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Answer :
Final answer:
This question involves the concept of conservation of momentum in the case of a collision between a bowling ball and pin. Prior to the collision, the momentum is carried only by the moving bowling ball. After the collision, momentum is shared by both the bowling ball and the pin, which is scattered at a 65° angle to the ball's initial direction.
Explanation:
This question involves the topic of conservation of momentum, which is a fundamental concept in physics. Momentum is conserved during collisions as long as no external forces are involved. In the case of the bowling ball and pin, we have a high school level physics problem involving a 5.50kg bowling ball moving at 9.0m/s colliding with a 0.85kg bowling pin.
Before the collision, the momentum of the bowling ball is the only momentum in the system since the pin is stationary. This momentum can be calculated by multiplying the mass of the bowling ball by its velocity (5.50kg * 9.0m/s).
After the collision, the bowling pin is scattered at an angle of 65° with a speed of 15.0m/s. This gives the pin a momentum of 0.85kg * 15.0m/s. The bowling ball will also have some remaining momentum.
To draw diagrams for the system before and after the collision, you must depict the initial motion of the bowling ball and the stationary pin. After the collision, draw the bowling ball continuing in roughly the same direction, but with a reduced speed, and the pin moving off at a 65° angle to the initial direction of motion of the bowling ball.
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