Question

70kg clown on ice skates is inatially at rest when he catches a 5kg ball moving towards him at 15m/sec. Neglecting friction, the speed of the clown and the ball is caught will most nearly be what

Answer 1

Conservation of momentum!

Momentum in has to equal the momentum out. momentum (p) = mass (m) times velocity (v)

You start out with momentum of the clown in one direction and the ball in the opposite direction. You end up with both the clown and the ball moving at the same speed in the same direction

Answer 2

here we have to consider the momentum of the system clown + ball
Let's suppose that we have a collision along one determined direction
the initial momentum has the subsequent magnitude:
\(m_C \cdot v_C=5 \cdot 15=75\) \(Kg \;m/s\)
final momentum is:
\((m_B+m_C) \cdot v_f\)
where \(m_B,m_C\) are the masses of the ball and clown respectively, \(v_f\) is the magnitude of the final celocity
Now, if we can neglect any friction force, then we can write:
\[\left( {{m_B} + {m_C}} \right){v_f} = 75\]
namely the conservation of total momentum, since the system ball + clown is an isolated system. Then we have:
\[\Large {v_f} = \frac{{75}}{{{m_B} + {m_C}}} = \frac{{75}}{{70 + 5}} = ...?\]
which is the requested speed

Answer 3

oops... velocity*