Inelastic Collision and Pendulums

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SUMMARY

The discussion centers on an inelastic collision involving two pendulums with masses m1 = 0.145 kg and m2 = 0.200 kg. The first pendulum is released from a height of d = 0.092 m, and the problem requires determining how high the center of mass rises after the collision. The key takeaway is that in a completely inelastic collision, mechanical energy is not conserved, but momentum is. The correct approach involves using conservation of momentum to find the velocity of the combined mass after the collision, which is essential for calculating the height reached by the center of mass.

PREREQUISITES
  • Understanding of inelastic collisions in physics
  • Knowledge of conservation of momentum principles
  • Familiarity with potential and kinetic energy equations
  • Basic algebra for solving equations
NEXT STEPS
  • Study the principles of conservation of momentum in inelastic collisions
  • Learn how to calculate the velocity of combined masses post-collision
  • Explore the relationship between kinetic energy and height in pendulum systems
  • Review examples of inelastic collisions in real-world scenarios
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of inelastic collisions and energy conservation principles.

seichan
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Homework Statement


Two 23-cm long pendulums (each made of a massless string and a ball) are initially situated as shown in the figure. The masses of the left and right balls are m1= .145 kg and m2= .200 kg , respectively. The first pendulum is released from a height d= .092 m and strikes the second. Assuming that the collision is completely inelastic and neglecting the mass of the strings and any frictional effects, how high does the center of mass rises after the collision?
prob05.gif


Homework Equations


U(x)=mgh
KE=1/2mv^2

The Attempt at a Solution


I started out by solving for the velocity of mass one when it strikes mass two. Because it is raised to a height d, its potential energy is mgd. Because energy is conserved, this is equal to the kinetic energy at its equilibrium point (where mass two is at rest). Thus, m1gd=.5m1V^2
sqrt(2gd)=V
I then used this velocity to calculate the Kinetic energy of mass two.
.5m2V^2=KE
.5m2(2gd)=KE
m2gd=KE
This, in turn, is equal to the potential energy at the height it goes to. Thus:
m2gh=m2gd
h=d

This, however is not the case (clearly). Whatever you could tell me about where my thinking has gone wrong would be greatly appreciated.
 
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seichan said:
[

The Attempt at a Solution


I started out by solving for the velocity of mass one when it strikes mass two. Because it is raised to a height d, its potential energy is mgd. Because energy is conserved, this is equal to the kinetic energy at its equilibrium point (where mass two is at rest).

It is mentioned explicitly that the collision is completely inelastic. In such collisions, mechanical energy is not conserved, and the two objects stick together. But the momentum of the system just before collision is equal to the momentum just afterward. From that you can find the speed just after impact.

Try it now.
 
The problem says that the collision is completely inelastic. This means that the bodies stick together after the collision. This means that V of m1 is NOT equal V of m2 after the collision. Use conservation of momentum to find the velocity of the combined mass after the collision and then figure how high it goes.
 

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