Question

A 2.0 kg arrow is fired at 10 m/s into a stationary 10kg block of ice what is the velocity of the arrow / ice combination after the collision?

Answer 1

Answer:

1.67 m/s

Explanation:

Since this is an inelastic collision (the block and the arrow stick together after the collision), we can solve the problem by using the law of conservation of momentum:

$p_i = p_f\\m_a v_a + m_b v_b = (m_a + m_b) v$

where

$m_a = 2.0 kg$ is the mass of the arrow

$v_a = 10 m/s$ is the initial speed of the arrow

$m_b = 10 kg$ is the mass of the block

$v_b = 0$ is the initial speed of the block

$v = ?$ is the final speed of the arrow+block

Substituting and re-arranging the equation, we find:

$v=\frac{m_a v_a + m_b v_b}{m_a +m_b}=\frac{(2.0 kg)(10 m/s)+0}{2.0 kg+10.0 kg}=1.67 m/s$