A truck and a car are brought to a hault by application of same braking force. Which one will come to stop in a shorter distance if they are moving with same (a) velocity (b) kinetic energy and (c) momentum ?

Explanation

By applying breakes the body is brought to rest, so `v = 0` and `a = (-F//m)` (as it is retardation). If s is the distance travelled in stopping (called stopping distance), from 3rd equation of motion,
`v^(2) = u^(2) + 2as`
we have `0 = u^(2) - 2 (F//m)s`
i.e., `s = (mu^(2))/(2F)`
But `KE = (1)/(2) mu^(2)` and also `KE = (p^(2))/(2m)`
So, `s = (mu^(2))/(2F) = (KE)/(F) = (p^(2))/(2mF)`
From this it is clear that:
(a) If u is same:
`S prop (m u^(2) //2F)`, i.e., `s prop m`
Now as mass of car is less than that of truck, so car will stop in shorter distance.
(b) If KE is same:
`s prop (KE//F)`
So, both will stop after travelling same distance.
(c) If p is same:
`s = (p^(2)//2mF)`, i.e., `s prop (1//m)`
Now as mass of truck is more than that of car, so truck will stop in a shorter distance.