A4.6 kg bowling ball sliding to the right at 7.43 m/s has an elastic head-on collision with another 4.6 kg bowling ball initially at rest. the first ball stops after the collision. find the velocity of the second ball after the collision. answer in units of m/s.

V₂ = 7.43 m/s

Explanation:

To find the velocity of the second ball after the collision, we will use the following formula:

where M₁ = mass of ball 1 = 4.6 kg  

M₂ = mass of ball 2 = 4.6 kg  

U₁ = velocity of ball 1 before collision = 7.43 m/s  

U₂ = velocity of ball 2 before collision = 0 m/s  

V₁ = velocity of ball 1 after collision = 0 m/s  

V₂ = velocity of ball 1 after collision = ?

So substituting the given values in the above formula to get:

(4.6 × 7.43) + (4.6 × 0) = (4.6 × 0) + (4.6 × V₂)

34.178 = (4.6)V₂

V₂ = 7.43 m/s

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