Question

A bullet of gun going horizontal with velocity of $400m/s$ embedded in a bag of sand and comes to rest. The masses of bullet and bag are respectively $0.025kg$ and $1.975kg$ respectively. Find velocity of (bullet + sand). How much loss of kinetic energy in this process?[Ans : $5m/s,1975J$]

Answer 1

Initial momentum of bag is zero.

From conservation of momentum

Final momentum = initial momentum

$(M+m)V=mu$

Where, $u=$ initial velocity of bullet, $V=$ final velocity of both (buller+bag)

$V=\frac{mu}{M+m}=\frac{0.025\times 400}{1.927+0.025}=5m{s}^{-1}$

Loss in kinetic energy

$=\frac{1}{2}m{u}^2-\frac{1}{2}(M+m)V^2$

$\Rightarrow \frac{1}{2}\times 0.025\times {400}^2-\frac{1}{2}(1.927+0.025)\times 5^2$

$\Rightarrow 1975J$

Hence, velocity (bag+bullet) is $5m{s}^{-1}$ and loss in kinetic energy is $1975J$ .