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 - General Problems - Page 196: 72

Answer

The block will rise up to a height of 0.831 meters.

Work Step by Step

Let $m$ be the mass of a bullet and let $M$ be the mass of the block. Let $v$ be the initial speed of the bullet. 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 relationship between $v'$ and the height $h$ reached by the block: $\frac{1}{2}(m+M)~(v')^2 = (m+M)~gh$ $v' = \sqrt{2gh}$ We can replace $v'$ in the first equation above. $m~v = (m+M)\sqrt{2gh}$ $h = (\frac{m}{m+M})^2~\frac{v^2}{2g}$ $h = (\frac{0.0250~kg}{0.0250~kg+1.40~kg})^2~\frac{(230~m/s)^2}{(2)(9.80~m/s^2)}$ $h = 0.831~m$ The block will rise up to a height of 0.831 meters.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.