Question

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 5.40 m before stopping. How far does the lighter fragment slide?

Part A: Assume that both fragments have the same coefficient of kinetic friction.

Answer 1

Answer:-

we can state that from given condition there is external force acting in x-direction, so x-componet of momentum will not be conserved for given system of two fragments, at the same time the two part start to fly away. Hence the momentum in x-direction will be conserved.

so consider p_ix = p_fx

0= 7*m *v_M -m*v_m

from above equation we will get the velcotiy for heavier

v_M=v_m/7

at the time of accelerartion friction force is resist to the body and it cause the zero speed.

now we can calculate the net work done with respect to change in kinetic energy for both object.

W = -$\mu$k *M*g*(d₁) = -1/2 *(7m )* v_M²

Hence

v_M2 = 2$\mu$k *M*g*(d₁)

Same for Small object

v_m2 = 2$\mu$k *m*g*(d₂)

by comparing both and the velocity is one seven time.

We will get the

d1 = 1/(7²) d₂

d₂ = 49*d₁

d₂= 49*5.4 = 264.6 m (Answer)

Hence the lighter fragment will dispalced 264.6 m