Question

A bowling ball of mass mb = 3.3 kg is rolled down the lane with a velocity of v1 = 2.4 m/s. it strikes a single remaining pin of mass mp = 0.71 kg head on. after the collision the bowling ball has a velocity (in the same direction) of v2 = 1.3 m/s.

Answer 1

This high school physics question requires applying conservation of momentum principles to a bowling ball and pin collision to determine the post-collision velocity of the pin.

Step-by-step explanation:

The subject of the question is identifying what happens after a collision in a bowling scenario where momentum and kinetic energy are key principles in play. This is a classic physics problem dealing with the topic of inelastic collisions in one dimension and conservation of momentum. The problem provided is a real-life application of the concepts learned in a physics course, usually encountered by students at the high school level.

The student is asked to determine the post-collision velocity of the pin, using the conservation of momentum, which states that the total momentum of a closed system is constant if no external forces are acting on it. Since the bowling ball continues in the same direction, it's implied that this is a head-on elastic collision (provided no other information suggests energy is not conserved). The mass of the bowling ball (mb) is 3.3 kg, its initial velocity (v1) is 2.4 m/s, and its velocity after collision (v2) is 1.3 m/s. The mass of the pin (mp) is given as 0.71 kg. With these values, we can use the conservation of momentum formula to find the velocity of the bowling pin after the collision.