How Does Bullet Speed Change After Hitting a Pendulum?

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AI Thread Summary
The discussion centers on calculating the final speed of a bullet after it passes through a pendulum in an inelastic collision. Participants clarify that while momentum is conserved, kinetic energy is not due to the nature of inelastic collisions. The pendulum's height after the impact can be used to determine its velocity, which is essential for solving the problem. The confusion arises from mixing the energies of the bullet and the pendulum, which should be treated separately. Ultimately, the key takeaway is that energy conservation applies to the pendulum post-collision, while momentum conservation applies to the system of the bullet and pendulum during the collision.
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Homework Statement



A 10g bullet moving at 300.m/s hits and passes through a 2.20kg pendulum target. After the impact, the 1.20m long pendulum rises up to an angle of 15 degree, find the final speed of the bullet.

Homework Equations


Ek= 1/2MV^2
Ep= MGH
Momentum = MV

The Attempt at a Solution



I assumed all energy before and after the question will be = because the collision is inelastic, both masses don't stay together. I found the height the pendulum rises by doing 1.2-(1.2Cos15). Then I put everything i knew in EK= Ep + EK

(300^2)*(0.01)*(0.5) = (2.20)(9.8)(0.04) + (0.5)(0.01)(v')^2

When I solve for velocity I get the answer wrong. What am I doing wrong, am I wrong that energy is conserved ?
 
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why are you using 300^2 ? use conservation of linear momentum to get the relationship between the initial and final velocities of the bullet.
 
I tried solving via energy.
 
Though when I try solving momentum i get. 300(0.01)=2.2(v)+0.01(v) 2 variables.
 
well you need BOTH energy and momentum to solve this... since bullet gets stuck inside the pendulum, this is an inelastic collision and kinetic energy is not conserved...
 
Though the bullet leaves the pendulum, do we still consider this inelastic ?
 
oh sorry. didn't read the question properly. let me think
 
Conservation of momentum works regardless of the nature of the collision. The trick here is to figure out how much momentum ended up in the pendulum bob...
 
well. the collision is still inelastic. use the conservation of energy to get the velocity of the
pendulum immediately after the collision. then use conservation of linear momentum to connect all three velocities
 
  • #10
this doesn't work.
 
  • #11
And if the collision is still inelastic you can't "use the conservation of energy to get the velocity of the pendulum immediately after the collision" because in a inelastic collision EK is not conserved.
 
  • #12
zaddyzad said:
And if the collision is still inelastic you can't "use the conservation of energy to get the velocity of the pendulum immediately after the collision" because in a inelastic collision EK is not conserved.

You were given other information about the pendulum involving its energy...
 
  • #13
Thats why I tried solving by... Ek(bullet) = Ep(pendulum) + EK(bullet)
Why doesn't that work ?
 
  • #14
well, let's say that velocity of the pendulum immediately after the collision is v_2. Now since the gravitational force is conservative, here, the total energy is
conserved and you can use conservation of energy to connect angle with this velocity and find v_2. since the bullet is momentarily inside the pendulum, the KE of the bullet is NOT conserved. but if we take system as bullet + pendulum, then there are no
external forces in x direction. so we can still use conservation of linear momentum.
 
  • #15
Yes I tried that 300(0.01) = 2.21(v'). But what do I do with that velocity.
I tried Ek(pendulum+bullet) = Ek + Ep yet got the wrong answer.
 
  • #16
Is there something wrong with this ?
 
  • #17
zaddyzad said:
Thats why I tried solving by... Ek(bullet) = Ep(pendulum) + EK(bullet)
Why doesn't that work ?

Because energy is not conserved over the collision. Before and after the collision it is, for both objects separately.

Momentum, on the other hand, is always conserved. What do you need to know about the pendulum in order to calculate its momentum immediately after the collision? What information do you have on hand that could tell you?
 
  • #18
zaddyzad said:
Yes I tried that 300(0.01) = 2.21(v'). But what do I do with that velocity.
I tried Ek(pendulum+bullet) = Ek + Ep yet got the wrong answer.

The bullet does not remain in the pendulum, and energy is not conserved over an inelastic collsion. Even so, there is other information available that will allow you to calculate the KE of the pendulum bob immediately after the collision. If you have its KE, what else can you find?
 
  • #19
300(0.01) = 2.21(v'), is that correct for the velocity of the moment the bullet strikes the pendulum? And if it is i can find its KE, which can also lead to finding its height ?
 
  • #20
But I already found its height using trig.
 
  • #21
before the collision, you just have the momentum of the bullet. after the collision, you have the momentum of the bullet plus the momentum of the pendulum (immediately after). so equate the two... you can get the velocity of the pendulum immediately after using the energy conservation for the pendulum ONLY. use that value in the above momentum equation to get the velocity of the bullet after it passes through
 
  • #22
zaddyzad said:
300(0.01) = 2.21(v'), is that correct for the velocity of the moment the bullet strikes the pendulum? And if it is i can find its KE, which can also lead to finding its height ?

No, in this case you should never consider the bullet and pendulum to be one object. It just passes through.

You're holding the wrong end of the stick here :smile: Start with the height that the pendulum reaches and work back to find its initial velocity.
 
  • #23
So, I did Ek(bullet) = Ek(pendulum)<-- found velocity, I used this for the momentum of the pendulum.
then did P(bullet) = P(pendulum) + P(bullet). I didn't get the answer
 
  • #24
AHHH, don't know who to listen to.
 
  • #25
what answers you have got for the problem so far ?
 
  • #26
For gneill. Ek(pendulum) = Ep(pendulum). Correct, and that will give me its intial velocity that turned into its height?
 
  • #27
zaddyzad said:
So, I did Ek(bullet) = Ek(pendulum)<-- found velocity, I used this for the momentum of the pendulum.
That is incorrect. The KE of the bullet is never equal to that of the pendulum. KE is not conserved over the collision. Forget mixing pendulum and bullet KE's.

Use the change in height of the pendulum to find its initial velocity.
 
  • #28
Ek(pendulum) = Ep(pendulum), isn't thzat what i said ?
 
  • #29
zaddyzad said:
For gneill. Ek(pendulum) = Ep(pendulum). Correct, and that will give me its intial velocity that turned into its height?

Yes.
 
  • #30
because with that I am not mixing bullet with the pendulum, and there is a form of energy that did give the pendulum height, and that's its initial velocity.
 
  • #31
zaddyzad said:
because with that I am not mixing bullet with the pendulum, and there is a form of energy that did give the pendulum height, and that's its initial velocity.

Yes.
 
  • #32
gneill said:
Because energy is not conserved over the collision. Before and after the collision it is, for both objects separately.?

Can you explain this please ?
 
  • #33
I got the answer though :D
 
  • #34
And what are some other possibilities of finding the initial velocity of the pendulum. why didnt 300(0.01) = (2.21)V' work? Because for a brief instant that the bullet does hit the block or the slight second its leaving the bullet is in the block, and momentum should be transferred no ?
 
  • #35
zaddyzad said:
Because energy is not conserved over the collision. Before and after the collision it is, for both objects separately.

Can you explain this please ?

Before the collision the bullet has a constant KE assuming that there's no air friction.
After the collision the bullet has a constant KE assuming that there is no air friction.

Before collision the pendulum bob has zero velocity and a constant height, so its KE and PE are constant.
After collision the pendulum is moving in a conservative field (gravitation), so total energy is conserved.
 
  • #36
zaddyzad said:
And what are some other possibilities of finding the initial velocity of the pendulum. why didnt 300(0.01) = (2.21)V' work? Because for a brief instant that the bullet does hit the block or the slight second its leaving the bullet is in the block, and momentum should be transferred no ?

The bullet and block are never one object moving with the same velocity. Simply occupying the same space does not count :wink:
 
  • #37
I don't fully understand why EK(bullet) = EK(bullet) + EP(bullet) doesn't work. The kinetic energy before and after the collision is constant, and so it the EP that the pendulum has. Why isn't the energy of the system being conserved.
 
  • #38
whats the answer you got finally ?
 
  • #39
zaddyzad said:
I don't fully understand why EK(bullet) = EK(bullet) + EP(bullet) doesn't work. The kinetic energy before and after the collision is constant, and so it the EP that the pendulum has. Why isn't the energy of the system being conserved.

Energy is only conserved for perfectly elastic collisions. Otherwise, energy is always lost in the collision to various "loss" pathways such as frictional heating, sound, plastic deformation of materials, breaking of atomic bonds(tearing, breaking), and so on.
 
Last edited:
  • #40
gneill said:
Energy is only conserved for perfectly inelastic collisions. .

Though kinetic energy is only conserved in perfect elastic collision?
 
  • #41
zaddyzad said:
Though kinetic energy is only conserved in perfect elastic collision?

D'oh! My bad. Of course it's ELASTIC collisions only. I fixed the text in the original.
 

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