## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 7 - Problems - Page 268: 89

#### Answer

(a) The change in momentum of the car due to the fly is $-0.003~kg~m/s$ (b) The change in momentum of the fly due to the car is $0.003~kg~m/s$ (c) A total of 101,010 flies are required to reduce the car's speed by 1 km/h.

#### Work Step by Step

(a) We can convert the ball's speed from units of km/h to units of m/s: $100~km/h \times \frac{1000~m}{1~km} \times \frac{1~hr}{3600~s} = 27.777778~m/s$ By conservation of momentum, the final momentum of the system is equal to the initial momentum. We can find the final speed $v_f$ of the car and fly: $m_f~v_f = m_0~v_0$ $v_f = \frac{m_0~v_0}{m_f}$ $v_f = \frac{(1000~kg)(27.777778~m/s)}{1000~kg+0.0001~kg}$ $v_f = 27.777775~m/s$ We can find the change in momentum of the car: $\Delta p = p_f-p_0$ $\Delta p = m~v_f-m~v_0$ $\Delta p = m~(v_f-v_0)$ $\Delta p = (1000~kg)~(27.777775~m/s-27.777778~m/s)$ $\Delta p = -0.003~kg~m/s$ The change in momentum of the car due to the fly is $-0.003~kg~m/s$ (b) We can find the change in momentum of the fly: $\Delta p = p_f-p_0$ $\Delta p = m~v_f-m~v_0$ $\Delta p = m~(v_f-v_0)$ $\Delta p = (0.0001~kg)~(27.777775~m/s-0~m/s)$ $\Delta p = 0.003~kg~m/s$ The change in momentum of the fly due to the car is $0.003~kg~m/s$ (c) We can find the number of flies required to reduce the car's speed by 1%: $m_f~v_f = m_0~v_0$ $m_f~(0.99~v_0) = m_0~v_0$ $m_f = \frac{m_0}{0.99}$ $m_f = \frac{1000~kg}{0.99}$ $m_f = 1010.1010~kg$ To reduce the car's speed by 1%, the final mass of the car and the flies must be $1010.1010~kg$. Then the mass of the flies must be $10.1010~kg$. We can find the number of flies required: $\frac{10.1010~kg}{0.0001~kg} = 101,010~flies$ A total of 101,010 flies are required to reduce the car's speed by 1 km/h.

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