Question

An object with total mass m_total = 6.3 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 = 44 J.

1)What is the speed of the smaller piece after the collision? (m/s)

2)What is the speed of the larger piece after the collision? (m/s)

3)If the explosion lasted for a time t = 0.028 s, what was the average force on the larger piece? (N)

4)What is the magnitude of the change in momentum of the smaller piece? (kg-m/s)

5)What is the magnitude of the velocity of the center of mass of the pieces after the collision? (m/s)

Answer 1

M_total = 6.3 kg   ,

V = 0   = initial velocity of object

m₁ = m

m₂ = 3 m

m₁ + m₂ = 6.3 kg

m + 3 m = 6.3 kg

m = 1.575 kg

m₁ = 1.575

m₂ = 3 m = 4.725 kg

using conservation of momentum ::

mV = m₁ V₁ + m₂ V₂
m (0) = m V₁ + 3m V₂

V₁ = - 3 V₂

Using conservation of energy ::

KE of m₁ + KE of m₂ = 44

(0.5) m₁ V₁² + (0.5) m₂ V₂² = 44

m (-3V₂)² + (3m) (V₂)² = 88

6 m V₂² = 44

6 (1.575) V₂² = 44

V₂ = 2.16 m/s
v₁ = - 3 V2 = -3 x 2.16 = -6.48 m/s

1) speed of smaller block = V₁ = - 6.48 m/s

2) speed of larger block = V₂ = 2.16 m/s

3) Using the formula , impulse = change in momentum

F x t = change in momentum

t = 0.028 sec

change in momentum of larger piece is given as ::

final momentum - initial momentum = (m₂) (V₂) - 0 = 4.725 x 2.16 = 10.21 kg m/s

F (0.028) = 10.21

F = 364.6 N

4) change in momentum of smaller mass = m1 V1 -0 = 1.575 x 6.48 = 10.21 kgm/s

5) velocity of center of mass = 0 m/s