How Does Momentum Change Affect Force on a Wall with Rebounding Ice Cubes?

[Touchme]

Small ice cubes, each of mass 5.00 g, slide down a frictionless ski-jump track in a steady stream, as shown in Figure P6.71. Starting from rest, each cube moves down through a net vertical distance of y = 1.75 m and leaves the bottom end of the track at an angle of 40.0° above the horizontal. At the highest point of its subsequent trajectory, the cube stikes a vertical wall and rebounds with half the speed it had upon impact. If 10.0 cubes strike the wall per second, what average force is exerted on the wall?

I have no idea how to solve this question. I know how to get the final velocity when the ice is lauched. mgh = 0.5mv^2 and the answer is 5.857 m/s. Can someone guide me in doing this? Hints or suggestion will help too.

 

[OlderDan]

Small ice cubes, each of mass 5.00 g, slide down a frictionless ski-jump track in a steady stream, as shown in Figure P6.71. Starting from rest, each cube moves down through a net vertical distance of y = 1.75 m and leaves the bottom end of the track at an angle of 40.0° above the horizontal. At the highest point of its subsequent trajectory, the cube stikes a vertical wall and rebounds with half the speed it had upon impact. If 10.0 cubes strike the wall per second, what average force is exerted on the wall?

I have no idea how to solve this question. I know how to get the final velocity when the ice is lauched. mgh = 0.5mv^2 and the answer is 5.857 m/s. Can someone guide me in doing this? Hints or suggestion will help too.

This question was posted elsewhere, but the thread died out I believe. First step after finding the velocity at the end of the ramp is to find the velocity at the highest point of the trajectory, and then the change in momentum of each cube when it bounces. The average force is related to the momentum changes of the cubes and the time interval between the bounces.

 

[andrevdh]

At the highest point of the trajectory the y-velocity component will be zero. This means that only the horizontal velocity component remains. This means that the force that the ice blocks will exert on the wall will be horizontal. The wall will respond with a reaction force of similar magnitude in the opposite direction sending it back horizontally in the opposite direction. What will the resulting momentum change be for a single ice cube?