Show all work for full credit. A 2.3 kg pendulum bob is pulled out at an angle s
ID: 3160773 • Letter: S
Question
Show all work for full credit.
A 2.3 kg pendulum bob is pulled out at an angle such that its center of mass is 0.26 m above its height when it passes through its lowest point. It is released and the pendulum swings down. At the lowest point in its swing it strikes a 1.5 kg block initially at rest on a frictionless surface.
a) What is the gravitational potential energy of the pendulum initially.
b) What is the speed of the pendulum bob just before striking the block?
c) If the pendulum bob continues forward with a speed of 1.1 m/s, what is the speed of the block after the collision?
d) Is the collision elastic? Explain your answer.
Explanation / Answer
a) gravitational potential energy = m g h
= 2.3 x 9.8 x 0.26 = 5.8604 J
b) Kinetic energy gained = potential energy
5.8604 = 1/2 mv 2
v= 2.25 m/s
c) from conversation of momentum
m1u1 = m2v2 + m1v1
= 2.3 x 2.25 = 2.3 x 1.1 + 1.5 x v2
V2 = 1.31 m/s
d) Kinetic energy is not conserved as we can see
initial kinetic energy is 1/2 m1 u1 2 = 5.82 J
Finat kinetci energy is 1/2 m1 V12 + 1/2 m2 V22 = 1.39 + 1.26 J = 2.65 J
So it is not elastic as kinetic energy is not conserved