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A ball has a mass of 4.65 kg and approximates a ping pong ball of mass 0.060 kg that is at rest by striking it in an elastic collision. The initial velocity of the bowling ball is 5.00 m / s, determine the final velocities of both masses after the collision. Use equations 9.21 and 9.22 from the textbook. The book is on WebAssign.

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Answer:

the final velocity of the ball is 4.87 m/s

the final velocity of the ping ball is 9.87 m/s

Explanation:

Given;

mass of the ball, m₁ = 4.65 kg

mass of the ping ball, m₂ = 0.06 kg

initial velocity of the ping ball, u₂ = 0

initial velocity of the ball, u₁ = 5 m/s

let the final velocity of the ball = v₁

let the final velocity of the ping ball, = v₂

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁  +  m₂u₂  = m₁v₁   +  m₂v₂

4.65(5)  +   0.06(0)   =   4.65v₁   +   0.06v₂

23.25 + 0 = 4.65v₁  +  0.06v₂

23.25 = 4.65v₁  +  0.06v₂  ------ (1)

Apply one-directional velocity equation;

u₁ + v₁ = u₂  +  v₂

5 + v₁ = 0  +  v₂

5 + v₁ = v₂

v₁ = v₂ - 5  -------- (2)

substitute equation (2) into (1)

23.25 = 4.65(v₂ - 5)  +  0.06v₂

23.25 = 4.65v₂  -  23.25   +   0.06v₂

46.5 = 4.71 v₂

v₂ = 46.5/4.71

v₂ = 9.87 m/s

v₁ = v₂ - 5

v₁ = 9.87 - 5

v₁ = 4.87 m/s

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