Question

A bowling ball with a mass of 7.0 kg strikes a pin that has a mass of 2.0 kg. The pin flies forward with a velocity of 6.0 m/s, and the ball continues forward at 4.0 m/s. What was the original velocity of the ball?
a
5.7 m/s
b
6.6 m/s
c
4.0 m/s
d
3.3 m/s

Answer 1

To find the original velocity of the bowling ball before the collision, we can use the principle of conservation of momentum. The total momentum before the collision must equal the total momentum after the collision.

Let:
- \( m_b = 7.0 \, \text{kg} \) (mass of the bowling ball)
- \( m_p = 2.0 \, \text{kg} \) (mass of the pin)
- \( v_b \) = original velocity of the bowling ball (unknown)
- \( v_p = 6.0 \, \text{m/s} \) (velocity of the pin after the collision)
- \( v'_b = 4.0 \, \text{m/s} \) (velocity of the bowling ball after the collision)

Using the conservation of momentum:
\[
\text{Initial Momentum} = \text{Final Momentum}
\]
\[
m_b \cdot v_b + m_p \cdot 0 = m_b \cdot v'_b + m_p \cdot v_p
\]

The pin is initially at rest, so its initial velocity is 0. Plugging in the values gives:
\[
7.0 \cdot v_b + 0 = 7.0 \cdot 4.0 + 2.0 \cdot 6.0
\]
Calculating the right side:
\[
7.0 \cdot 4.0 = 28.0 \, \text{kg m/s}
\]
\[
2.0 \cdot 6.0 = 12.0 \, \text{kg m/s}
\]
Now add these two results together:
\[
28.0 + 12.0 = 40.0 \, \text{kg m/s}
\]

So our equation becomes:
\[
7.0 \cdot v_b = 40.0
\]

Solving for \( v_b \):
\[
v_b = \frac{40.0}{7.0} \approx 5.71 \, \text{m/s}
\]

Thus, the original velocity of the bowling ball is approximately \( 5.7 \, \text{m/s} \).

The correct answer is:
**a) 5.7 m/s**