1. ## A big car...

A big car of mass 2m travels at speed v, and a small car of mass m travels with a speed 2v. Both skid to a stop with the same coefficient of friction.
(a) The small car will have:
(1) a longer stopping distance
(2) the same stopping distance
(3) a shorter stopping distance

(b) Calculate the ratio of the stopping distance of the small car to that of the large car. (Use the work-energy theorem, not Newton's laws.)

2. Originally Posted by LLELLA
A big car of mass 2m travels at speed v, and a small car of mass m travels with a speed 2v. Both skid to a stop with the same coefficient of friction.
(a) The small car will have:
(1) a longer stopping distance
(2) the same stopping distance
(3) a shorter stopping distance

(b) Calculate the ratio of the stopping distance of the small car to that of the large car. (Use the work-energy theorem, not Newton's laws.)
We have to do a little work with Netwon's 2nd. For both cars I will use a coordinate system with +y upward.

Big car (BC): The free body diagram has a weight wBC downward, a normal force NBC upward, and a friction force, say, to the left. (I am taking the car to be moving to the right. The choice doesn't matter.) So in the y direction:
$\sum F_y = N_{BC} - w_{BC} = ma_y = 0$
Thus NBC = (mBC)g. Thus the friction force on BC is fBC = (mu)NBC = (mu)(mBC)g = 2(mu)mg. (mBC = 2m)

Now the work done by friction to stop the car will be equal in magnitude to the change in kinetic energy of the car. Specifically:
$W = \Delta KE$
$-f_{BC}d_{BC} = -(1/2)m_{BC}v_{BC}^2$
(The work done by friction is always negative, do the dot product if you aren't convinced. Here vBC is the initial speed of the BC, dBC is the stopping distance, and the final speed is, of course, zero.)
So:
$d_{BC} = \frac{(1/2)m_{BC}v_{BC}^2}{f_{BC}}$
In terms of the given variables: mBC = 2m, vBC = v
$d_{BC} = \frac{2mv^2}{2 \mu mg} = \frac{v^2}{2 \mu g}$

Small car: Virtually identical to the BC, except now mSC = m. So fSC = (mu)mg. So using the work formula once again, and putting mSC = m and vSC = 2v:
$d_{SC} = \frac{(1/2)m_{SC}v_{SC}^2}{f_{SC}} = \frac{m(2v)^2}{2 \mu mg} = 4 \left ( \frac{v^2}{2 \mu g} \right )$

So the small car takes 4x the distance to stop.

-Dan