Question

A bullet of mass 50 kg is fired from a riffle of mass 2 kg and the total kinetic energy produced by the explosion in 2050 J.The kinetic energy of the bullet is

• A. 50 J
• B. 2000 J
• C. 2050 J
• D. 587.7 J

Answer 1

The correct option is B 2000 J

Mass of bullet and that of gun are $m_b=0.05kg$ and $2kg$ respectively.
Kinetic energy of bullet $K_b=\frac{P_b^2}{2m_b}$
Kinetic energy of riffle $K_r=\frac{P_r^2}{2m_r}$
Energy of explosion $E=K_b+K_r=\frac{P_b^2}{2m_b}+\frac{P_r^2}{2m_r}$
But according to conservation of law of momentum, $P_b=P_r=P$
$\therefore$ $2050=\frac{P^2}{2\left(0.05\right)}+\frac{P^2}{2\left(2\right)}$
$⟹P^2=200$
Thus kinetic energy of bullet $K_b=\frac{P^2}{2m_b}=\frac{200}{2\left(0.05\right)}=2000J$