Calculating Average Force on a Vertical Wall from Ice Cube Collisions

[Pulk]

1. Small ice cubes, each of mass 5.60 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 = 1.70 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 strikes 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?

The attempt at a solution
So I was able to find the Vx (because when the cube hits the wall it has Vy=0) with ease using sqrt(2*g*h) * cos 40.
Then in order to find the force my goal was to find the momentum and divide it by the time it took the cube to hit the wall which I could solve using the Vyinitial & Vyfinal. This did not work and now I'm stuck in a bit of a rut and am out of ideas.

 

[robbondo]

Just throwing out ideas. You can determine the velocity when the ice cube is flying off the ramp, because at the top of the ramp kinetic energy is zero, and at the bottom of the ramp, the potential is zero, ie height is zero. Also force is a vector quantity so, you probably have to take that into consideration as far as the angle when the cube hits the wall? I dunno, I'm just giving you something else to consider.

 

[saket]

Well done.. u did the tough part of the problem! Remaining is easy.
I guess, u are confusing with the time-thing.
Find the change in momentum of one cube as it hits the wall. Get total for 10 cubes. Now, this total change in momentum is brought about in "1 second"!

 

[Pulk]

So, I tried it again and examined the change in momentum. I realized that since the v of the cube after the collision the change in momentum would be 1/2 of the original for each cube. Then I multiplied this value by 10, to take into account the 10 cubes. Unfortunately after this work I still can't figure out the correct answer.

 

[saket]

So, I tried it again and examined the change in momentum. I realized that since the v of the cube after the collision the change in momentum would be 1/2 of the original for each cube. Then I multiplied this value by 10, to take into account the 10 cubes. Unfortunately after this work I still can't figure out the correct answer.

"the change in momentum would be 1/2 of the original for each cube"

no.. u r dealing with momentum, a vector quantity... please take care of the sign.

 

[Pulk]

Yup, I realized that mistake and got the answer right, thanks for your help!