Question

A 960-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.6 m before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be 0.80, calculates the speed of the sports car at impact. What was that speed? Express answer in two significant figures.

Answer 1

Answer:

the speed of the sports car at impact=  21.66 m/s

Explanation:

maximum Acceleration of the car moving on a rough road is given by

= -μg= 0.8×9.81= -7.848 m/s^2

if v is the initial velocity Let vf be the velocity of car after collision and d=2.6 m be the displacement of the car after collision

we know that

$v_f^2-v^2= 2ad$

⇒$(v_f)^2-0^2= 2\times7.848\times2.6$

v_f= 6.38 m/s

now using the law of conservation of momentum before and after collision

we have

$mv_{ic}+Mv_{isuv}= (M+m)v$

v_{isuv}= 0 m/s

$mv_{ic}=(M+m)v_f$

$v_{ic}= \frac{(M+m)v_f}{m}$

$v_{ic}= \frac{(2300+960)6.38}{960}$

v_{ic} =21.66 m/s

the speed of the sports car at impact=  21.66 m/s

Answer 2

Answer:

speed of car is 21.6 m/s

Explanation:

Given speed:

sport car weight is 960 kg

SUV car weight is 2300 kg

skid distance is 2.6 m

coefficient of friction is 0.80

Taking After impact:  

force of kinetic friction = µk = µ(m1 +m2) g

F = (0.80)(960+2400)(9.81) = 25,584.48 N

acceleration$a = \frac{F}{M} = \frac{25584.48}{(960+2300)}$

a = 7.84 m/s^2

from equation of motion

d =1/2 at^2

siolving for t

$t^2  = \frac{d}{0.5 a}$

$t^2 = \frac{2.6}{0.5 \times 7.84} = 0.662$

t = 0.813 s

momentum after collision MV

$m = (960 + 2300) \times at$

$m = (960+2300) \times 7.84 \times 0.813 = 20778.97 kgm/s$

mometum of car before collision = MV = 20778.97 KGM.S

Speed of car before collision = $v = \frac{m}{M1} = \frac{20778.97}{960} = 21.64 m/s$