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

5.5 m/s

Step-by-step 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 = (m_b u_b - m_B v_B)/(m_b)=((0.4 kg)(18 m/s)-(0.2 kg)(25 m/s))/(0.4 kg)=5.5 m/s`