Question

(a) A ball of mass m is thrown vertically upward from the ground with an initial speed v, its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attains the speed again before striking the ground. It implies that the magnitude of initial and final momentum of the ball are same. Yet, it is not an example of conservation of momentum. Explain why?
(b) A bullet of mass 20 g is horizontally fired with a velocity $150m{s}^1$ from a pistol of mass 2 kg. What is the recoil velocity of the pistol?

Answer 1

(a) Yes, it is not an example of conservation of momentum because momentum remains conserved when no external force is acting on the object. In this case, force of gravity is acting on the ball.

(b) $m_1=20g$ $=$ $\frac{20}{1000}$ $=$ $\frac{1}{50}$ kg

$v_1=150m{s}^{-1}$

$m_2=2kg$

According to coservation of momentum:

$m_1{v}_1=m_2{v}_2$

$\therefore$ $\frac{1}{50}\times 150=2\times v_2$

Therefore the recoil velocity of the pistol, $v_2=\frac{150}{50\times 2}=1.5m{s}^{-1}$