## Physics (10th Edition)

Published by Wiley

# Chapter 7 - Impulse and Momentum - Problems - Page 196: 67

#### Answer

i) Total kinetic energy stays the same. ii) The total momentum in part b is smaller. In both a and b, $\sum KE=1160J$ In a, $\sum\vec{p}=+580kg.m/s$ in b, $\sum\vec{p}=+140kg.m/s$

#### Work Step by Step

i) The total kinetic energy of Jim and Tom is $$\sum KE=\frac{1}{2}(m_{Jim}v_{Jim}^2+m_{Tom}v_{Tom}^2)$$ As we can see, KE is a scalar quantity and works only with the magnitude of velocity. Therefore, in both cases, because the magnitudes of Jim's and Tom's velocities do not change, total KE stays the same. ii) The total momentum of Jim and Tom is $$\sum \vec{p}=m_{Jim}\vec{v_{Jim}}+m_{Tom}\vec{v_{Tom}}$$ Unlike energy, momentum is a vector and works with both magnitude and direction of velocity. Therefore, because Jim and Tom go in opposite directions in b), the total momentum in b) is smaller. 1) Find $\sum KE$ As mentioned, in both a) and b), $\sum KE$ is similar and equals $$\sum KE=\frac{1}{2}(m_{Jim}v_{Jim}^2+m_{Tom}v_{Tom}^2)$$ $$\sum KE=\frac{1}{2}(90\times4^2+55\times4^2)=1160J$$ 2) Find $\sum\vec{p}$ - Case a): $$\sum \vec{p}=m_{Jim}\vec{v_{Jim}}+m_{Tom}\vec{v_{Tom}}$$ $$\sum\vec{p}=90\times(+4)+55\times(+4)=+580kg.m/s$$ - Case b): $$\sum \vec{p}=m_{Jim}\vec{v_{Jim}}+m_{Tom}\vec{v_{Tom}}$$ $$\sum\vec{p}=90\times(+4)+55\times(-4)=+140kg.m/s$$

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