Question

john throws .4 kg ball with velocity of 18 m/s. Hits .2 kg bottle and bottle flies 25 m/s. How fast is ball traveling after hitting the bottle?

Answer 1

Answer:

5.5 m/s

Explanation:

Assuming the bottle is initially stationary, we can write the law of conservation of momentum as follows:

$p_i = p_f\\m_b u_b = m_b v_b + m_B v_B$

where

$m_b = 0.4 kg$ is the mass of the ball

$u_b = 18 m/s$ is the initial velocity of the ball

$m_B = 0.2 kg$ is the mass of the bottle

$v_B = 25 m/s$ is the final velocity of the bottle

$v_b$ is the final velocity of the ball

Solving for $v_b$,

$v_b = \frac{m_b u_b - m_B v_B}{m_b}=\frac{(0.4 kg)(18 m/s)-(0.2 kg)(25 m/s)}{0.4 kg}=5.5 m/s$