Conservation of energy/momentum in a pendulum?

• mattpd1
In summary, the conversation discusses a problem involving a pendulum with a bob of mass 200g and a block of mass 1kg at rest at the bottom of the pendulum's swing. The first question asks for the speed of the bob right before it makes impact with the block, which is found to be 2.4 m/s. The second question asks for the velocity of the bob and block right after impact, which are calculated to be -1.6 m/s and 0.8 m/s respectively. The final question asks for the maximum rebound angle of the bob after impact, which can be found by calculating the height it reaches in its upward swing using its velocity at the bottom.
mattpd1

Homework Statement

http://www.phy-astr.gsu.edu/Hsiao-Ling/quiz2211-F10.ppt"

You have a pendulum:
The bob's mass is 200g.
The string length is 1m.
The angle the bob is raised to is 45 degrees.

You also have a block that is at rest at the bottom of the pendulum's swing:
The block's mass is 1kg.
The block is at rest.
No friction.

What is the speed of the bob right before it makes impact with the block?
What is the velocity of the bob and block right after impact?
What is the maximum rebound angle of the bob after impact?

Homework Equations

Total energy of system = U + K
Initial momentum of the system = Final momentum of the system
Total initial energy = total final energy

The Attempt at a Solution

I think I have solved the first question:
By calculating the change in height of the bob, to be .293m, I found the potential energy to be .574 joules. The kinetic energy right before impact should also equal .574, therefore velocity = 2.4 m/s. So right before impact the bob's speed is 2.4 m/s, is this right?

For the second questions:
Initial momentum = final momentum
.48 = .2Vf(bob) + Vf(block)

This is as far as I can get. How can I solve these velocities individually, and how to I go about finding the rebound angle?

Last edited by a moderator:
I think I may have answered the second part too.

Vf(block)=.8 m/s
Vf(bob)= -1.6 m/s

Does this look right?

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mattpd1 said:

The Attempt at a Solution

I think I have solved the first question:
By calculating the change in height of the bob, to be .293m, I found the potential energy to be .574 joules. The kinetic energy right before impact should also equal .574, therefore velocity = 2.4 m/s. So right before impact the bob's speed is 2.4 m/s, is this right?
Yes.
For the second questions:
Initial momentum = final momentum
.48 = .2Vf(bob) + Vf(block)

This is as far as I can get. How can I solve these velocities individually, and how to I go about finding the rebound angle?
Looks like you got the correct velocities below.

For rebound angle, you now have the "reverse" problem from before: when you knew the initial height of the bob, you were able to calculate it's velocity at the bottom of the swing. Now you know it's velocity at the bottom, so how high will it go in its upward swing?
mattpd1 said:
I think I may have answered the second part too.

Vf(block)=.8 m/s
Vf(bob)= -1.6 m/s

Does this look right?
Yes.

Got it, thank you much.

Your approach to solving the first question is correct. The potential energy at the highest point of the pendulum's swing is equal to the kinetic energy right before impact. Therefore, the velocity of the bob right before impact is 2.4 m/s.

For the second question, you have correctly stated that the initial momentum of the system (before impact) is equal to the final momentum of the system (after impact). Using the conservation of momentum equation, you can solve for the final velocities of the bob and the block. However, since the block is at rest before impact, its final velocity will be 0. This means that the final velocity of the bob can be calculated by solving the equation 0.48 = 0.2Vf(bob) + 0.

To find the maximum rebound angle of the bob after impact, you can use the conservation of energy equation. The initial total energy of the system (before impact) is equal to the final total energy of the system (after impact). The initial total energy includes both the potential energy and the kinetic energy of the bob, while the final total energy only includes the potential energy of the bob (since its kinetic energy becomes 0 after impact). By setting these two equal to each other and solving for the rebound angle, you can find the maximum angle the bob will reach after impact.

In summary, the key concepts to keep in mind when solving this problem are the conservation of energy and momentum. By using these principles, you can solve for the velocities and angles in the system before and after impact.

1. What is the law of conservation of energy/momentum in a pendulum?

The law of conservation of energy/momentum in a pendulum states that the total energy or momentum of a pendulum system remains constant, regardless of the changes in the motion of the pendulum. In other words, the energy/momentum cannot be created or destroyed, but it can be converted from one form to another.

2. Why is the conservation of energy/momentum important in a pendulum?

The conservation of energy/momentum is important in a pendulum because it helps us understand and predict the motion of the pendulum. Knowing that the total energy/momentum remains constant allows us to calculate the potential and kinetic energy/momentum at any point in the pendulum's swing.

3. How is energy/momentum conserved in a pendulum?

Energy/momentum is conserved in a pendulum because the pendulum's motion is governed by the law of conservation of energy/momentum. As the pendulum swings back and forth, the potential energy/momentum is converted into kinetic energy/momentum and vice versa, but the total energy/momentum remains the same.

4. What factors affect the conservation of energy/momentum in a pendulum?

The main factors that affect the conservation of energy/momentum in a pendulum are the length of the pendulum, the mass of the bob, and the angle at which the pendulum is released. A longer pendulum, a heavier bob, and a larger release angle will result in a higher conservation of energy/momentum.

5. How can the conservation of energy/momentum be applied in real-life situations?

The conservation of energy/momentum in a pendulum has many real-life applications, such as in clocks, amusement park rides, and energy harvesting devices. It is also used in physics experiments and simulations to study the behavior of pendulums and other systems that follow the law of conservation of energy/momentum.

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