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This is the second of our many quizzes on Newton's second law. Click on this to access more such competitive quizzes on Mechanics.

Newton's second law Quiz 2

physics

A varying force as a function of time given by $F=a{{t}^{2}}+bt$ is applied on a block of mass 5 kg. The momentum of block at t=5 sec will be-$[\text{initial block is at rest}]$ 

Momentum function with time is calculated and value can be obtained by putting t=5sec in this function. 

  $F=a{{t}^{2}}+bt\\  \Rightarrow P=\int{F.dt}\\  \Rightarrow P=\int{a{{t}^{2}}+bt.dt}\\  \Rightarrow P =\dfrac{a{{t}^{3}}}{3}+\dfrac{b{{t}^{2}}}{2}\\  \therefore P =\dfrac{2a{{t}^{3}}+3b{{t}^{2}}}{6}$

 A cart of mass 4.5 Kg is moving on a smooth horizontal surface. The velocity time graph of its motion is shown in figure, the change in momentum of cart in t=1 sec to t= 2 sec will be-<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/da4d9647-4492-4386-82b3-9f038c84aebd1189596003023926765.png"/>

 Velocity at t=0 sec is zero so momentum at t=1 sec is equal to zero. 

  $\text{Momentum at t=1 sec,}\\  p=mv=4.5\times 0=0\\  \text{Momentum at t=2 sec,}\\p=mv=4.5\times 5=22.5\,kgm/s\\  \text{Change in momentum,}\\  \therefore 22.5kgm/s-0=22.5\,kgm/s$

 A force is applied on a block of $5\,Kg$ present on an inclined plane as shown in figure, so that the acceleration of the block in up the inclined surface is $1\,m/{{s}^{2}}$. Now this force is removed and applied opposite to the block down the inclined. The acceleration of block in new condition will be-<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e86829e3-b585-4d35-83ec-b91a273134111347463551386551888.png"/> 

 The angle inclined to the plane will distribute the weight of block in normal direction and down the inclined plane. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/dbd7f742-f68c-4d86-b351-78a3020595cf878706698314903731.png"/>$ \\ \text{Net upward inclined force is,}\\{{F}_{u}}=F-mg\sin (30)\\  \Rightarrow {{F}_{u}} =F-5\times 10\times\dfrac{1}{2}\\  \Rightarrow {{F}_{u}} =F-25\\  \text{Acceleration up the inclined will be-}\\  a=\dfrac{{{F}_{u}}}{m}\\  \Rightarrow a =\dfrac{F-25}{5}\\  \Rightarrow a =1\\  \Rightarrow F-25=5\\  \Rightarrow F=30N\\  \text{Now, down the inclined force-}\\  {{F}_{d}}=F+mg\sin (30)\\  \Rightarrow {{F}_{d}} =30+5\times 10\times \dfrac{1}{2}\\\Rightarrow {{F}_{d}} =55N\\  \text{Acceleration will be-}\\  a=\dfrac{{{F}_{d}}}{m}\\  \Rightarrow a =\dfrac{55}{5}\\  \therefore a =11m/{{s}^{2}}$

Ramesh throws a ball of mass 5 kg in upward direction from a height of 4 m from the ground and with a speed of 4 m/s, after some time the ball came back and hits the ground with a momentum $P$.The value of $P$ calculated by Ramesh will be- 

Final velocity at the time of collision with ground is calculated by applying energy conservation in order to calculate momentum. 

 $\text{Height achieved by ball will be-}\\  {{v}^{2}}={{u}^{2}}+2ah\\  \Rightarrow {{0}^{2}}={{4}^{2}}-2\times 10\times h\\  \Rightarrow h=0.8\\  \text{Now total height above the ground = 0.8 + 4= 4.8m}\\  \text{By apply energy conservation we have,}\\  mgh=\dfrac{1}{2}m{{v}^{2}}\\ \Rightarrow 5\times 10\times 4.8=\dfrac{1}{2}\times 5\times {{v}^{2}}\\  \Rightarrow v=4\sqrt{6}\\ \text{Momentum,}\\  mv=5\times 4\sqrt{6}\\  \therefore mv =20\sqrt{6}$

A train of 400 m long is travelling on a railway track with a speed of 5 m/s, on 100 m before the station it starts de-accelerating and eventually comes to rest after the train fully reaches the station of length 400 m. The acceleration of train will be- 

When a train fully covers the station its engine will cover a whole of 500 m distance during the course of retarding motion of the train. 

 $\text{Total distance covered by train before coming to rest=100 m + 400 m=500 m.}\\  \text{By applying 3rd equation of motion we have,}\\  {{v}^{2}}={{u}^{2}}+2as\\  \Rightarrow {{0}^{2}}={{5}^{2}}-2a\times 500\\  \therefore a=0.025\,m/{{s}^{2}}$