Chapter 1: Q20E (page 264)

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a $0.600 kg$ ball that is traveling horizontally at $10.0 m / s$ . Your mass is $70.0 k g$ . (a) If you catch the ball, with what speed do you and the ball move afterward? (b) If the ball hits you and bounces off your chest, so afterward it is moving horizontally at $8.0 m / s$ in the opposite direction, what is your speed after the collision?

Expert verified

(a) The speed of you and ball move afterward is $0.0849 m / s$

(b) The speed after collision is $0.154 m / s$ .

Step by step solution

Mass of ball is $m_{A} = 0.400 kg$

Mass of yours $m_{B} = 70.0 kg$

Let x be in the direction the ball is traveling initially.

When two bodies collide with each other then the total momentum before the collision is equal to the total momentum after the collision.

The speed of you and the ball move afterward is written as:

$P_{1 x} = P_{2 x} m_{A} v_{1} = (m_{A} + m_{B}) v_{2} 0.600 \times 10 = (0.600 + 70) v_{2} 6 = 70.6 v_{2}$

$v_{2} = \frac{6}{70.6} = 0.0849 m / s$

Hence,the speed of you and ball move afterward is $0.0849 m / s$ .

Here, the horizontal speed of ball when it hit you and bounces off your chest is $v = 8.0 m / s$ .

The speed after collision is calculated as follows:

$P_{1 x} = P_{2 x} m_{A} v_{1} = (m_{A} v + m_{B} v_{B 2}) 0.600 \times 10 = (0.600) (- 8) + 70 \times v_{B 2}$

$\begin{array}{rcl} v_{B 2} & = & \frac{6 + 4.8}{70} \\ = & 0.154 m / s \end{array}$

Hence, the speed after collision is $0.154 m / s$ .

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