Question

Two particles A and B have masses 2 kg and 5 kg respectively. Initially the particles are both
travelling in
the same direction with A behind B. A Is travelling with speed 10ms and B Is
travelling with
speed 4ms. The particles collide and after the collision both particles continue
In the same direction. Given that the coefficient of restitution between the particles is 1/6 find
the speed of both particles after the collision.

Answer 1

Answer:

Particle A has a final speed of 5 m/s, and particle B has a final speed of 6 m/s.

Explanation:

Momentum is conserved:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(2 kg) (10 m/s) + (5 kg) (4 m/s) = (2 kg) v₁ + (5 kg) v₂

40 = 2v₁ + 5v₂

Coefficient of restitution is negative of the ratio of the difference in final velocities over the difference in initial velocities.

e = -(v₂ − v₁) / (u₂ − u₁)

1/6 = -(v₂ − v₁) / (4 − 10)

1 = v₂ − v₁

Solve the system of equations using substitution.

40 = 2v₁ + 5 (v₁ + 1)

40 = 2v₁ + 5v₁ + 5

35 = 7v₁

v₁ = 5

Since v₂ = v₁ + 1:

v₂ = 6

Particle A has a final speed of 5 m/s, and particle B has a final speed of 6 m/s.