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A bullet through a pendulum.

  • Thread starter zaddyzad
  • Start date
149
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1. 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.

2. Homework Equations
Ek= 1/2MV^2
Ep= MGH
Momentum = MV

3. The Attempt at a Solution

I assumed all energy before and after the question will be = because the collision is inelastic, both masses dont 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 ?
 
875
17
why are you using 300^2 ? use conservation of linear momentum to get the relationship between the initial and final velocities of the bullet.
 
149
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I tried solving via energy.
 
149
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Though when I try solving momentum i get. 300(0.01)=2.2(v)+0.01(v) 2 variables.
 
875
17
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.....
 
149
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Though the bullet leaves the pendulum, do we still consider this inelastic ?
 
875
17
oh sorry. didn't read the question properly. let me think
 

gneill

Mentor
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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...
 
875
17
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
 
149
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this doesn't work.
 
149
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And if the collision is still inelastic you cant "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.
 

gneill

Mentor
20,486
2,612
And if the collision is still inelastic you cant "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...
 
149
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Thats why I tried solving by.... Ek(bullet) = Ep(pendulum) + EK(bullet)
Why doesnt that work ?
 
875
17
well, lets say that velocity of the pendulum immediately after the collision is [itex]v_2[/itex]. 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 [itex]v_2[/itex]. 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.
 
149
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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.
 
149
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Is there something wrong with this ?
 

gneill

Mentor
20,486
2,612
Thats why I tried solving by.... Ek(bullet) = Ep(pendulum) + EK(bullet)
Why doesnt 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?
 

gneill

Mentor
20,486
2,612
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?
 
149
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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 ?
 
149
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But I already found its height using trig.
 
875
17
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
 

gneill

Mentor
20,486
2,612
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.
 
149
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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
 
149
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AHHH, dont know who to listen to.
 
875
17
what answers you have got for the problem so far ?
 

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