Question

An object with total mass m_total = 8.9 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 = 49 J.

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

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

3) If the explosion lasted for a time t = 0.02 s, what was the average force on the larger piece?

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

Answer 1

Total Mass M = 8.9 kg

Mass of small piece m = M/ 4 = 8.9 / 4= 2.225 kg

Mass of larger piece m ' = 3 m = 3 M / 4 = 6.675 kg

Initial velocity of the object u = 0

Initial kinetic energy of the object K = 0

Total energy of the pieces after collision K ' = (1/2) m v ² + ( 1/ 2)m' v' ²

According to problem K ' = 49 J

So, (1/2) m v ² + ( 1/ 2)m' v' ² = 49 J

1.1125 v ² +  3.3375 v' ² = 49 J -------( 1)

From law of conservation of linear momentum , Mu = m v + m ' v '

0 = ( M / 4 ) v + (3M /4) v '

From this   ( M / 4 ) v = - (3M /4) v '

v = -3v' -----( 2)

Substitute equation ( 2) in equation ( 1) you get ,

1.1125 (-3v') ² +  3.3375 v' ² = 49 J

10.0125  v' ² + 3.3375 v' ²=49

13.35  v' ² = 49

  v' ² = 3.67

v ' = 1.915 m / s

(1).The speed of the smaller piece after the collision v = -3v '

   = -5.747 m / s

2) The speed of the larger piece after the collision v ' = 1.915 m / s

3) The explosion lasted for a time t = 0.02 s

The average force on the larger piece F = Change in momentum / time

= m'(v'-0) / t

= 12.78 / 0.02

= 639.13 N

4) The magnitude of the change in momentum of the smaller piece = m (v - 0 )

= m v

= 12.78 kg m/s