Question

A 16.0kg canoe moving to the left at 12.5m/s makes an elastic head-on collision with a 14.0kg raft moving to the right at 16.0m/s. After the collision, the raft moves to the left at 14.4m/s. Disregard any effects of the water. Find the velocity of the canoe after the collision.

Answer 1

The canoe is moving at 14.1 m/s to the right after the collision.

Explanation:

According to the law of conservation of momentum, in absence of external forces the total momentum of the system must be conserved before and after the collision. So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 16.0 kg$ is the mass of the canoe

$u_1 = -12.5 m/s$ is the initial velocity of canoe (we take right as positive direction, and since the canoe is moving to the left, its velocity is negative)

$v_1$ is the final velocity of the canoe

$m_2 = 14.0 kg$ is the mass of the raft

$u_2 = +16.0 m/s$ is the initial velocity of the raft

$v_2 = -14.4 m/s$ is the final velocity of the raft

Re-arranging the equation and substituting the values, we find: the final velocity of the canoe:

$v_1 = \frac{m_1 u_1 + m_2 u_2-m_2 v_2}{m_1}=\frac{(16.0)(-12.5)+(14.0)(16.0)-(14.0)(-14.4)}{16.0}=+14.1 m/s$

So, the canoe is moving at 14.1 m/s to the right after the collision.

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