Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 11 - Impulse and Momentum - Exercises and Problems: 52

Answer

After the explosion, the speed of the heavier fragment is 20 m/s and it is moving straight down.

Work Step by Step

We first find the velocity of the rocket after it accelerates upward for 2.00 seconds. $v = a~t$ $v = (10.0~m/s^2)(2.00~s)$ $v = 20.0~m/s$ We then find the momentum of the rocket just before it explodes. We can call this momentum $p_0$. $p_0 = m~v$ $p_0 = (1500~kg)(20.0~m/s)$ $p_0 = 30,000~N~s$ Next, we find the height $h$ of the rocket just before it explodes: $h = \frac{1}{2}at^2$ $h = \frac{1}{2}(10.0~m/s^2)(2.00~s)^2$ $h = 20~m$ After the explosion, the lighter fragment (with mass 500 kg) reaches a height that is 510 meters above the point of the explosion. We can find the lighter piece's velocity $v_L$ just after the explosion. $v_f^2 = v_L^2+2gy$ $v_L^2 = 0-2gy$ $v_L = \sqrt{-2gy}$ $v_L = \sqrt{-(2)(-9.80~m/s^2)(510~m)}$ $v_L = 100~m/s$ Just after the rocket explodes, the sum of the momenta of the two fragments will be equal to $p_0$ (by conservation of momentum). We can find the velocity $v_H$ of the heavier piece (with mass 1000 kg) just after the explosion. $p_f=p_0$ $(1000~kg)~v_H+(500~kg)(100~m/s) = 30,000~N~s$ $(1000~kg)~v_H = -20,000~N~s$ $v_H = -20~m/s$ After the explosion, the speed of the heavier fragment is 20 m/s and it is moving straight down.
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