Question

A 910-kg sports car collides into the rear end of a 2500-kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.5 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 the speed sports car at impact?

Express your answer to two significant figures and include the appropriate units.

Answer 1

First we need to know how fast the BOTH cars were moving immediately after the impact, then we can use conservation of momentum two work out how fast the sports car must have been going immediately before the impact.
We can use the equations of motion and Newton's second law (F=ma) to work out the speed of the joined vehicles immedieatly after the collision.
F=ma
a=F/m
F=uN where u is the coefficient of friction and N is the normal reaction force between the cars and the road, equal to mg (mass times gravity)
N=mg
F=umg
substitute this in to a=F/m
a=umg/m
a=ug
a=0.8*9.8
a=7.84 m/s^2

Now we use the equation of motion v^2=u^2+2ax
where:
v is the final velocity (0)
u is the initial velocity (unknown)
a is the acceleration (-7.84) (negative because the car is slowing down)
x is the distance (3)
v^2=u^2+2as
u^2=v^2-2as
u^2=0-(2*-7.84*2.5)
u^2=39.2
u=6.26 m/s

So now we know that the speed of both cars immediately after the collision was 6.26 m/s
Consevation of energy tells us that
m(1)v(1)=m(2)v(2) where the (1) and (2) refer to initial and final states.
910*v(1) = (2500+910)*6.26
910 v(1) = 21346.6
v(1) = 23.45 m/s
the velocity of the sports car just before impact was 23.45 m/s.