## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 7 - Problems - Page 266: 71

#### Answer

After the collision, puck B's speed is $1.73~m/s$ and it moves at an angle of $30.0^{\circ}$ below the +x-axis.

#### Work Step by Step

Let $M$ be the mass of each puck. By conservation of momentum, the final momentum of the system is equal to the initial momentum. We can find the horizontal component $v_x$ of the final velocity of puck B: $M~v_x +M~(1.0~m/s)~cos~60.0^{\circ} = M~(2.0~m/s)$ $v_x = (2.0~m/s)-(1.0~m/s)~cos~60.0^{\circ}$ $v_x = 1.5~m/s$ We can find the vertical component $v_y$ of the final velocity of puck B: $M~v_y + M~(1.0~m/s)~sin~60.0^{\circ} = M~(0)$ $v_y = -(1.0~m/s)~sin~60.0^{\circ}$ $v_y = -0.866~m/s$ We can find the magnitude of puck B's velocity after the collision: $\sqrt{v_x^2+v_y^2} = \sqrt{(1.5~m/s)^2+(-0.866~m/s)^2} = 1.73~m/s$ We can find the angle $\theta$ below the +x-axis: $tan~\theta = \frac{0.866}{1.5}$ $\theta = tan^{-1}(\frac{0.866}{1.5})$ $\theta = 30.0^{\circ}$ After the collision, puck B's speed is $1.73~m/s$ and it moves at an angle of $30.0^{\circ}$ below the +x-axis.

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