Question

A target of mass $M$ is moving with a velocity $u$. Lead shots, each of mass $m$, are fired on it in the opposite direction with a velocity $v$. Find the number of bullets required to stop the target.

• A. $\frac{Mu}{mv}$
• B. $\frac{mv}{Mu}$
• C. $\frac{(M+m)u}{v}$
• D. $\frac{mu(M+m)}{v}$

Answer 1

The correct option is A $\frac{Mu}{mv}$

Let $n$ be the number of lead shots required to stop the target.

Given:

Mass of the target = $M$

Velocity of the target = $u$

Mass of each shot = $m$

Velocity of each shot = $v$

Initial momentum of the target and the shots is $P_1=Mu-nmv$ (the minus sign indicates that the shots are moving in the opposite direction)

Final momentum of the target and the shots is $P_2=0$

From the conservation of momentum,

$P_1=P_2$

$\Rightarrow Mu-nmv=0$

$\Rightarrow n=\frac{Mu}{mv}$