Question

A 0.3kg softball has a velocity of 15m/s at an angle of35^o below the horizontal just before making contact withthe bat. What is the magnitude of the change in momentum ofthe ball while in contact with the bat if the ball leaves with avelocity of 20m/s, vertically downward and 20m/s, horizontally backtoward the pitcher?

Answer 1

Initial momentum of the ball is p_i= (0.3kg)(15m/s)(cos(-35))i +(0.3kg)(15m/s)(sin(-35))j

                                                =(3.69kg.m/s)i -(2.58kg.m/s)j

If the ball leaves the bat vetrically down ward the the finalmomentum is p_f =(0.3kg)(20m/s)(-j)

                                                                                                               = (-6kg.m/s)j

Then the change in momentum isΔp =(-6kg.m/s)j -[(3.69kg.m/s)i -(2.58kg.m/s)j]

                                                     = (-3.69kg.m/s)i -(3.42kg.m/s)j

Then the magnitude of change in momentum is Δp =√[(-3.69kg.m/s)²+(-3.42kg.m/s)²]

                                                                           = 5.03kg.m/s

If the ball leaves the bat horizontally back to thepitcher, the the final momentum is p_f =(0.3kg)(20m/s)(-i)

                                                                                                               = (-6kg.m/s)i Then the change in momentumisΔp = (-6kg.m/s)i -[(3.69kg.m/s)i -(2.58kg.m/s)j]

                                                     = (-9.69kg.m/s)i +(2.58kg.m/s)j

Then the magnitude of change in momentum is Δp =√[(-9.69kg.m/s)²+(-2.58kg.m/s)²]

                                                                           = 10.03kg.m/s