A pendulum consists of a compact 1.00 kg bob attached to a string of length 2.40
ID: 1324153 • Letter: A
Question
A pendulum consists of a compact 1.00 kg bob attached to a string of length 2.40 m. A block of mass m rests on a horizontal frictionless surface. The pendulum is released from rest at an angle of 53? with the vertical. The bob collides elastically with the block at the lowest point in its arc. Following the collision, the maximum angle of the pendulum with the vertical is 5.73?. Determine the mass m (there are two possible values).
larger value kg smaller value kg A pendulum consists of a compact 1.00 kg bob attached to a string of length 2.40 m. A block of mass m rests on a horizontal frictionless surface. The pendulum is released from rest at an angle of 53½ with the vertical. The bob collides elastically with the block at the lowest point in its arc. Following the collision, the maximum angle of the pendulum with the vertical is 5.73½. Determine the mass m (there are two possible values).Explanation / Answer
Note that the total energy of the bob before collision is
E_final = MgH = mgL(1 - cos(53))
As
m = 1 kg
g = 9.8 m/s^2
L = 2.4 m
Thus,
E_initial = 9.37E+00 J
Also, the velocity of the bob at impact is
Vi = sqrt(2gh) = sqrt(2 g L (1 - cos (53))
Thus,
Vi = 4.327888782 m/s
Now, we get get the bob's velocity after the collision, as
Vf = 0.484807858 m/s
Using conservation of momentum,
m2 v2f = 3.843080924 kg*m/s [For the case where the bob went to the left after the collision]
m2 v2f = 4.81269664 kg*m/s [For the case where the bob went back to the right after the collision]
Using conservation of energy,
1/2 m2 v2f^2 = 9.247791326 J
Solving for m2 in the first case:
m2 = 0.799 kg [ANSWER, smaller value]
For the second case,
m2 = 1.25 kg [ANSWER, larger value]
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