A 0.28 kg puck is released from rest at the top of a 3.0 m high ramp. It slides
ID: 1704129 • Letter: A
Question
A 0.28 kg puck is released from rest at the top of a 3.0 m high ramp. It slides down the frictionless ramp and slides 1.5 m across a horizontal surface until hitting a 0.15 kg puck at rest. The two pucks stick together and continue sliding. They then encounter a spring with a spring constant of 262 N/m.a) What was the initial gravitational potential energy of the puck?
b) What was the velocity of the puck as it reached the horizontal surface?
c) What was velocity of the two pucks after the collision?
d) How far is the spring compressed?
Explanation / Answer
The mass of the first puck m1 = 0.28kg
the height of the ramp h = 3.0m
the mass of the second puck m2 = 0.15kg
The spring constant k = 262 N/m
(a) The initial gravitational energy
E = mgh = (0.28)(9.8)(3)
= 8.23 J
(b) from law of conservation of energy
mgh = 1/2mv^2
then the velocity of the puck
v = 2gh
= 2*9.8*3 = 7.67 m/s
(c) from law of conservation of momentum
m1v1 = (m1 + m2) vf
therefore the velocity
vf = m1v1/(m1+m2)
= (0.28)(7.67) / (0.28+0.15)
= 4.99 or 5m/s
(d) Again from law of conservation of energy
1/2 (m1+ m2) vf^2 = 1/2 kx^2
Therefore the compression
x = [(m1+m2)/k ] vf
= [(0.28+0.15)/262 ] (5)
= 0.20m or 20cm
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