Question

A spring clamped to the table shoots a 130 g block directly into another 90 g steel block, this perfectly elastic collision will launch the second block up a ramp. When the spring is compressed 16 cm, the ball travels vertically up a 76 cm ramp and then compresses another (100 N/m) at the top, 8 cm. What then is the spring constant for the first spring? Assume that all surfaces are frictionless.

Answer 1

Soln: By conservation of energy, Potential Energy in the first spring is equal to the potential enegy of the second spring.

Given, For the second Spring , K = 100 N/m and compression = 8 cm = 0.08 m

Therefoe Potential energy stored = 1/2 k x2 =  1/2 x 100 x .08 x0.08 = 32 x 10-2 J

Potential Energy of the first spring = 1/2 k x2

Here compression x= 16 cm = 0.16 m

By conservation of energy

1/2 k x2  = 32 x 10-2 J

1/2 x k x 0.16 x0.16 = 32 x 10-2

k = (2x 32 x 10-2) /  0.16 x0.16 = 2500 x 10-2 = 25 N/m