A torry and a car moving with the same kinetic energy are brought to rest by the application of brakes. Which provide equal retarding forces. Which of them will come to rest in shorter time? Which will come to rest in less distance?

A torry and a car moving with the same kinetic energy are brought to r

1.	A torry and a car moving with the same kinetic energy are brought to rest by the application of brakes. Which provide equal retarding forces. Which of them will come to rest in shorter time? Which will come to rest in less distance?
Answer» Solution :(a) Initial kinetie energy of each vehicle is K and when they come to rest their final kinetic energy is zero. The change in kinetic energy of both the car and lorry is K. By work -energy theorem FS= Change in kinetic energy `FS= K, S = (K)/(F)` Where F is the retarding force. As K and F are same for both the vehicles, they come to rest after travelling the same distance (b) To find the time in which each vehicle comes to rest, we have to consider the momentum, `p = sqrt(2m xx K) " since " K= (p^(2))/(2m)` Th initial momentum of the car is less than that of the lorry as K is same for both but mass of the car is less than that of the lorry. Final momentum is zero for both the vehicles. So, change in momentum is less for the car. By Newton.s second LAW, `F= (Delta P)/(Delta t), Delta t= (Delta p)/(F)` Since .F. is same, `Delta t PROP Delta p` As car has smaller momentum, the car comes to rest EARLIER.