Question

An object with total mass m_total = 6.4 kg is sitting at rest when it explodes into two pieces. The two pieces, after the explosion, have masses of m and 3m. During the explosion, the pieces are given a total energy of E = 45 J.

What is the speed of the smaller piece after the collision?

What is the speed of the larger piece after the collision?

If the explosion lasted for a time t = 0.029 s, what was the average force on the larger piece?

What is the magnitude of the change in momentum of the smaller piece?

What is the magnitude of the velocity of the center of mass of the pieces after the collision?

Answer 1

m+3m = 6.4

:. 4m = 6.4

m= 1.6 and 3m = 4.6

momentum should be conserved.

m*v1 = 3m*v2

:. v1 = 3v2

and

1/2*m*v1^2 + 1/2*3m*v2^2 = 45

m*v1^2 + 3m*v2^2 = 90

m*9v2^2 + 3m*v2^2 = 90

12m*v2^2 = 90

m*v2^2 = 7.5

1.6*v2^2 = 7.5

:. v2 = 2.165 ( speed of lager piece)

and v1 = 3*v2

= 6.495 (speed of smaller piece)

average force = change in momentum / time

=3m*v2 - 0 / time

= 4.6*2.165 / 0.029

= 343.41

change in momentum = m*v1 - 0

= 1.6*6.495

= 10.392

velocity of centre of mass will be zero even after the explosion, as the pieces will explode in other directions, so that the momentum is conserved.

Answer 2

total mass=6.4 kg

after explosion ,two pieces have masses m,3m

i.e m1=1.6 kg m2=4.8 kg

let v1 ,v2 are velocities of m1,m2 after explosion

by principle of moments

m1v1+m2v2=0

=>m1v1=-m2v2

- sign indicatesopp direction

v1/v2=3

the energy given=45 J=1/2m1v1^2+1/2m2v2^2

=>45=9.6V2^2

=>V2=2.165 m/sec

=>v1=6.495 m/sec

the average force on the larger piece=m2v2/t=348.34 N

magnitude of change in momentum of smaller piece=m1v1=10.392 kg.m/sec

magnitude of the velocity of the center of mass of the pieces after the collision is v=3.4275 m/sec