Lonnie pitches a baseball of mass 0.500 kg.The ball arrives at home plate with a speed of 35.0 m/s and is batted straight back to Lonnie with a return speed of 50.0 m/s.If the bat is in contact with the ball for 0.055 s,what is the impulse experienced by the ball?

Question

Quizplus · Accepted Answer

The impulse experienced by the ball is given by the change in momentum of the ball, which can be calculated as follows:

Initial momentum of the ball = m*v1 = 0.500 kg * 35.0 m/s = 17.5 kg*m/s
Final momentum of the ball = m*v2 = 0.500 kg * (-50.0 m/s) = -25.0 kg*m/s

Change in momentum = final momentum - initial momentum = (-25.0 - 17.5) kg*m/s = -42.5 kg*m/s

The negative sign indicates that the direction of the impulse is opposite to the initial momentum of the ball.

Since the bat is in contact with the ball for 0.055 s, the average force exerted by the bat on the ball can be calculated as follows:

Average force = change in momentum / time = (-42.5 kg*m/s) / (0.055 s) = -773 N

Again, the negative sign indicates that the force is in the opposite direction to the initial momentum of the ball.

Therefore, the impulse experienced by the ball is 42.5 N*s (choice B).

Lonnie Pitches a Baseball of Mass 0