Question

26. Adisc of mass 2 kg is kept floating horizontally in air
by firing bullets of mass 0.1 kg each vertically at it
at the rate of 5 per second. If the bullets rebound
with the same speed, the speed with which these are
fired will be
V
Mg
(1) 1.96 m/s
(3) 9.8 m/s
(2) 0.98 m/s
(4) 19.6 m/s​

Answer 1

Given,

• Mass of disc, $\sf{M=2\ kg}$

• Mass of one bullet, $\sf{m=0.1\ kg}$

• No. of bullets fired per second, $\sf{n=5}$

Let the initial velocity of the bullet be $\sf{u.}$

The bullet rebounds with same speed, hence its final velocity, $\sf{v=-u.}$

Total mass of bullets fired per second $\sf{=nm=0.5\ kg}$

By second law, the change in momentum of the bullets provide necessary force to the disc to prevent it from falling down.

Change in momentum of the bullets fired in 1 second,

$\longrightarrow\sf{\Delta p=nm(v-u)}$

$\longrightarrow\sf{\Delta p=0.5(-u-u)}$

$\longrightarrow\sf{\Delta p=0.5\times-2u}$

$\longrightarrow\sf{\Delta p=-u}$

Hence force acting on the disc by the bullets is,

$\longrightarrow\sf{F=\dfrac{\Delta p}{\Delta t}}$

$\longrightarrow\sf{F=\dfrac{-u}{1}}$

$\longrightarrow\sf{F=-u}$

Negative sign shows that force is acting upwards. Its magnitude is,

$\longrightarrow\sf{F=u}$

This force is equal to weight of disc as it is necessary to balance the disc.

Therefore,

$\longrightarrow\sf{F=Mg}$

$\longrightarrow\sf{u=2g}$

$\longrightarrow\underline{\underline{\sf{u=19.6\ m\,s^{-1}}}}$

Hence (4) is the answer.