Question

During a circus act, a performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of m1 = 10.5 kg and the horizontal component of its velocity is v = 8.85 m/s when the performer, of mass m2 = 61.5 kg, catches it. If the performer is on nearly frictionless roller skates, what is his recoil velocity?

Answer 1

The recoil velocity is 1.29 m/s

Step-by-step explanation:

In a collision, the momentum is conserved:

m₁v₁ + m₂v₂ = V(m₁ + m₂)

Where

m₁ = mass of ball = 10.5 kg

m₂ = mass of performer = 61.5 kg

v₁ = velocity of the ball = 8.85 m/s

v₂ = velocity of performer = 0

V = velocity after the collision = ?

Clearing V:

`V=(m_(1)v_(1)+m_(2)m_(2))/(m_(1)+m_(2) ) =((10.5*8.85)+(61.5*0))/(10.5+61.5) =1.29m/s`

Answer 2

1.512 m/s

Step-by-step explanation:

M₁ = 10.5

M₂ = 61.5kg

v₁ = 8.85m/s

v₂ = ?

This question involves conservation of momentum and law of conservation of momentum states that, the total momentum of a system does not change before and after impact and it's always constant.

Momentum = mass * velocity = M*V

total momentum of the system before caught the ball = total momentum after he caught the ball.

M₁V₁ = M₂V₂

10.5 * 8.85 = 61.5 * V₂

V₂ = 92.925 / 61.5

V₂ = 1.5109m/s ≈ 1.512m/s