Question

A bomb at rest explodes into two fragments of mass 3 kg and 1 kg The total kinetic energy of fragments is 6×10⁴ J. Calculate kinetic energy of bigger fragment.

Answer 1

m1 = 3 kg
m2 = 1 kg
So by conservation of momentum
m1v1 +m2v2 = (m1+m2)0 m1v1 +m2v2 = 0
m1v1 =−m2v2
3v1 =−1v2
−3v1 =v2
Total kinetic energy :
1/2* m1v1^2+1/2* m2v2^2 =6 *10^4
1/2( 3*v1^2+v2^2) = 6 *10^4
3*v1^2+(-3v1)^2= 12 *10^4
3*v1^2+9v1^2= 12 *10^4
12v1^2= 12 *10^4
v1^2= 10^4
v1= 100 m/s
so v2= −300 m/s
In elastic collision the momentum is conserved as there is no loss in the body deterioration!