A 7300 N elevator is to be given an acceleration of 0.150g by connecting it to a cable of negligible weight wrapped around a tur
1 answer:
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The final velocity of the skater is 2.34 m/s forward
Explanation:
We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.
Before the ball is thrown, the total momentum is:
where
M = 61 kg is the mass of the skater
m = 4.0 kg is the mass of the ball
u = 2.5 m/s (forward) is the combined velocity of the skater and the ball
After, the ball is thrown at twice the velocity, so the final total momentum is
where
V is the final velocity of the skater
v = 2(2.5) = 5.0 is the final velocity of the ball
Since the total momentum must be conserved, we can write
So, the skater is moving at 2.34 m/s (forward) after the shot.
Learn more about momentum:
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