You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo: there is negligible friction between your feet and the ice. A friend throws you a 0.400 kg ball that is traveling horizontally at 10.0 m/s. Your mass is 70.0 kg. (a) If you catch the ball, with what speed do you and the ball move afterward? (b) If the ball hits you and bounces off your chest, so afterward it is moving horizontally at 8.0 m/s in the opposite direction, what is your speed after the collision?
Solution 20E Problem (a) Step 1: Mass of the person m = 70Kg 1 Mass of the ball m = 0.402 Velocity of the person before collision v = 0 1 Velocity of the ball v = 10.0m/s 2 Step 2: If the person catches the ball, the momentum is conserved. The total momentum before collision is equal to the total momentum after collision. After collision both the person and the ball move together along the same direction with same velocity. Step 3: Therefore m v + m v = (m +m )v 1 1 2 2 1 2 f Where v is fe velocity of the ball and the person Since v =1 m v = (m +m )v 2 2 1 2 f Step 4: Applying known values m v v = 2 2 f (m 1m ) 2 0.4 10 vf (70+0.4) v = 0.0568 m/s f Velocity of the person and the ball after the catch is 0.0568 m/s Problem (b) Step 1: The ball hits the person and moves with 8.0m/s (v ) in the opposite2irection. Therefore v = -20m/s So the ball and the person moves in opposite direction. Step 2: From conservation of momentum m2 2 m v1 1 v 2 2 Initial momentum of the person is zero. Where v is1 e velocity of the person after collision u2 s the initial velocity of the ball (10m/s)