Question 14: A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s 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.

**Answer:**Given, Mass of bullet, m = 10 g = 10/1000 kg = 0.01 kg

Initial velocity of bullet, u = 150 m/s

Since bullet comes to rest, thus final velocity, v =0

Time, t = 0.03 s

Distance of penetration, i.e. Distance, covered (s)=?

Magnitude of force exerted by wooden block =?

We know that, `v=u+at`

`=>0=150ms^(-1)+axx0.03s`

`=>-150m//s = axx0.03s`

`=>a=-(150m//s)/(0.03s) = -5000ms^(-2)`

We know that, `s=ut+1/2at^2`

`=>s=150m//sxx0.03s` `+1/2(-5000ms^(-2))xx(0.03s)^2`

`=>s=4.5m-2500ms^(-2)xx0.0009s^2`

`=>s=4.5m-2.25m`

`=>s=2.25m`

Magnitude of force exerted by wooden block

We know that, Force = mass x acceleration

Or, `F = 0.01 kg xx – 5000 m s^-2 =-50 N`

Therefore, Penetration of bullet in wooden block = 2.25 m

Force exerted by wooden block on bullet = – 50 N. Here negative sign shows that force is exerted in the opposite direction of bullet.

Question 15: An object of mass 1 kg travelling in a straight line with a velocity of 10 m/s collides with, 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.

**Answer:** Given, mass of moving object, m_{1} = 1 kg

Mass of the wooden block, m_{2} = 5kg

Initial velocity of object, u_{1} = 10 m/s

Initial velocity of wooden block, u_{2} = 0

Final velocity or moving object and wooden block, v =?

Total momentum before collision and after collision =?

We know that,

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

`=>1\ kg xx 10m//s+5\ kgxx0` `=1\ kg xx v + 5\ kgxxv`

`=>10\ kg\ m//s=v(1\ kg+5\ kg)`

`=>10\ kg m//s=vxx 6\ kg`

`=>v=(10\ kgms^(-1))/(6\ kg)`

`=>v= 1.66m//s` ---(i)

Total momentum of object and wooden block just before collision

`=m_1u_1+m_2u_2`

`=1\ kg xx 10 ms^(-1)+5\ kg xx0`

`=10\ kg ms^(-1)`

Total momentum just after collision

`=m_1v_1+m_2v_2`

`=m_1v+m_2v = v(m_1+m_2)`

[∵ both the objects move with same velocity `v` after collision]

`=(1\ kg + 5\ kg) xx 10/6 m//s` (from eq^{n}(i)

`=6\ kg xx 10/6 m//s = 10\ kg ms^(-1)`

Thus, Velocity of both the object after collision = 1.66 m/s

Total momentum before collision = 10 kg m/s

Total momentum after collision = 10 kg m/s

Question 16: An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

**Answer:**Given, Initial velocity, u = 5 m/s

Final velocity, v = 8 m/s

Mass of the given object, m = 100 kg

Time, t = 6 s

Initial momentum and Final momentum =?

Magnitude of force exerted on the object =?

We know that, Momentum = mass x velocity

Therefore, initial momentum = mass x initial velocity

`= 100 kg xx 5 m//s = 500 kg m//s`

Final momentum = mass x final velocity

`= 100 kg xx 8 m//s = 800 kg m//s`

We know that,`v=u+at`

`=>8m//s=5m//s+axx6s`

`=>8m//s-5m//s=axx6s`

`=>3m//s=axx6s`

`=>(3m//s)/(6s)=0.5ms^(-2)`

Now, Force exerted on object = Mass x Acceleration

= 100 kg 0.5 m/s/s

= 50 N

Question 17: 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.

**Answer:** We know, that as per the Law of Conservation of Momentum; total momentum of a system before collision is equal to the total momentum of the system after collision.

In this case, since the insect experiences a greater change in its velocity so it experiences a greater change in its momentum. From this angle, Kiran’s observation is correct.

Motorcar is moving with a larger velocity and has a bigger mass; as compared to the insect. Moreover, the motorcar continues to move in the same direction even after the collision; which suggests that motorcar experiences minimal change in its momentum, while the insect experiences the maximum change in its momentum. Hence, Akhtar’s observation is also correct.

Rahul’s observation is also correct; because the momentum gained by the insect is equal to the momentum lost by the motorcar. This also happens in accordance to the law of conservation of momentum.

Question 18: 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}.

**Answer:**Given, Mass of dumb-bell = 10 kg

Distance, s = 80 cm = 80/100 = 0.8 m

Acceleration, a = 10 m/s/s

Initial velocity of dumb-bell, u = 0

Momentum =?

We know that,

`v^2=u^2+2as`

`=>v^2=0+2xx10ms^(-2)xx0.8m`

`=>v^2=16m^2s^(-2)`

`=>v=sqrt(16m^2s^(-2))=4m//s`

Now, we know that, momentum = mass x velocity

`= 10 kg xx 4 m//s = 40 kg m//s`

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