a 2.5 kg ball moving at 7.50 m/s is caught by a 70.0 kg man while the man is standing on ice. how fast will the man / ball combination be moving after the ball is caught by the man?
To find the final speed of the man/ball combination after the ball is caught, we can apply the principle of conservation of momentum.
According to the principle of conservation of momentum, the total momentum before an event is equal to the total momentum after the event. In this case, the momentum before the ball is caught is given by the product of the mass and velocity of the ball, and the momentum after the ball is caught is given by the product of the mass and velocity of the man/ball combination.
The equation we can use to express this conservation of momentum is:
(mass of ball * velocity of ball) + (mass of man * velocity of man) = (mass of ball + mass of man) * final velocity
Let's plug in the given values:
Mass of ball (m1) = 2.5 kg
Velocity of ball (v1) = 7.50 m/s
Mass of man (m2) = 70.0 kg
Velocity of man (v2) = ?
Using the conservation of momentum equation:
(2.5 kg * 7.50 m/s) + (70.0 kg * 0 m/s) = (2.5 kg + 70.0 kg) * final velocity
(18.75 kg*m/s) = (72.5 kg) * final velocity
Dividing both sides by the mass of the entire system:
final velocity = (18.75 kg*m/s) / (72.5 kg)
final velocity ≈ 0.2586 m/s
Therefore, the man/ball combination will be moving at approximately 0.2586 m/s after the ball is caught by the man.