Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 9 - Impulse and Momentum - Exercises and Problems: 42

Answer

The speed of the first ball after the collision is 1.73 m/s at an angle of $54.7^{\circ}$ south of east.

Work Step by Step

Let $M$ be the mass of each ball. We can find the east-west component of the initial momentum of the system of the two balls. Let east be the positive direction. $p_{0x} = (M)(2.00~m/s)+(M)(-1.00~m/s)$ $p_{0x} = (M)(1.00~m/s)$ By conservation of momentum, the east-west component of the momentum after the collision must be equal to $p_{0x}$. We can find the east-west component $v_x$ of the first ball's velocity after the collision. $p_{fx} = p_{0x}$ $(M)~v_x + (M)(0)= (M)(1.00~m/s)$ $v_x = 1.00~m/s$ The east-west component of the first ball's velocity after the collision is 1.00 m/s to the east. The north-south component of he initial momentum is zero. By conservation of momentum, the north-south component of the momentum after the collision must be zero. We can find the north-south component $v_y$ of the first ball's velocity after the collision. Let north be the positive direction. $p_{fy} = p_{0y}$ $p_{fy} = 0$ $(M)~v_y + (M)(1.41~m/s)= 0$ $v_y = -1.41~m/s$ The north-south component of the first ball's velocity after the collision is 1.41 m/s to the south. We can find the speed of the first ball. $v = \sqrt{(v_x)^2+(v_y)^2}$ $v = \sqrt{(1.00~m/s)^2+(-1.41~m/s)^2}$ $v = 1.73~m/s$ We can find the angle $\theta$ south of east. $tan(\theta) = \frac{v_y}{v_x}$ $\theta = arctan(\frac{1.41~m/s}{1.00~m/s})$ $\theta = 54.7^{\circ}$ The speed of the first ball after the collision is 1.73 m/s at an angle of $54.7^{\circ}$ south of east.
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