Question

A ball of mass $0.2kg$ rests on a vertical post of height $5m$ . A bullet of mass $0.01kg$ , travelling with a velocity $Vm{s}^{-1}$ in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of $20m$ and the bullet at a distance of 100 m from the foot of the post. The velocity $V$ of the bullet is

• A. $250m{s}^{-1}$
• B. $350m{s}^{-1}$
• C. $400m{s}^{-1}$
• D. $500m{s}^{-1}$

Answer 1

Answer: (D)

Time of flight $\left(t_f\right)=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 5}{10}}=1s$
Horizontal range (R) = horizontal velocity $\times$ time of flight
$\therefore$ Horizontal velocities of the bullet and of the ball after the collision respectively are
${\left(v\right)}_{bullet}=\frac{100}{1}=100m{s}^{-1}$
${\left(v\right)}_{ball}=\frac{20}{1}=20m{s}^{-1}$
From the conservation of momentum, Total initial momentum = total final momentum
$\Rightarrow {\left(m\right)}_{bullet}\times V={\left(m\right)}_{bullet}\times {\left(v\right)}_{bullet}+{\left(m\right)}_{ball}\times {\left(v\right)}_{ball}$
$\Rightarrow 0.01V=0.01\times 100+0.2\times 20$
$\Rightarrow V=500m{s}^{-1}$