## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 7 - Problems - Page 267: 78

#### Answer

Since the car moving west was traveling at a speed of 30.8 m/s, which is well over the speed limit of 19.4 m/s, the car that was traveling west was speeding.

#### Work Step by Step

We can find the rate of deceleration while the two cars were sliding: $F_N~\mu_k = ma$ $mg~\mu_k = ma$ $a = g~\mu_k$ $a = (9.80~m/s^2)(0.80)$ $a = 7.84~m/s^2$ We can find the speed $v_0$ of the two cars just after the collision: $v_f^2 = v_0^2+2ad$ $v_0^2 = v_f^2 - 2ad$ $v_0 = \sqrt{v_f^2 - 2ad}$ $v_0 = \sqrt{0 - (2)(-7.84~m/s^2)(17~m)}$ $v_0 = 16.3~m/s$ By conservation of momentum, the final momentum of the system is equal to the initial momentum in both the north and west directions. We can use the north component of momentum to find the initial velocity of the car that was moving north: $p_{0n} = p_{fn}$ $(1300~kg)~v= (1100~kg+1300~kg)(16.3~m/s)~sin~30^{\circ}$ $v= \frac{(1100~kg+1300~kg)(16.3~m/s)~sin~30^{\circ}}{1300~kg}$ $v = 15.0~m/s$ We can use the west component of momentum to find the initial velocity of the car that was moving west: $p_{0w} = p_{fw}$ $(1100~kg)~v= (1100~kg+1300~kg)(16.3~m/s)~cos~30^{\circ}$ $v= \frac{(1100~kg+1300~kg)(16.3~m/s)~cos~30^{\circ}}{1100~kg}$ $v = 30.8~m/s$ We can convert the speed limit of 70 km/h to units of m/s: $70~km/h \times \frac{1000~m}{1~km} \times \frac{1~h}{3600~s} = 19.4~m/s$ Since the car moving west was traveling at a speed of 30.8 m/s, which is well over the speed limit of 19.4 m/s, the car that was traveling west was speeding.

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