# Ball, spring and curved incline - is this right?

Ball, spring and curved incline -- is this right?

Picture a smooth, level, frictionless surface (A) that leads into a curved incline (B-C). The top of the incline (C) is 0.4 meter above the flat surface. At the start of A is a spring with a spring coefficient of 10, with one end fixed and a ball weighing 10 g placed at the other end (facing the curve). The spring is then compressed by 10 cm. When released, the ball flies into the air from point C. Determine the ball's speed when it leaves C.

I first determined the height the ball reaches, using:
0.5kx^2 = mgh
h = 0.51 m

Then determined the velocity the ball would have if it were dropped from a height of 0.11 meter (that is, the height above point C), so that
v^2 = 2ah
v^2 = 2 (9.8)0.11
v = 1.46 m/s

I'm not sure about the second calculation, because the ball could have been moving at an angle when it left C.

HallsofIvy
Homework Helper
The angle doesn't matter because you are asked for "speed", not "velocity".
This is, as you saw, a question of conservation of energy. The ball starts with the energy given to it by the spring (the same as the work done to compress the spring: 0.5kx2 as you calculated. The maximum height the ball can reach, which you calculate is irrelevant- it's the energy you need.

When the ball is at point C, mgh= (10)(9.8)(4) is the potential energy. Subtract that from the ball's total energy to find the kinetic energy at that point and use
(1/2)mv2 to find the speed.

On second thought your method of finding the highest point and then the speed of a ball dropped from that height to C should give exactly the same answer. Use my method as a check!

Yes, i got the same answer. Since this is a conservation of energy problem, obviously your approach is the more elegant of the two. Makes sense: the PE that has been consumed has been converted to speed.
Thanks