A bowling ball has an initial momentum of [tex][tex]$+30 \, \text{kg} \cdot \text{m/s}$[/tex][/tex] and hits a stationary bowling pin. After the collision, the bowling ball leaves with a momentum of [tex][tex]$+13 \, \text{kg} \cdot \text{m/s}$[/tex][/tex].

What is the magnitude of the final momentum of the bowling pin if it has a mass of [tex]1.5 \, \text{kg}[tex]$[/tex]?

A. [tex]0 \, \text{kg} \cdot \text{m/s}$[/tex][/tex]
B. [tex]11 \, \text{kg} \cdot \text{m/s}[tex]$[/tex]
C. [tex]17 \, \text{kg} \cdot \text{m/s}$[/tex][/tex]
D. [tex]43 \, \text{kg} \cdot \text{m/s}$[/tex]



Answer :

To solve this problem, we need to use the principle of conservation of linear momentum, which states that the total momentum of a system remains constant if no external forces are acting on it.

Here’s the detailed step-by-step solution:

1. Identify the initial and final momenta of the bowling ball:
- Initial momentum of the bowling ball = [tex]\( +30 \, \text{kg} \cdot \text{m/s} \)[/tex]
- Final momentum of the bowling ball = [tex]\( +13 \, \text{kg} \cdot \text{m/s} \)[/tex]

2. Calculate the change in momentum of the bowling ball:
- Change in momentum of the bowling ball = Initial momentum - Final momentum
- Change in momentum of the bowling ball = [tex]\( 30 \, \text{kg} \cdot \text{m/s} - 13 \, \text{kg} \cdot \text{m/s} \)[/tex]
- Change in momentum of the bowling ball = [tex]\( 17 \, \text{kg} \cdot \text{m/s} \)[/tex]

3. Use the principle of conservation of momentum:
- According to the conservation of momentum, the change in momentum of the bowling ball is equal and opposite to the momentum gained by the bowling pin (assuming an isolated system with no external forces).

4. Calculate the final momentum of the bowling pin:
- Since the bowling pin was initially stationary, its initial momentum was [tex]\(0 \, \text{kg} \cdot \text{m/s} \)[/tex].
- Final momentum of the bowling pin = Change in momentum of the bowling ball
- Final momentum of the bowling pin = [tex]\( 17 \, \text{kg} \cdot \text{m/s} \)[/tex]

Therefore, the magnitude of the final momentum of the bowling pin is [tex]\( \boxed{17 \, \text{kg} \cdot \text{m/s}} \)[/tex].

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