Question

A pool ball of 0.17 kg moves at 10 m/s and strikes a stationary ball (at rest) also at 0.17 kg.

Answer 1

The final velocity of the second ball is +10 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum. In fact, 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:

$m_1 = 0.17 kg$ is the mass of the first ball

$u_1 = 10 m/s$ is the initial velocity of the first ball (we take its direction as positive direction)

$v_1 = 0$ is the final velocity of the first ball

$m_2 = 0.17 kg$ is the mass of the second ball

$u_2 = 0$ is the initial velocity of the second ball

$v_2$ is the final velocity of the second ball

Re-arranging the equation and substituting the values, we find:

$v_2 = \frac{m_1 u_1}{m_2}=\frac{(0.17)(10)}{0.17}=10 m/s$

And since the sign is positive, the direction is the same as the initial direction of the first ball.

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