Potential energy stored in spring = (1/2) k s^2
That is the work done on the ball by the spring
I am not sure at what height the ball is when the spring is compressed.
Then the ball speeds from the ramp at speed (1/2)m Vr^2 = (1/2)k s^2
so calculate Vr
It now has horizontal speed =Vr cos θ which remains constant until we hit the floor
The vertical speed is Vr sin θ, call that Vi
Its height is then H + y above the floor
Now do the vertical problem
v = Vi - g t
h = (H+y) + Vi t - (1/2) g t^2
h = 0 at floor, solve quadratic for t
then you have v at the floor and the horizontal speed u = Vr cos θ
speed at floor = sqrt(u^2+v^2
A child's toy consists of a block that attaches to a table with a suction cup, a spring connected to that block, a ball, and a launching ramp. The spring has a spring constant k, the ball has a mass m, and the ramp rises a height y above the table, the surface of which is a height H above the floor.
Initially, the spring rests at its equilibrium length. The spring then is compressed a distance s, where the ball is held at rest. The ball is then released, launching it up the ramp. When the ball leaves the launching ramp its velocity vector makes an angle θ with respect to the horizontal.
Throughout this problem, ignore friction and air resistance.
1) Calculate vr, the speed of the ball when it leaves the launching ramp and Express the speed of the ball in terms of k, s, m, g, y, and/or H.
2) With what speed will the ball hit the floor?
Express the speed in terms of k, s, m, g, y, and/or H.
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