Simplified
mAvA−mAuA=−(mBvB−mBuB)
By rearranging is
mAvA+mBvB=mAuA+mBuB
where
mAuA+mBuB
is the sum of the moments of A and B before the collision and
mAvA+mBvB
is the sum of the moments of A and B after the collision.
This equation states that the final momentum of the two objects after the collision equals the initial momentum of the two objects before the collision.
So the law of conservation of momentum for an isolated system is that the sum of the total initial momentum for an event equals the total initial momentum.
Question 76. An astronaut has a mass of 80 kg on Earth.
- i) What is its weight on Earth?
- ii) What will be its mass and weight on Mars with gm=3.7ms-2?
Answer 76. i) Given:
Mass of the astronaut: m=80kg
Found: his weight on the ground.
We already know that
acceleration of gravity: ge=10ms−2
acceleration of Mars: gm=3.7ms−2
weight: w=m×g
weight on Earth: we =m × ge
we=80×10
we=800N
- ii) What will be its mass and weight on Mars with gm=3.7ms-2?
Answer 76. ii) Given:
Mass of the astronaut:m=80kg
Find his mass and weight on Mars.
We already know that,
acceleration of gravity: ge=10ms−2
acceleration of Mars: gm=3.7ms−2
weight: w=m×g
weight on Mars: wm= m× gm
wm=80×3.7
wm=296Nw
The mass of the astronauts remains the same on Mars because it is a constant value.
The mass on Mars is m=80kg.
Question 77. Which of the following has more inertia:
a). A rubber ball and a stone of the same size?
Answer 77. The inertia depends entirely on the mass of the object. The greater mass, the greater inertia and vice versa. In the case of the rubber ball and the stone of the same size, it is clear that the stone would have greater inertia than the ball. Despite the same size, the stone weighs more than the rubber ball.
b). A bike and a train?
Answer 77. Inertia depends entirely on the object’s mass: the greater mass, the greater inertia and vice versa. In the case of the bicycle and a train, it is clear that since the train weighs more than the bicycle, the train will have relatively greater inertia than the bicycle.
c). A five rupee coin and a one rupee coin?
Answer 77. The inertia depends entirely on the object’s mass: the greater mass, the greater inertia and vice versa. In the case of the 5 rupee coin and the 1 rupee coin, the 5 rupee coin will have relatively higher inertia than the 1 rupee coin because the 5 rupee coin weighs more than the 1 rupee coin. Coin.
Question 78. Try to determine how often the speed of the ball changes in the following example:
“A soccer player kicks a soccer ball to another player on his team, who kicks the ball into the goal. The opposing team’s goalkeeper picks up the ball and kicks it towards a player on his team.
Also, indicate the agent who will provide the force in each case.
Answer 78. In the example, the number of times the soccer ball changes speed is four.
(i) The speed of the ball changes when the first player kicks the ball towards another player on his team. Where the agent exerting the force is the soccer player’s foot kicking the ball.
(ii) Second, the speed of the ball changes when the other player kicks the ball towards the goal, where the agent applying the force is the foot of the other player who is now kicking the ball towards the goal. (iii) The speed of the ball changes a third time when the opposing team’s Goalkeeper stops the ball. Where the agent exerting the force are the hands of the Goalkeeper picking up the ball.
(iv) The speed of the ball changes for the fourth time when the Goalkeeper shoots toward his team player. The agent applying the force is the Goalkeeper’s foot, who now kicks the ball towards his teammate.
Question 79. Explain why some leaves may fall from a tree when its branch is violently shaken.
Answer 79. Some of the leaves may detach from the tree if the branch is vigorously shaken since the components of the tree would be set in motion while the leaves tend to continue in their dormant state. That is due to the inertia of rest. The force of the shaking would quickly act on the leaves as the direction changed, causing the leaves to break off and fall off the tree.
Question 80. Why do you fall forward when a moving bus stops and backwards when it accelerates from a standstill?
Answer 80. Passengers seated in the moving bus are pushed forward by inertia when the bus suddenly stops because the passenger’s upper body continues to move while the lower body which is in. contact with the seat remains at rest. That pushes the passenger’s upper body forward in the direction the bus was moving before it stopped.
Similarly, when the bus accelerates from a standstill, passengers on the bus are pushed backwards by inertia because the passenger’s upper body remains at rest while the lower body is in contact with the seat. As a result, the passenger’s upper body is pushed back against the direction of the bus.
Question 81 If action is always equal to reaction, explain how a horse can pull a cart.
Answer 81. In such cases, we are dealing with unbalanced forces. It is true that the horse exerts an action force on the carriage and experiences a reaction force from the carriage. But the horse also generates an action force on the ground it is walking on and experiences a reaction force from the ground.
When pulling the cart, the action force of the horse pulling the coach is more significant than the reaction force of the cart resisting the pull. Therefore, the carriage moves in the direction of the horse’s power.
By stepping on the ground, the horse creates an active force on the earth in a backward direction with each step, while the floor creates a reaction force that pushes the horse forward. This helps the horse to pull a cart.
Question 82 Explain why it is difficult for a firefighter to hold a hose that is discharging large volumes of water at high speed.
Answer 82. Newton’s third law of motion makes it difficult for a firefighter to hold onto a hose expelling large volumes of water at high speed. In this case, the water being flung forward with great force (action) creates a backward force that causes the hose to move backwards (reaction). As a result of this force and backward movement, it becomes difficult for the firefighter to hold the hose correctly and be stable. .
Question 83. According to the third law of motion, after we press on an object, the item pushes on us with an equal and opposite force. When the thing is a huge truck parked on the side of the road, it probably won’t move. One student justifies this by saying that the two equal and opposite forces cancel each other out. Comment on this logic and explain why the truck is not moving.
Answer 83. According to the third law for motion, the object repels us with an equal and opposite force when we encounter object. This pair of forces is called an action-reaction pair.
In the case of a huge truck parked at the side of the road, the action-reaction pair is the weight of any truck exerting the force on the road in the downward direction (action) and the friction of the road in the upward direction (reaction) that keeps the truck s steady. .. These two equal and opposite forces cancel each other out so that the truck does not move.
For it to move, we must apply an additional external force to overcome the friction used along the way.
Hence, as the student explained, the truck does not move because the two equal and opposite forces of the car and road cancel each other out.
Question 84. Give Newton’s third law of motion, and how does it explain human walking on the ground?
Answer 84. Newton’s third law of motion shows that for every action, there is an equal and opposite reaction acting on two different bodies. This establishes that the existence of the action-reaction force pair, i.e. for each action force produced, would produce an equal and opposite reaction force.
Newton’s third law of motion could explain humans walking on the ground. By walking on the ground, the man creates an active force in a backward direction toward the ground with each step, while the ground makes the reaction force that pushes the man forward and allows him to walk.
Question 85. Why does someone hold the gun firmly against his shoulders while firing a bullet?
Answer 85. If the shot is fired, a person firing the bullet holds the gun firmly against his shoulder due to the gun’s recoil. That is consistent with Newton’s third law of motion, which states that for every action, there is an equal and opposite reaction acting on different bodies. When the bullet is fired, the forward motion of the shell (action) creates the recoil or backward movement of the weapon (reaction). If the action force is much larger, an equivalent recoil force would be generated in the reverse order. .
When the person wielding the weapon does not hold it on his shoulder correctly it could result in injury, the shoulder absorbs most of the force throughout recoil, permitting the shooter to get a steady shot.
So if it is not held firmly by his shoulders, the shot would not be accurate, which could cause the gun to fly out of your hands.
Question 86. A bullet with a mass of 50g is fired from a rifle with a mass of 4kg with an initial velocity of 35ms−1. Calculate the rifle’s initial recoil velocity.
Answer 86. Given:
Mass of the rifle: m1=4kg
Mass of the bullet: m2=50g=0.05kg
the initial velocity of the rifle: u1=0ms−1 (it is stationary during the shot)
the initial velocity of the bullet: u2=0ms−1 (part of rest, in the barrel of the rifle)
firing velocity of the bullet:v2 =35ms−1
Find: Rifle Recoil Velocity:v1
Using the law of conservation of momentum, for an isolated system, the sum of the total initial momentum for an event is equal to the magnitude of the total initial movement,
m1v1+m2v2=m1u1+m2u2
where
m1u1+m2u2
is the total of the momentum of the gun and bullet before the shot is and
m1v1+m2v2
is the total of the momentum of the gun and bullet after the shot.
Substituting the values in
m1v1+m2v2=m1u1+m2u2
we get
(4×v1)+(0.05×35)=(4×0)+(0.05×0)
(4×v1)+ ( 17.5 ) =0
(4×v1)=−17.5
v1=−4.375 ms−1
(The negative sign indicates the backwards direction in which the rifle moves during recoil)
So the recoil speed of the rifle is 4.375 ms−1.
Question 87. An 8000 kg locomotive pulls a train of 5 cars on a horizontal track, each weighing 2000 kg. If the engine exerts a force of 40000N and the trail provides a frictional force of 5000N, then calculate
(a) The net acceleration force
(b) The acceleration of the train
Answer 87. Assume:
Force exerted by the engine on the cars: F= 40000N
Friction exerted on the wagons: f=5000N
Mass of the motor: me=8000kg
Mass of each wagon: mw=2000kg
Mass of the five wagons: mW=5×mw=5×2000=10000kg
Mass of the complete train: mT =me+mW=8000+10000=18000kg
Found: G-force.
The net G-force can iinD by subtracting the frictional force from the engine’s force on the cars.
So, Net Accelerating Force=Force Of Engine-Frictional Force
Net Accelerating Force=F-f
Net Accelerating Force=40000-5000
Net Accelerating Force=35000N
(b) The acceleration of the train
Answer 87. Given:
Force exerted by the machine on the wagons: F=40000N
Friction exerted on the wagons: f=5000N
Mass of the machine: me=8000kg
Mass of one each Wagon:mw=2000kg
Mass of the five wagons:mW=5×mw=5×2000=10000kg
Mass of the entire train: mT=me+mW=8000+10000=18000kg
Acceleration of the train.
We already know that
F=ma
a=Fm
a=Net Accelerating Force Mass Of Train
a=3500018000
a=1.944ms−2
(c) The force from car 1 on car 2.
answer 89. Suppose:
Force exerted by the engine on the wagons: F=40000N
Friction exerted on the wagons: f=5000N
Mass of the engine: me=8000kg
Mass of each wagon: mw=2000kg
Mass of the five wagons: mW=5× mw=5×2000=10000kg
Mass of the whole train: mT=me+mW=8000+10000=18000kg
To find the force exerted by wagon 1 on wagon 2
Where wagon 1 exerts the pulling force on the remaining 4 cars
F=maF=ma
F21=(4mw)×a
F21=(4×2000)×1.944
F21=15552N
Question 88. Two objects with a mass of 1.5 kg each drive in the same straight line but in opposite directions. The speed of each object is 2.5 ms−1 before the collision during which they stick together. What will be the speed of the combined object after the collision?
Answer 88. Given:
mass of object 1: m1=1.5kg
mass of object 2: m2=1.5kg
The initial velocity of object 1: u1=2.5ms−1
The initial velocity of object 2: u2=−2.5ms−1 (negative sign because it is moving in the opposite direction)
Mass of combined object after a collision: m = m1+m2=1.5+1.5=3kg
Wanted: Final velocity of the combined object after the impact: v
Using the law of conservation of momentum, for an isolated system, the sum of the initial momentum for an event is equal to the total initial momentum,
mv =m1u1+m2u2
Where
m1u1+m2u2
is the total momentum of the objects before the collision and
mv
is the total momentum of the combined objects after the collision.Plug in the values – mv=m1u1+m2u2
(3×v)=(1.5×2.5)+(1.5×−2.5)
(3×v)=(3.75)+(−3.75s)
(3×v)=0
v=0ms−1
So the velocity of the combined object after the collision is 0ms−1
Question 89. A hockey ball with a mass of 200 g and moving at 10 ms−1 is hit by a hockey stick to return it to its original path at a speed of 5 ms−1. Calculate the change in momentum as the hockey ball moves due to the force exerted by the hockey stick.
Answer 89. Given:
Mass of the hockey ball: m=200g=0.2kg
The initial speed of the hockey ball: u=10ms−1
The final speed of the hockey ball: v=−5ms−1 when running back to its original direction
Found: Change of momentum of the ball due to the force of the hockey stick
Change Of Momentum = mv − mu
Where Mu is the initial momentum of the hockey ball and mv is the final momentum of the hockey ball.
Set the values in
Change of Momentum=mv−mu
Change of Momentum=(0.2×−5)−(0.2×10)
Change of Momentum=(−1)−(2)
Change of Momentum=−3kgms−1
So, the change of The momentum of the hockey ball due to the force of the hockey stick is −3kgms−1
Benefits of Solving Force and Laws of Motion Class 9 Questions
Following are some of the benefits for students to practise from Extramarks NCERT curriculum-based important questions Class 9 Science Chapter 9:
- The class 9 science chapter 9 important questions are made based on the NCERT book while adhering to the CBSE’s latest curriculum.
- The answers are provided in a detailed manner for students to have a clear understanding of the topic and answer any question asked in their examinations.
- Questions are covered from all the major topics of the chapter, ensuring no topic has been spared. These important questions have been prepared by subject matter experts with years of experience to provide the students with a smooth and deep learning experience so that they need not look elsewhere for any assistance.
- Force and Laws of Motion Class 9 Extra Questions enhance students’ confidence to face their examinations efficiently and excel in academics.
Extramarks credibility lies in providing reliable and trusted study material related to NCERT from Classes 1 to 12 for all the subjects.
Q.1 A ship sinks to different levels, if its journey takes it to different regions. This is due to the
i. Variation in amount of mineral salts dissolved in sea water and the temperature due to the climatic conditions
ii. Constancy in amount of mineral salts dissolved in sea water and the temperature due to the climatic conditions
iii. The variation in the weight of the ship with temperature
iv. The consistency in the density of seawater at various regions
Marks:1
Ans
variation in amount of mineral salts dissolved in sea water and the temperature due to the climatic conditions
Q.2 A piece of iron is totally immersed in water. If its density is 7.8 x 10^{3} kg/m^{3} and volume is 10^{-4 }m^{3} then calculate
(i) The upthrust
(ii) Apparent weight of iron piece in water.
Marks:5
Ans
Density of iron, d = 7.8 x 10^{3} kg/m^{3};
Volume of iron piece, V = 10^{-4} m^{3}
Mass of iron piece, M = V x d
= (10^{-4} m^{3}) x (7.8 x 10^{3} kg/m^{3})
= 0.78 kg
(i) Upthrust, = weight of displaced water
So, = Volume of displaced water x density of water x g
[Where g is the acceleration due to gravity]
So, = Volume of iron piece x density of water x g
[Volume of displaced water = volume of iron piece]
Hence, = (10^{-4} m^{3}) (1000 kg/m^{3}) (10 m/s^{2})
= 1 N
(ii) Apparent weight,
= true weight of iron piece upthrust on iron piece in water
True weight of iron piece = M x g
= (0.78 kg) (10 m/s^{2})
= 7.8 N
Thus, = 7.8 N 1 N
= 6.8 N
Q.3 The density of gold is 19.3 × 10^{3 }kg/m^{3}. Find its relative density.
Marks:1
Ans
CBSE Class 9 Science Important Questions
FAQs (Frequently Asked Questions)
1. Why should I refer to the important questions in Class 9 Science Chapter 9?
You should refer to the important questions in Class 9 Science Chapter 9 for practising various questions after completing the chapter from the NCERT textbook and other resources. It would be best if you also revise notes from your class lectures to develop a solid conceptual understanding before starting to solve them. Solving multiple problems will help you to analyse your weaker topics. Based on it you can study the theory of those topics again to strengthen your overall understanding of the chapter.
2. What is the chapter 'Force and Laws of Motion about?
The chapter ‘Force and Laws of Motion studies different types of forces we come across daily. Also, one learns about the various laws of motion that exist in the universe for working consistently.. The important topics included in the chapter are as follows:
- Balanced and unbalanced focus
- First law of motion
- Inertia and the mass
- Second law of motion
- Third law of motion
- Conservation of momentum