a 400 g rubber ball is dropped a vertical distance of 12 m onto the pavement. it is in contact with the pavement for 0.01 s and rebounds to a height of 10 m. what is the total change in momentum? what average force is exerted by the pavement on the ball?

Question

this is what i have so far. 12/.01 and 10/.01 to get the velocities. so (.4)1000-.4(1200)= -80 kg. m/s. I think this is the total change in momentum if i did that correctly. i do not know how to get the force.

Answer 1

The velocity from the fall is
vf= sqrt(2g*h)=sqrt (2*9,8*12)= sbout 15m/s

The velocity at rebound is given by
vi=sqrt(2gh)=sqrt(2*9.8*10=about 14m/s

Force*timecontact=changemomentum
force= changemomentum/time
=about (.4*15-(.4)(-14))/.01
about 1160 N.

check my thinking. Remember, when the ball reverses, the sign of the momentum changes also.