Question

A 2.90-kg ball, moving to the right at a velocity of +1.53 m/s on a frictionless table, collides head-on with a stationary 6.20-kg ball. Find the final velocities of (a) the 2.90-kg ball and of (b) the 6.20-kg ball if the collision is elastic. (c) Find the magnitude and direction of the final velocity of the two balls if the collision is completely inelastic.

Answer 1

(a) Let us consider the initial velocity of 2.9 kg ball is u₁ = 1.53 m/s
Initial velocity of 6.2 kg ball is u₂ = 0 m/s
After the collision
the velocity of 2.9 kg ball is V₁
the velocity of ball 6.2 kg is V₂
Therefore initial momentum of the system = m₁u₁ = 2.9*1.53 = 4.437 -------(1)
Final momentum after the collision
= m₁V₁ + m₂V₂ = (2.9*V₁) + (6.2*V₂)-------------(2)
Since the momentum will remain conserved
2.9V₁ + 6.2V₂ = 4.437 -------------(3)
Now the collision is elastic therefore
Velocity of approach = Velocity of seperation
u₁ - u₂ = V₂ - V₁
1.53 - 0 = V₂ - V₁
V₂ - V₁ = 1.53 ---------(4)
On solving 3 and 4
V₁ = -0.554 m/s (-ve sign indicates that the ball rebound in the -ve direction)
V₂ = 0.975 m/s
(b) The velocity of 6.2 kg ball will be 0.975 m/s
(c) If the collision is completely inelastic then both ball will move together after the collision
Initial momentum = m₁u₁ = 2.9*1.53 = 4.437 -----------(1)
Final momentum = (m₁ + m₂)V = (2.9+ 6.2)V -------------(2)
Applying the conservation of momentum
(2.9+ 6.2)V = 4.437
V = 0.486 m/s
the direction will be in + x direction