Force of a Baseball Swing.? A baseball has mass 0.145 kg. (a) If the velocity of a pitched ball has a magnitude of 45.0 m/s and the batted ball’s velocity is 55.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat. (b) If the ball remains in contact with the bat for 2.00 ms, find the magnitude of the average force applied by the bat.

Solution 8E Step 1: a) initial momentum of the ball, P initialu Where, m - mass of the ball and u - initial velocity Provided, mass of the ball, m = 0.145 kg Initial speed of the ball, u = 45 m/s Therefore, P = 0.145 kg × 45 m/s = 6.525 kg m/s initial Final momentum of the ball, P = mv final Where, m - mass of the ball and v - final velocity Provided, mass of the ball, m = 0.145 kg Final speed of the ball, u = 55 m/s Therefore, P final0.145 kg × 55 m/s = 7.975 kg m/s We should not take signs here because, we are dealing only with magnitudes. Change in momentum of the ball, P = P finalPinitial P = 7.975 kg m/s - 6.525 kg m/s = 1.45 kg m/s Step 2: a) Impulse is the large force acting for a short time. So, Impulse = Force × t We know that, force, F = ma = m (v-u)/t Or, F = ( mv- mu) / t That is, F = (P finalPinitialt Or, F = P / t So, we can write, Impulse = (P / t) × t = P Therefore, the impulse = 1.45 kg m/s