Potential Energy - 2D inelastic collision - ballistic pendulum

[potatogirl]

Homework Statement

Consider a projectile (0.1 kg) fired into a ballistic pendulum (0.5 kg, r_cm = 0.3m), the resulting collision is inelastic and the pendulum swings to a maximum angular deflection of 25º. Assume r_cm is the same with or without the projectile attached. What is the change in potential energy for the pendulum due to the collision?

Relevant Equations

ΔPE = mgΔh
ΔPE = ΔKE

I really can't find anything in my textbook about how PE relates to 2D collisions, but here is what I do know...

ΔPE = mgΔh

And I know that since it is a completely inelastic equation kinetic energy is not conserved but momentum is conserved. The change in kinetic energy should be equal to the change in potential energy. So I thought about using:

ΔPE = ΔKE

And then solving for PE...but I don't know how to go about that since ΔKE is 1/2mv^2 - 1/2mv_i^2 and I don't have the velocity (or time variable to solve for it).

There is a given of r_cm = 0.3m which I thought would be the center of mass, and already resolved for x and y components. Is that wrong?

Finally, there are some additional equations about the conservation of momentum, where final momentum equals initial momentum. Since momentum is conserved, I could totally see how that may apply. But again, without knowing velocity I wouldn't know how to apply that.

Anyways...I have a feeling this is really simple but I am missing something. The question also says "see figure 2" but there is no figure 2 included in the document.

Edit: Also, if the answer is just ΔPE = mgΔh - do I simply just plug in total mass (0.6 kg) times gravity (-9.8m/s^2) time Δh (25º)? That doesn't feel like the way....

Thank you for any insight! This isn't a homework problem, just a "pre-lab" question.

 

[hutchphd]

Homework Statement: Consider a projectile (0.1 kg) fired into a ballistic pendulum (0.5 kg, r_cm = 0.3m), the resulting collision is inelastic and the pendulum swings to a maximum angular deflection of 25º. Assume r_cm is the same with or without the projectile attached. What is the change in potential energy for the pendulum due to the collision?
Relevant Equations: ΔPE = mgΔh
ΔPE = ΔKE

And I know that since it is a completely inelastic equation kinetic energy is not conserved but momentum is conserved. The change in kinetic energy should be equal to the change in potential energy.

This statement is not true. During the collision some of the energy goes to deforming the participants and ending as heat

1. You need to assume the collision itself is very quick and conserves momentum.
2. Then assume the remainng KE is given to gravitational potential

 

[kuruman]

ΔPE = ΔKE

This statement is incorrect. Energy conservation means that the sum of kinetic and potential energy is constant. This means that if one form gains a certain amount of Joules, the other form must lose that same amount. In other words the sum of changes must be zero,$$\Delta KE + \Delta PE =0 \implies \Delta PE=-\Delta KE.$$As @hutchphd suggested, you can use this idea after the bullet is fully embedded in the pendulum. At that point, the pendulum + bullet system has kinetic energy (how much?) that is converted to potential energy as it rises to maximum height.

 

[potatogirl]

This statement is incorrect. Energy conservation means that the sum of kinetic and potential energy is constant. This means that if one form gains a certain amount of Joules, the other form must lose that same amount. In other words the sum of changes must be zero,$$\Delta KE + \Delta PE =0 \implies \Delta PE=-\Delta KE.$$As @hutchphd suggested, you can use this idea after the bullet is fully embedded in the pendulum. At that point, the pendulum + bullet system has kinetic energy (how much?) that is converted to potential energy as it rises to maximum height.

Ah, okay good to know. Thanks.

So from here, I am having a hard time determining how to calculate anything without velocity. Also, I am not sure I understand the given variable r_cm

I am used to seeing x_cm for the location of center of mass, or v_cm for velocity of center of mass. Is it just the velocity of the center of mass that I am given there? The r is throwing me off.

Thanks again!

 

[kuruman]

So from here, I am having a hard time determining how to calculate anything without velocity.

What velocity do you think you need?

Also, I am not sure I understand the given variable r_cm

The pendulum rises to maximum angle of 25°. How does that translate into a maximum height from the lowest point of the motion if r_cm = 0.3 m ? What if r_cm = 0.6 m?

 

[hutchphd]

How much (vector) momentum is there just before the collision? How much immediately after? What is the mass of the bullet-pendulum lump? So what is its velocity immediately after the collision? and therefore what is its KE?

 

[potatogirl]

Momentum = mass * velocity. So without velocity how do I know momentum? Similarly, conservation of momentum says the final momentum (m_f1v_f1 + m_f2v_f2) is equal to initial momentum (m_i1v_i1 + m_i2v_i2) but again without velocity, how would I go about that?

The mass of the bullet-pendulum lump should be the sum of masses, so 0.6 kg.

Calculating KE I would normally do with 1/2mv²
but I just keep running into not having a way to get velocity...

 

[potatogirl]

What velocity do you think you need?

The pendulum rises to maximum angle of 25°. How does that translate into a maximum height from the lowest point of the motion if r_cm = 0.3 m ? What if r_cm = 0.6 m?

I assume I use trig for the y-comp to get the height. Is that right? Like 0.3m * cos25 . But I am running into a fundamental confusion about what r_cm represents I think. Does it represent the rest height? AKA the lowest point?

 

[kuruman]

r_cm represents the length of the string that forms the pendulum. For purposes of calculation assume that the bullet + pendulum center of mass is at length r_cm from the point of support of the string.

Yes, use trig but draw a drawing to figure out the change in height from the lowest point of the motion to maximum height. Don't guess.

 

[potatogirl]

Thanks!

 

[Herman Trivilino]

Have you looked up ballistic pendulum in the index of a textbook? There should be a worked example.