Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 9 - Center of Mass and Linear Momentum - Problems - Page 251: 55a

Answer

The velocity of the 5.0-kg block immediately after the collision is $~~2.0~m/s$

Work Step by Step

We can use conservation of momentum to find the velocity $v_f$ of the 5.0-kg block immediately after the collision: $p_f = p_i$ $(5.0~kg)~v_f+(10.0~kg)(2.5~m/s) = (5.0~kg)(3.0~m/s)+(10.0~kg)(2.0~m/s)$ $(5.0~kg)~v_f = (5.0~kg)(3.0~m/s)+(10.0~kg)(2.0~m/s)-(10.0~kg)(2.5~m/s)$ $v_f = \frac{(5.0~kg)(3.0~m/s)+(10.0~kg)(2.0~m/s)-(10.0~kg)(2.5~m/s)}{5.0~kg}$ $v_f = 2.0~m/s$ The velocity of the 5.0-kg block immediately after the collision is $~~2.0~m/s~~$ in the original direction.
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