Conservation of Momentum in an Inelastic Collision

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  • #1
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Homework Statement


A bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length L and negligible mass.

What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle.

Basically, what occurs is a bullet m is shot through a mass M which is suspended by a rod and pinned on the wall (it pivots in a circle). The problem asks for the minimum value of v so that M will spin in a circle once if the bullet emerges with a speed of v/2.


Homework Equations



m1v1+m2v2=m1v1'+m2v2'

Maybe a work/energy equation?

The Attempt at a Solution



I understand that the momentum of the bullet is transferred to the pendulum which is why the pendulum swings around and the bullet leaves with half the velocity it entered with, but I don't know how to find the amount of energy it takes to spin the pendulum.

Any help would be greatly appreciated.
 

Answers and Replies

  • #2
What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle

Do you know what that statement means?
 
  • #3
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Do you know what that statement means?

It means that the pendulum swings around in an perfect circle (it stops exactly where it starts).
 
  • #4
Yes, but it can swings with different velocities. What would be the minimum one, at the upper vertical position?

P.S. Actually the pendulum never stops!
 
  • #5
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I don't think it uses the equation I listed above with a set velocity. They give me L (the radius of the spin), which leads me to believe that I need to calculate the work done by the pendulum in the spin (W=FD and d is 2Lpi (circumference), but I'm not sure about force) and correlate the work done to the energy lost by the bullet when it leaves the pendulum. Also, there really is no displacement, so I'm not sure if I can even use the work equation for that.

What do you mean by the pendulum never stops? I'm assuming there is gravity even though they don't mention it.
 
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  • #6
You don't have to use a set velocity in the conservation of momentum!

But you do have to understand from the problem's statement that the pendulum must have zero velocity in the upper vertical position.
That's why it states minimum and barely.
 
  • #7
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Are you sure? I understand a=0 at that point, but the pendulum is moving the entire rotation.
 
  • #8
You are asked to calculate the minimum value, so think like this.

Let A be the point where the collision took place and let B the upper point.

In order for the pendulum to make the entire rotation it must reach point B. The velocity of M at B depends from the velocity at A, [itex]v_B[/itex] increases if [itex]increases[/itex]. Thus if you find the [itex]v_A[/itex] in order [itex]v_B=0[/itex], then you can say that this [itex]v_A[/itex] is the mimimum one, in the sence that every velocity above that ensures that the pendulum makes an entire rotation.

P.S. At point B the accerelation is not zero.
 
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  • #9
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Ohhh, I'm sorry if I incorrectly explained the problem, but the bullet is traveling at v and it asks for the minimum speed of the bullet as it passes through the pendulum in order for the pendulum to go around one rotation.
 
  • #10
:smile: My English are pretty poor! I will try again to explain.

  • From the conservation of momentum you can find the velocity [itex] V_A[/itex] of M, in terms of the velocity v of m.
  • From the conservation of energy for the mass M, from A to B, you can find the velocity [itex]V_B[/itex] in terms of [itex]V_A[/itex], i.e. in terms of v.
  • Now demanding [itex]V_B=0[/itex] in order the pendulum barely swings you can calulate v.

Does it makes any sence?
 
  • #11
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Doesn't Va of M is equal to zero since the pendulum is initially not moving? I don't quite understand what to do when the initial and final velocities of an object is zero, but it moves in between that period.
 
  • #12
Mass M:
Velocity before collision zero
Veloity after collision [itex]V_A[/itex] not zero! (zero at point B)

Mass m:
Velocity before collision v
Velocity after collision v/2

Can you find [itex]V_A[/itex] from the conservation of momentum?
 
  • #13
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Ahhhh, I simply solve for the variable Va. Let me try that! :) I really appreciate your help Rainbow Child.
 
  • #14
Ok! The equation you wrote (conservation of momentum) reads

[tex]p_{initial}=p_{final}\Rightarrow m\cdot u+M\cdot 0=m\cdot \frac{u}{2}+M\cdot V_A[/tex]

so you have for [itex]V_A[/itex] ...
 
  • #15
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Correct, that is what I have. I solved for Va and got Va=mv/2M. I'm not sure what to do after this. How would I find Vb and v from this?
 
  • #16
Now write the conservation of energy for M, from the point A to the point B, to calculate [itex]V_B[/itex] in terms of v.
 
  • #17
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I don't quite understand. Is that the 1/2mv^2 equation?
 
  • #18
The conservation of energy is

Potential EnergyA+Kinetic EnergyA=Potential EnergyB+Kinetic EnergyB (1)

For mass M,

[tex]PE_A=0,\, KE_A=\frac{1}{2}\,M\,V_A^2,\,PE_B=M\,g\,2\,L,\, KE_B=0[/tex]

Plug the above values in (1) in order to find [itex]V_A[/itex]
 
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  • #19
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I just solved for Va and got sqrt(2gL) (the mass canceled out). This equation gives me the speed of if the pendulum before the collision. I also have the speed of the pendulum after collision, which I found earlier from the conservation of momentum equation. Is this correct?
 
  • #20
Oupps! My mistake! [tex]PE_B=M\,g\,2\,L[/tex]. The mass M is at height [itex]2\,L[/itex].
Now equate the two results for [itex]V_A[/itex] in order to find v.
 
  • #21
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Ahhh! I should've caught that mistake. I now have Va=sqrt(4gL). I understand how to get v, but I am still unclear why mv/2M and sqrt(4gL) are equal? Are they just two different ways to write Va?
 
  • #22
We have computed [itex]V_A[/itex] by two ways.

The velocity [itex]V_A[/itex] is the final velocity of M after the collision, and simultaneously is the initial velocity for the movement of the pendulum.
 
  • #23
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Ah, everything finally makes sense now. I re-read everything you've said and it's all very clear now. :)

Thanks again Rainbow Child! You're a physics genius! I extremely appreciate you sticking with me the 3 or so hours it took for me to complete this fairly simple problem.
 
  • #24
Every problem fails to the category "fairly simple problem" when it is solved!!! :rofl:

Glad I helped! :smile:
 
  • #25
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Why don't you get the right answer if you only use conservation of energy->

initial KE = final KE + final U

[tex]\frac{1}{2} mv^2 = 2LMG + \frac{1}{2} m(v/2)^2[/tex]
 

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