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 - Page 240: 11

Answer

(a) The impulse exerted on the rocket by the engine is $15,000~N~s$ (b) The rocket reaches its maximum speed at the end of the 30.0-second period when the rocket exerts a force on the rocket. After the force ends, the rocket reaches its maximum speed of 110 m/s

Work Step by Step

(a) We can use the graph to find the impulse exerted on the object. The impulse is equal to the area under the force versus time graph. $J = \frac{1}{2}~F_{max}~t$ $J = \frac{1}{2}(1000~N)(30.0~s)$ $J = 15,000~N~s$ The impulse exerted on the rocket by the engine is $15,000~N~s$. (b) The rocket reaches its maximum speed at the end of the 30.0-second period when the rocket exerts a force on the rocket. We can use the impulse to find the final momentum $p_f$. $p_f = p_0+J$ $p_f = m~v_0+J$ $p_f = (425~kg)(75.0~m/s)+15,000~N~s$ $p_f = 46,875~N~s$ We can use the final momentum to find the velocity $v_f$ after the force ends. $m~v_f = p_f$ $v_f = \frac{p_f}{m}$ $v_f = \frac{46,875~N~s}{425~kg}$ $v_f = 110~m/s$ After the force ends, the rocket reaches its maximum speed of 110 m/s.
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