Question

After being struck by a bowling ball, a 1.5 kg bowling pin moving to the right at +3.0 m/s collides with another 1.5 kg bowling pin at rest. Find the velocity of the second pin when:
A)The first pin moves to the right after the collision at +0.5 m/s.
b)The first pin stops moving when it hits the second pin.

Answer 1

1) Velocity of the second pin after the collision: +2.5 m/s

2) Velocity of the second pin after the collision: +3.0 m/s

Explanation:

1)

We can solve the firs part of the problem by using the principle of conservation of momentum: the total momentum of the system (consisting of thw two pins) 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

$m_1 = 1.5 kg$ is the mass of the first pin

$u_1 = +3.0 m/s$ is the initial velocity of the first pin

$v_1 = +0.5 m/s$ is the final velocity of the first pin

$m_2 = 1.5 kg$ is the mass of the second pin

$u_2 = 0$ is the initial velocity of the second pin, which is at rest

$v_2$ is the final velocity of the second ball

Re-arranging the equation and solving the equation for v2, we find:

$v_2 = \frac{m_1 u_1-m_1 v_1}{m_2}=\frac{(1.5)(+3.0)-(1.5)(+0.5)}{1.5}=+2.5 m/s$

And the positive sign means that the second pin is moving in the same direction as the first pin.

2)

In this second part we apply again the law of conservation of momentum, but in this case we have:

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

where

$m_1 = 1.5 kg$

$u_1 = +3.0 m/s$

$v_1 = 0 m/s$, because in this case, the first pin stops moving when it hits the second pin

$m_2 = 1.5 kg$

$u_2 = 0$

$v_2$

And so, solving again for v2, we find:

$v_2 = \frac{m_1 u_1-m_1 v_1}{m_2}=\frac{(1.5)(+3.0)-0}{1.5}=+3.0 m/s$

This means that the second pin starts moving at +3.0 m/s after the collision.

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