# Professional Application A 5.50-kg bowling ball moving at

## Problem 48PE Chapter 8

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition

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Problem 48PE

Professional Application A 5.50-kg bowling ball moving at 9.00 m/s collides with a 0.850-kg bowling pin, which is scattered at an angle of 85.0º to the initial direction of the bowling ball and with a speed of 15.0 m/s. (a) Calculate the final velocity (magnitude and direction) of the bowling ball. (b) Is the collision elastic? (c) Linear kinetic energy is greater after the collision. Discuss how spin on the ball might be converted to linear kinetic energy in the collision.

Step-by-Step Solution:

Step-by-step solution In this problem we have to calculate the final velocity of the bowling ball ,the collision is elastic or not and kinetic energy after the collision and also explain how spin on the ball might be converted to linear kinetic energy in the collision. Step 1 of 8 The momentum is conserved in collision, therefore use conservation of momentum. Step 2 of 8 (a) The expression of conservation of momentum along x-axis is, Here, are masses of bowling ball and bowling pin respectively, and is scattered angles is initial velocity of the bowling ball and is the final speed of the bowling ball and is the final velocity of the bowling pin. Step 3 of 8 Substitute for , for , for , for , for in the above expression, Hence the final speed of the bowling ball is along x-axis 8.80 m/s Step 4 of 8 The expression of the conservation of momentum along y-axis is: Substitute for , for , for , for and for in above expression of conservation of momentum along y-axis. Hence the final speed of the bowling ball is along y-axis 8.80 m/s

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