Q:

A:

\begin{array}{l}\text{Given,massofthedumbell,m=10kg}\\ \text{distancecoveredbythedumbell,}\\ \text{s=80cm=0}\text{.8m}\\ {\text{accelerationinthedownwarddirection,a=10ms}}^{\text{-2}}\\ \text{initialvelocityofthedumbell,u=0}\\ \text{finalvelocityofthedumbell,v=?}\\ \text{Onusingthethirdequationofmotion,}\\ {\text{v}}^{\text{2}}{\text{=u}}^{\text{2}}\text{+2as}\\ \text{Onputtingthegivenvalues,}\\ {\text{v}}^{\text{2}}{\text{=0}}^{\text{2}}{\text{+2(10ms}}^{\text{-2}}\text{)(0}\text{.8m)}\\ {\text{=16m}}^{\text{2}}{\text{s}}^{-2}\\ \text{or,v=}\sqrt{{\text{16m}}^{\text{2}}{\text{s}}^{-2}}{\text{=4ms}}^{\text{-1}}\\ \text{Therefore,themomentumwithwhichthedumbellhitsthe}\\ {\text{flooris=mv=10kg\xd74ms}}^{\text{-1}}{\text{=40kgms}}^{\text{-1}}\text{.}\end{array}

Q:

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:

A:

$\begin{array}{l}\mathrm{Given},{\text{initalvelocity,u=5ms}}^{\text{-1}}\\ {\text{finalvelocity,v=8ms}}^{\text{-1}}\\ \text{mass,m=100kg}\\ \text{timetakenbytheobjecttoaccelerate,t=6s}\\ \\ \therefore \text{Intialmomentum=mu=100}\times {\text{5=500kgms}}^{\text{-1}}\\ \text{Finalmomentum=mv=100}\times 8{\text{=800kgms}}^{\text{-1}}\\ \mathrm{Force}\text{exertedontheobject,F=}\frac{\mathrm{mv}-\mathrm{mu}}{\mathrm{t}}\\ \text{=}\frac{800-500}{6}\text{=}\frac{300}{6}\text{=50N.}\end{array}$

Q:

A:

$\begin{array}{l}\mathrm{Given},\text{massoftheobject,m=1kg}\\ \text{massofthewoodenblock,M=5kg}\\ \mathrm{velocity}{\text{oftheobjectbeforecollision,u}}_{\text{1}}{\text{=10ms}}^{\text{-1}}\\ \mathrm{velocity}{\text{ofthewoodenblockbeforcollision,u}}_{\text{2}}\text{=0}\\ \\ \mathrm{Total}\text{momentumbeforecollision,}\\ {\text{P}}_{\text{inital}}={\text{mu}}_{\text{1}}{\text{+Mu}}_{\text{2}}\text{=1}\times \text{10}+\text{5}\times 0=10{\text{kgms}}^{\text{-1}}\\ \mathrm{After}\text{collision,boththeblockssticktogether,hence}\\ \mathrm{the}\text{thecombinedmass=M+m=5+1=6kg}\\ \text{Let,thevelocityofthecombinedmassisV}\\ \text{Then,asperconservationlawofmomentum,}\\ {\text{P}}_{\text{inital}}{\text{=P}}_{\text{final}}\\ \text{10=6}\times \text{V}\\ \text{orV=}\frac{10}{6}{\text{=1.67ms}}^{\text{-1}}\\ \text{Hence,thevelocityofthecombinedmassafter}\\ {\text{collisionwouldbe1.67ms}}^{\text{-1}}\text{.}\\ {\text{Now,finalmomentum,P}}_{\text{final}}\text{=(M+m)V=6}\times \text{1.67}\\ {\text{=10kgms}}^{\text{-1}}\text{.}\end{array}$

Q:

A:

\begin{array}{l}Given,{\text{initalvelocityofthebullet,u=150ms}}^{\text{-1}}\\ \text{finalvelocityofthebullet,v=0}\\ \text{(asthebulletcomestorest)}\\ \text{timetakentocometorestforbullet,t=0}\text{.03s}\\ \text{massofthebullet,m=0}\text{.01kg}\\ \text{Onusingthefirstequationofmotion,}\\ \text{v=u+at}\\ \text{or,0=150+a}\times \text{0}\text{.03}\\ {\text{or,a=-5000ms}}^{\text{-2}}\text{}\\ \text{(Here,negativesignshowsthatthevelocityofthe}\\ \text{bulletisdecreasing)}\\ \text{Onusingthethirdequationofmotion,}\\ {\text{v}}^{\text{2}}={\text{u}}^{\text{2}}+2as\\ or,{\text{0}}^{\text{2}}={\text{(150)}}^{\text{2}}+2\times (-5000)\times s\\ or,\text{s=}\frac{22500}{10000}=2.25\text{m}\\ \text{Hence,thedistanceofpenentrationofthebullet}\\ \text{intotheblockis2}\text{.25m}\text{.}\\ On\text{usingNewton'ssecondlawofmotion,}\\ \text{Forceexertedbythewoodenblockonthebullet,}\\ \text{F=ma=0}\text{.01}\times 5000=50\text{N}\text{.}\end{array}

Q:

momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

A:

\begin{array}{l}\text{Given,massofthehockeyball,m=200g=0}\text{.2kg}\\ {\text{velocityoftheball, v}}_{\text{1}}{\text{=10ms}}^{\text{-1}}\\ {\text{Initialmomentum,P}}_{\text{1}}{\text{=mv}}_{\text{1}}\\ \text{Now,hockeyballtravelsintheoppositedirection}\\ {\text{withvelocity,v}}_{\text{2}}{\text{=-5ms}}^{\text{-1}}\text{}\\ {\text{Finalmomentum,P}}_{\text{2}}{\text{=mv}}_{\text{2}}\\ \therefore {\text{Changeinmomentum,\Delta P=P}}_{\text{1}}{\text{-P}}_{\text{2}}\text{=m}\left[{\text{v}}_{\text{1}}{\text{-v}}_{\text{2}}\right]\\ \text{=0}\text{.2kg}\left[{\text{10ms}}^{\text{-1}}{\text{-(-5ms}}^{\text{-1}}\text{)}\right]{\text{=3kgms}}^{\text{-1}}\text{.}\\ {\text{Thechangeinmomentumcomesouttobe3kgms}}^{\text{-1}}\text{andthenegativesignindicates}\\ \text{thattheballstartedmovinginoppositedirection}\text{.}\end{array}

Q:

A:

Q:

A:

\begin{array}{l}{\text{Given,massesoftheobjects,m}}_{\text{1}}{\text{=m}}_{\text{2}}\text{=1}\text{.5kg}\\ {\text{velocitiesoftheobjects,v}}_{\text{1}}\text{=2}{\text{.5ms}}^{\text{-1}}\\ {\text{v}}_{\text{2}}\text{=-2}{\text{.5ms}}^{\text{-1}}\\ {\text{Here,velocityofm}}_{\text{2}}\text{isconsiderednegativeasitis}\\ \text{movinginoppositedirectionbeforecollision}\text{.}\\ \text{Boththeobjectssticktogetheraftercollision}\text{.}\\ {\text{Hence,totalmassM=m}}_{\text{1}}{\text{+m}}_{\text{2}}\text{=1}\text{.5kg+1}\text{.5kg=3kg}\\ \text{Let,velocityofthecombinedmassaftercollision=V}\\ \text{Asperconservationlawoflinearmomentum,}\\ {\text{m}}_{\text{1}}{\text{v}}_{\text{1}}{\text{+m}}_{\text{2}}{\text{v}}_{\text{2}}\text{=MV}\\ \therefore \text{(1}\text{.5kg\xd72}{\text{.5ms}}^{\text{-1}}\text{)+(1}\text{.5kg\xd7-2}{\text{.5ms}}^{\text{-1}}\text{)=3V}\\ \text{or,V=0}\\ \text{Thus,thevelocityofthecombinedobjectaftercollision}\\ \text{willbezero}\text{.}\end{array}

Q:

A:

Q:

- (mv)
^{2} - mv
^{2} - ½
^{ }mv^{2} - mv

A:

The correct option is (d).

Explanation:

Q:

A:

\begin{array}{l}\text{Given,m=1500kg}\\ \text{v=0}\\ \text{a=-1}{\text{.7ms}}^{\text{-2}}\\ \text{OnusingNewton\u2019ssecondlawofmotion,}\\ \text{Force,F=m \xd7 a=1500kg \xd7}\left(\text{-1}{\text{.7ms}}^{\text{-2}}\right)\text{=-2550N}\\ \text{Therefore,theforcebetweentheautomobileandthe}\\ \text{roadis2550N,inthedirectionoppositetothe}\\ \text{motionoftheautomobile}\text{.}\end{array}

Q:

(a) the net accelerating force and

(b) the acceleration of the train.

A:

\begin{array}{l}(a)\text{\hspace{0.17em}}\text{\hspace{0.17em}Given,}\text{forceexertedbytheengine,F=40000N}\\ \text{frictionalforceduetotracks,F'=5000N}\\ \text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.17em}}\therefore {\text{Netforce,F}}_{\text{net}}=40000-5000\text{=35000N}\text{}\\ \text{(b)}\text{\hspace{0.17em}}\text{Netforceonthewagons},\text{}{F}_{a}=\text{35}000\text{N}\\ \text{Massofthe5wagons},\text{}m=\text{2}000\times \text{5}=\text{1}0000\text{kg}\\ \text{Massoftheengine},\text{}m\prime =\text{8}000\text{kg}\\ \therefore \text{Totalmass},\text{M}=m+m\prime =\text{18}000\text{kg}\\ \text{UsingNewton}\u2019\text{ssecondlawofmotion,}\\ {\text{F}}_{a}=Ma\\ \text{or, a}\text{=}\frac{{\text{F}}_{a}}{M}\text{=}\frac{35000}{18000}=\text{1}{\text{.944ms}}^{\text{-2}}\\ \text{Hence, theaccelerationofthewagonsandthetrainis}\\ 1.944{\text{ms}}^{\text{-2}}.\\ \end{array}

Q:

A:

\begin{array}{l}{\text{Given, u=20ms}}^{\text{-1}}\\ \text{v=0}\\ \text{s=50m}\\ \text{massofthestone,m=1kg}\\ \text{Onusingthethirdequationofmotion,}\\ {\text{v}}^{\text{2}}{\text{=u}}^{\text{2}}\text{+2as(a=accelerationofthestone)}\\ \therefore {\text{0}}^{\text{2}}{\text{=(20ms}}^{\text{-1}}{\text{)}}^{\text{2}}\text{+2\xd7a\xd750m}\\ \text{or,a=-}\frac{{\text{400m}}^{\text{2}}{s}^{-2}}{\text{100m}}{\text{=-4ms}}^{\text{-2}}\\ \text{Here,negativesignshowsthataccelerationisacting}\\ \text{againstthemotionofthestone}\text{.}\\ \text{}\\ \text{ByusingNewton'ssecondlawofmotion,}\\ \text{Force,F=ma}\\ {\text{or,=1kg\xd7(-4ms}}^{\text{-2}}\text{)=-4N}\\ \text{Therefore,theforceoffrictionbetweenthestoneand}\\ \text{theiceis-4N}\text{.}\end{array}

Q:

A:

\begin{array}{l}\text{Given,intialvelocityofthetruck,u=0}\hfill \\ \text{distancetravelled,s=400m}\hfill \\ \text{timetaken,t=20s}\hfill \\ \text{acceleration,a=?}\hfill \\ \text{Onusingtherelation}\hfill \\ \text{s=ut+}\frac{\text{1}}{\text{2}}{\text{at}}^{\text{2}}\hfill \\ \text{400m=0(20s)+}\frac{\text{1}}{\text{2}}{\text{a(20s)}}^{\text{2}}\hfill \\ {\text{or,a=2ms}}^{\text{-2}}\hfill \\ \hfill \\ \text{Now,1metrictonne=1000kg}\hfill \\ \therefore \text{7metrictonnes=7\xd71000kg=7000kg}\hfill \\ \text{ByusingNewton'ssecondlaw,}\hfill \\ \text{F=ma}\hfill \\ {\text{=(7000kg)(2ms}}^{\text{-2}}\text{)=14000N}\hfill \\ {\text{Thus,theaccelerationofthetruckis2ms}}^{\text{-2}}\text{andthe}\hfill \\ \text{forceactingonitis14000N}\text{.}\hfill \end{array}

Q:

(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:

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:

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:

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.

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