Question

A pistol of mass 2 kg fires a bullet of mass 50 g.
The bullet strikes a stationary block of mass 12 kg

If the block, with the bullet embedded in it, moves
with a velocity of 4 m/s, the recoil velocity of the
pistol will be (Ignore friction and air resistance)
(1) –1.10 m/s (2) –0.5 m/s
(3) –0.55 m/s (4) –1 m/s

Answer 1

Let v be the velocity of bullet of mass m.

The momentum of the bullet will be mv.

Now, m=50 g=0.05 kg. Then,

momentum of the bullet=0.05v…………(1)

The collision between bullet and block is inelastic.

Using law of conservation of momentum, we have,

mv=(m+Mb)V, where, Mb is mass of the block and V is velocity of ( block + bullet) system.

Mb=0.5 kg and V=4 m/s. Then,

0.05v=(0.05 +0.5)V=0.55V. Therefore ,

v=(0.55/0.05)V or

v=11 V……….(2) or

v=44 m/s……(3)

Now, when bullet is fired, the momentum of the system of pistol and the bullet is conserved. Therefore ,

MpVp=mv , where Vp is recoil velocity of the pistol. Or

(2)Vp=(5x10^-2)(44).

Therefore,

Vp=1.1 m/s.

It means the ans will be -1.10m/s

Hope it will help you

Answer 2

Answer:

Explanation:

Let v be the velocity of bullet of mass m.

The momentum of the bullet will be mv.

Now, m=50 g=0.05 kg. Then,

momentum of the bullet=0.05v……

The collision between bullet and block is inelastic.

Using law of conservation of momentum, we have,

mv=(m+Mb)V, where, Mb is mass of the block and V is velocity of ( block + bullet) system.

Mb=0.5 kg and V=4 m/s. Then,

0.05v=(0.05 +0.5)V=0.55V. Therefore

v=(0.55/0.05)V or

v=11 V……….(2) or

v=44 m/s……(3)

Now, when bullet is fired, the momentum of the system of pistol and the bullet is conserved. Therefore ,

MpVp=mv , where Vp is recoil velocity of the pistol. Or

(2)Vp=(5x10^-2)(44).

Therefore,

Vp=1.1 m/s.

It means the ans will be -1.10m/s

Hope it will help you