Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 7 - Linear Momentum - Problems - Page 194: 42

Answer

The smallest value of $v$ which will cause the pendulum to swing over the top of the arc is $v = \frac{m+M}{m}\sqrt{4gl}$

Work Step by Step

Let $m$ be the mass of the projectile and let $M$ be the mass of the block on the pendulum. Let $v$ be the initial speed of a projectile. We can use conservation of momentum to find the speed $v'$ just after the collision. $m~v = (m+M)~v'$ We can use conservation of energy to find the height $h$ reached by the pendulum as it swings up. $\frac{1}{2}(m+M)~(v')^2 = (m+M)~gh$ $v' = \sqrt{2gh}$ We can replace $v'$ in the first equation above. $m~v = (m+M)\sqrt{2gh}$ $v = \frac{m+M}{m}\sqrt{2gh}$ Since we want the pendulum to swing all the way over the top, we can let $h = 2l$, where $l$ is the length of the pendulum. $v = \frac{m+M}{m}\sqrt{2g(2l)} = \frac{m+M}{m}\sqrt{4gl}$ The smallest value of $v$ which will cause the pendulum to swing over the top of the arc is: $v = \frac{m+M}{m}\sqrt{4gl}$
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