Question

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 0.370kg ball that is traveling horizontally at 9.00m/s . Your mass is 67.5kg .

Part A:If you catch the ball, with what speed do you and the ball move afterwards?

Part B:If the ball hits you and bounces off your chest, so that afterwards it is moving horizontally at 8.25

m/s in the opposite direction, what is your speed after the collision?

Answer 1

mass of the ball = 0.370 kg =m

velocity with which the ball approaches = 9.0 m/s=u

mass of the person to whom it is thrown = 67.5 kg = M

part (a) .when the poerson catches the ball ,the ball and man moves with same velocity that can be given by conservation of momentum

let V be the velocity of combined mass

and the initial direction of velocity of ball is taken to be negative ,

initial momentum of the ball = final momentum of ball+man together as a system

-m*u= (m+M)*V or V= -m*u/(m+M) = -0.049 m/s

hence the velocity with which the man and the ball would move = 0.049 m/s

part (b) when the ball hits me and rebounds back

after the rebound the velocity of ball = 8.25 m/s along +x direction =v(r)

so i will for sure move opposite to the ball along negative x direction ,

let my velocity after being striken by the ball = v

again applying conservation of momentum

initial momentum of ball before striking = final momentum of ball + my momentum

-m*u= m*v(r) - M*v or M*v= m*v(r)+m*u = 0.370*8.25 +0.370*9.0 =6.3825 or v=6.3825/M= 0.0981 m/s