Question

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 1.50 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?

Answer 1

how to approach

1 calc the potential energy (mgh) of each cube relative to the base - using the 1.5m

2 when the cubes exit the track, all that energy can be assumed to be kinetic energy - 1/2 m v^2

3 you can thus calc the velocity of each cube, and you can use the 40deg to resolve that into horiz and vertical components

4 use the horizontal Vcos40 one to calc the change in momentume of each cube

5 NB velocity just before hitting wall is v, velocity after is -v/2 so change in momentum is 3v/2

6 tricky bit coming up?! hopefully not. force = (A) mass of object * acceleration of object or (B) rate of change of momentum of object ie f = d(mv)/dt = ma, they're the same idea. ERGO as 10 cubes hit every second, the Force is 10 x change in momentum of 1 cube.

hope this helps.

Answer 2

thanks a lot