## University Physics with Modern Physics (14th Edition)

Published by Pearson

# Chapter 8 - Momentum, Impulse, and Collision - Problems - Exercises - Page 268: 8.73

#### Answer

The chunks of ice will go up to a height of 1.84 meters.

#### Work Step by Step

We can use conservation of momentum to find the speed after the collision. $m_2v_2 = m_1v_1$ $v_2 = \frac{m_1v_1}{m_2}$ $v_2 = \frac{(5.00~kg)(12.0~m/s)}{10.00~kg}$ $v_2 = 6.00~m/s$ When the blocks reach maximum height, the potential energy will be equal to the kinetic energy at the bottom of the hill. $PE = K$ $m_2~gh = \frac{1}{2}m_2~v_2^2$ $h = \frac{v_2^2}{2g}$ $h = \frac{(6.00~m/s)^2}{(2)(9.80~m/s^2)}$ $h = 1.84~m$ The chunks of ice will go up to a height of 1.84 meters.

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