E-Learning, E-Tutoring, School Education Support, Online Education, Digital Learning, Smart Learning

Q:

How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s⁻².

A:

\begin{array}{l} Given, mass of the dumbell, m = 10 kg \\ distance covered by the dumbell, \\ s = 80 cm = 0 .8 m \\ {acceleration in the downward direction, a = 10 ms}^{-2} \\ initial velocity of the dumbell, u = 0 \\ final velocity of the dumbell, v = ? \\ On using the third equation of motion, \\ v^{2} {= u}^{2} +2as \\ On putting the given values, \\ v^{2} {= 0}^{2} {+ 2(10 ms}^{-2})(0 .8 m) \\ {= 16 m}^{2} s^{- 2} \\ or,      v = \sqrt{{16 m}^{2} s^{- 2}} {= 4 ms}^{-1} \\ Therefore, the momentum with which the dumbell hits the \\ {floor is = mv= 10 kg × 4 ms}^{-1} {= 40 kgms}^{-1} . \end{array}

Q:

Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.

A:

As per the conservation of momentum,

(momentum of motorcar + momentum of insect)_before

= (momentum of motorcar + momentum of insect)_after

∴ Change in in momentum = 0

The direction of the insect gets reversed when it stuck on the windscreen and the velocity of the insect changes to a great extend. The motorcar continues moving with a constant velocity in the forward direction.

As, Kiran suggested that the insect suffers a greater change in momentum as compared to the motorcar hence, it is correct. The momentum of the insect after collision becomes very high as the motorcar is moving at a very high speed. Therefore, the momentum gained by the insect is equal to the momentum lost by the motorcar.

Secondly, Akhtar is also correct as the mass of the motorcar is very large as compared to the mass of the insect and the speed of the motorcar is also very high.

Now, Rahul’s explanation is correct as both the car and the insect experienced equal forces caused by the Newton’s action-reaction law. But Rahul’s statement becomes incorrect when he said that the system suffers a change in momentum because the momentum before the collision is equal to the momentum after the collision.

Q:

An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms^-1to 8 ms^-1in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

A:

\begin{array}{l} G i v e n, {inital velocity, u = 5 ms}^{-1} \\ {final velocity,  v = 8 ms}^{-1} \\ mass ,           m = 100 kg \\ time taken by the object to accelerate, t = 6 s \\ ∴ Intial momentum = mu = 100 \times {5 = 500 kgms}^{-1} \\ Final momentum = mv = 100 \times 8 {= 800 kgms}^{-1} \\ F o r c e exerted on the object,  F = \frac{m v - m u}{t} \\ = \frac{800 - 500}{6} = \frac{300}{6} = 50 N . \end{array}

Q:

An object of mass 1 kg travelling in a straight line with a velocity of 10 ms^-1collideswith, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

A:

\begin{array}{l} G i v e n, mass of the object, m = 1 kg \\ mass of the wooden block, M = 5 kg \\ v e l o c i t y {of the object before collision, u}_{1} {= 10 ms}^{-1} \\ v e l o c i t y {of the wooden block befor collision, u}_{2} = 0 \\ T o t a l momentum before collision, \\ P_{inital} = {mu}_{1} {+Mu}_{2} = 1 \times 10 + 5 \times 0 = 10 {kgms}^{-1} \\ A f t e r collision, both the blocks stick together, hence \\ t h e the combined mass = M+m = 5+1 = 6 kg \\ Let, the velocity of the combined mass is V \\ Then, as per conservation law of momentum, \\ P_{inital} {= P}_{final} \\ 10 = 6 \times V \\ or            V = \frac{10}{6} = 1 {.67 ms}^{-1} \\ Hence, the velocity of the combined mass after \\ collision would be 1 {.67 ms}^{-1} . \\ {Now, final momentum, P}_{final} = (M+m)V =6 \times 1 .67 \\ {=10 kgms}^{-1} . \end{array}

Q:

A bullet of mass 10 g travelling horizontally with a velocity of 150 ms⁻¹ strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

A:

\begin{array}{l} G i v e n, {inital velocity of the bullet, u = 150 ms}^{-1} \\ final velocity of the bullet, v = 0 \\ (as the bullet comes to rest) \\ time taken to come to rest for bullet, t = 0 .03 s \\ mass of the bullet, m = 0 .01 kg \\ On using the first equation of motion, \\ v = u +at \\ or,           0 = 150 + a \times 0 .03 \\ {or,           a = -5000 ms}^{-2} \\ (Here, negative sign shows that the velocity of the \\ bullet is decreasing) \\ On using the third equation of motion, \\ v^{2} = u^{2} + 2 a s \\ o r, 0^{2} = {(150)}^{2} + 2 \times (- 5000) \times s \\ o r, s = \frac{22500}{10000} = 2.25 m \\ Hence, the distance of penentration of the bullet \\ into the block is 2 .25 m . \\ O n using Newton's second law of motion, \\ Force exerted by the wooden block on the bullet, \\ F = ma = 0 .01 \times 5000 = 50 N . \end{array}

Q:

A hockey ball of mass 200 g travelling at 10 ms⁻¹ is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms⁻¹. Calculate the magnitude of change of
momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

A:

\begin{array}{l} Given, mass of the hockey ball, m = 200 g = 0 .2 kg \\ {velocity of the ball, v}_{1} {= 10 ms}^{-1} \\ {Initial momentum,  P}_{1} {= mv}_{1} \\ Now, hockey ball travels in the opposite direction \\ {with velocity,         v}_{2} {= - 5 ms}^{-1} \\ {Final momentum, P}_{2} {= mv}_{2} \\ ∴ {Change in momentum, ΔP = P}_{1} {-P}_{2} = m [v_{1} {- v}_{2}] \\ = 0 .2 kg [{10 ms}^{-1} {- (-5 ms}^{-1})] {= 3 kg ms}^{-1} . \\ {The change in momentum comes out to be 3 kgms}^{-1} and the  negative sign indicates \\ that the ball started moving in opposite direction . \end{array}

Q:

According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

A:

The static friction acting between the truck and the road is very large as the truck is very massive. Thus, truck does not move when someone pushes it. Therefore, it can be concluded that the applied force in one direction is cancelled out by the frictional force of equal amount acting in the opposite direction.

Q:

Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms^-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?

A:

\begin{array}{l} {Given, masses of the objects,   m}_{1} {=m}_{2} = 1 .5 kg \\ {velocities of the objects,  v}_{1} = 2 {.5 ms}^{-1} \\ v_{2} =- 2 {.5 ms}^{-1} \\ {Here, velocity of m}_{2} is considered negative as it is \\ moving in opposite direction before collision . \\ Both the objects stick together after collision . \\ {Hence, total mass M = m}_{1} {+m}_{2} = 1 .5 kg + 1 .5 kg = 3 kg \\ Let, velocity of the combined mass after collision = V \\ As per conservation law of linear momentum, \\ m_{1} v_{1} {+ m}_{2} v_{2} = MV \\ ∴ (1 .5 kg × 2 {.5 ms}^{-1})+(1 .5 kg × -2 {.5 ms}^{-1}) = 3V \\ or,            V = 0 \\ Thus, the velocity of the combined object after collision \\ will be zero . \end{array}

Q:

Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

A:

When the force of 200 N is applied on wooden cabinet, an equal and opposite force acts on it. This opposite force is the frictional force. Therefore, a frictional force of 200 N is exerted on the cabinet.

Q:

What is the momentum of an object of mass m, moving with a velocity v?

(mv)²

mv²

½mv²

mv

A:

The correct option is (d).

Explanation:

\begin{array}{l} Given, mass of the object = m \\ velocity of the object = v \\ Momentum, P = mass \times velocity \\ = mv \end{array}

Q:

An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms⁻²?

A:

\begin{array}{l} Given, m = 1500 kg \\ v = 0 \\ a = -1 {.7 ms}^{-2} \\ On using Newton’s second law of motion, \\ Force,  F = m × a =1500 kg × (-1 {.7 ms}^{-2}) = -2550 N \\ Therefore, the force between the automobile and the \\ road is 2550 N, in the direction opposite to the \\ motion of the automobile . \end{array}

Q:

A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force and (b) the acceleration of the train.

A:

\begin{array}{l} (a) Given, force exerted by the engine, F = 40000 N \\ frictional force due to tracks,  F' = 5000 N \\ ∴ {Net force,   F}_{net} = 40000 - 5000 = 35000 N \\ (b) Net force on the wagons, F_{a} = 35000 N \\ Mass of the 5 wagons, m = 2000 \times 5 = 10000 kg \\ Mass of the engine, m' = 8000 kg \\ ∴ Total mass, M = m + m' = 18000 kg \\ Using Newton ’ s second law of motion, \\ F_{a} = M a \\ or,                a = \frac{F_{a}}{M} = \frac{35000}{18000} = 1 {.944 ms}^{-2} \\ Hence, the acceleration of the wagons and the train is \\ 1.944 {ms}^{-2} . \end{array}

Q:

A stone of 1 kg is thrown with a velocity of 20 ms⁻¹ across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

A:

\begin{array}{l} {Given,      u = 20 ms}^{-1} \\ v = 0 \\ s = 50 m \\ mass of the stone, m = 1 kg \\ On using the third equation of motion, \\ v^{2} {= u}^{2} + 2as      (a = acceleration of the stone) \\ ∴ 0^{2} {= (20 ms}^{-1})^{2} +2×a×50 m \\ or, a = - \frac{{400 m}^{2} s^{- 2}}{100 m} {= - 4 ms}^{-2} \\ Here, negative sign shows that acceleration is acting \\ against the motion of the stone . \\ By using Newton's second law of motion, \\ Force,  F = ma \\ {or,              = 1 kg ×(-4 ms}^{-2}) = -4 N \\ Therefore, the force of friction between the stone and \\ the ice is - 4 N . \end{array}

Q:

A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)

A:

\begin{array}{l} Given,   intial velocity of the truck, u = 0 \\ distance travelled, s = 400 m \\ time taken, t = 20 s \\ acceleration, a = ? \\ On using the relation \\ s = ut + \frac{1}{2} {at}^{2} \\ 400 m = 0(20 s) + \frac{1}{2} {a(20 s)}^{2} \\ {or,              a = 2 ms}^{-2} \\ Now, 1 metric tonne = 1000 kg \\ ∴ 7 metric tonnes = 7×1000 kg = 7000 kg \\ By using Newton's second law, \\ F = ma \\ {= (7000 kg) (2 ms}^{-2}) = 14000 N \\ {Thus, the acceleration of the truck is 2 ms}^{-2} and the \\ force acting on it is 14000 N . \end{array}

Q:

A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop 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 on the ball opposing the motion. (d) there is no unbalanced force on the ball, so the ball would want to come to rest.

A:

The correct option is (c)

Explanation: Here, the ball comes to rest due to frictional force which opposes its motion. Frictional force always acts in the direction opposite to the direction of motion. Therefore, this force tries to stop the ball.

Q:

An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

A:

Yes. It is possible that an object is moving with a non-zero velocity even when the object experiences a net zero external unbalanced force. It is possible only when the object has been moving with a constant velocity in a certain direction. In order to change the state of motion, a net non-zero external unbalanced force must be applied on the object.

Q:

Why is it advised to tie any luggage kept on the roof of a bus with a rope?

A:

When the bus starts accelerating in forward direction, it comes in the state of motion but the luggage kept on the roof remains in the state of rest. Therefore, with the forward movement of the bus, the luggage remains at its original position and finally falls from the roof of the bus. Hence, it is advised to tie any luggage kept on the roof of a bus.

Q:

When a carpet is beaten with a stick, dust comes out of it. Explain.

A:

Due to inertia, an object opposes any change in its state of rest or state of motion. As soon as the carpet is beaten with a stick, the carpet comes into motion while the dust particles oppose their state of rest. As per Newton’s first law of motion, the dust particles remain stationary, while the carpet moves. Therefore, the dust particles come out of the carpet.

NCERT Solutions For Class 9 Science Chapter 9 Force and Laws of Motion

Connect with us

Download Extramarks – The Learning App

NCERT Solutions

CBSE Class

ICSE Class

Quick Links

Board Paper Solution 2020

Sample Paper

Exam Weightage

Solved Board Papers

Test Prep

Other Services

About Us

Terms & Conditions