A 90 kg man is standing still on frictionless ice. His friend tosses him a 10 kg ball, which has a horizontal velocity of 20 m/s. After catching the ball, what is the man's velocity?

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A 90 kg man is standing still on frictionless ice. His friend tosses him a 10 kg ball, which has a horizontal velocity of 20 m/s. After catching the ball, what is the man's velocity?

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Answer 1 · 2016-01-15T00:15:12

2 $\frac{m}{s}$ $m/s$

Explanation:

Using Law of Conservation of momentum.
Initial momentum is equal to final momentum
Momentum = $M_{m a n} . v_{m a n} + m_{b a l l} . v_{b a l l}$ $M_(man). v_(man)+m_(ball).v_(ball)$

Initial momentum = $90 \times 0 + 10 \times 20$ $90 times 0+ 10 times 20$

After the man catches the ball both ball and man move with the same velocity. Therefore
Final Momentum = $(M_{m a n} + m_{b a l l}) . v_{f i n a l}$ $(M_(man)+m_(ball)). v_(fi nal)$
Equating the two

$200 = 100 \times v_{f i n a l}$ $200= 100 times v_(fi nal)$

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A 90 kg man is standing still on frictionless ice. His friend tosses him a 10 kg ball, which has a horizontal velocity of 20 m/s. After catching the ball, what is the man's velocity?

1 Answer

Explanation:

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