Question

A bowling ball of mass mb = 3.9 kg is rolled down the lane with a velocity of v1 = 2.3 m/s. It strikes a single remaining pin mp = 0.83 kg head on. After the collision, the bowling ball has a velocity (in the same direction) of v2 = 1.55 m/s. What is the change in momentum of the bowling ball?

1) 0.83 kg·m/s
2) 1.55 kg·m/s
3) 2.3 kg·m/s
4) 3.9 kg·m/s

Answer 1

The change in momentum of the bowling ball after striking a pin is found by calculating the difference between the final and initial momentum, resulting in a loss of 2.925 kg·m/s. None of the provided options is correct.

Step-by-step explanation:

The change in momentum of the bowling ball can be determined by taking the difference between the final and initial momentum of the bowling ball. Momentum is calculated as the product of mass and velocity. Therefore, the change in momentum (Δp) is given by:

Δp = mb * v2 - mb * v1

Where:

• mb is the mass of the bowling ball
• v1 is the initial velocity of the bowling ball
• v2 is the final velocity of the bowling ball

Substitute the given values:

Δp = 3.9 kg * 1.55 m/s - 3.9 kg * 2.3 m/s

Δp = (3.9 kg * 1.55 m/s) - (3.9 kg * 2.3 m/s)

Δp = 6.045 kg·m/s - 8.97 kg·m/s

Δp = -2.925 kg·m/s

The negative sign indicates that the bowling ball has lost momentum in the direction it was originally traveling. The absolute value of the change in momentum is 2.925 kg·m/s. Thus, none of the provided options is correct.