Impossible Jmb A level question

In summary, the question is impossible to do. It's from an A level physics paper from 1980. A 5 gram pellet is fired vertically into 95 grams of clay at a velocity of v metres per second. Assume perfectly inelastic collision, and it should be solvable. I expect the height to be a function in v.
  • #1
eddie
26
0
Would you agree with me that the following question is impossible to do ,it's from A level physics paper 1980 multiple choice question no. 4 .A 5 gram pellet is fired vertically into 95 grams of clay at a velocity of v metres per second calculate the height the clay will reach
 
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  • #2
Assume perfectly inelastic collision, and it should be solvable. I expect the height to be a function in v.

The odd thing here is that the question doesn't specify the position of the gun. I assume the clay is somehow held by a sufficiently long rope such that the pellet is fired from underneath.
 
  • #3
Ahmad Kishki said:
Assume perfectly inelastic collision, and it should be solvable. I expect the height to be a function in v.

The odd thing here is that the question doesn't specify the position of the gun. I assume the clay is somehow held by a sufficiently long rope such that the pellet is fired from underneath.
The pellet is fired from underneath the clay no mention was made of the type of collision .I agree with you that it would be solvable if the collision was perfectly inelastic.In this case you cannot assume that momentum is conserved as there is a resultant foce on the system .
 
  • #4
The key word is "clay".
 
  • #5
eddie said:
The pellet is fired from underneath the clay no mention was made of the type of collision .I agree with you that it would be solvable if the collision was perfectly inelastic.In this case you cannot assume that momentum is conserved as there is a resultant foce on the system .

Clay assumes the collision will be perfectly inelastic. Momentum is conserved here, there are no external forces on the ssystem. Notice: it is not resultant force that matters, it is external forces. I think this might be where you are confused. Momentum is not conserved if for example the collisionwas restricted in one way or another such that an external force is present. Here we have NO external forces.

I think this might be helpful: http://www.physicsclassroom.com/class/momentum/Lesson-2/Isolated-Systems
 
  • #6
Vanadium 50 said:
The key word is "clay".
Thanks for taking the trouble to make such a pithy reply.I agree totally with your comment.
 
  • #7
Ahmad Kishki said:
Clay assumes the collision will be perfectly inelastic. Momentum is conserved here, there are no external forces on the ssystem. Notice: it is not resultant force that matters, it is external forces. I think this might be where you are confused. Momentum is not conserved if for example the collisionwas restricted in one way or another such that an external force is present. Here we have NO external forces.

I think this might be helpful: http://www.physicsclassroom.com/class/momentum/Lesson-2/Isolated-Systems
Thanks for your reply Mr.Kishki ,there will be a resultant force on the bullet and clay during the period of momentum transfer before they reach a common velocity.Momentum is only conserved when the resultant force on a system is zero .Note resultant force equals rate of change of momentum.PS. there would be no problem if the bullet was fired horizontally as in the Ballistic Pendulum experiment.
 
  • #8
Why are you doing this using "forces" or even conservation of momentum? Why not solve this directly using conservation of energy? The bullet has an initial kinetic energy, and the bullet plus clay has a final potential energy since at the maximum height, it has no KE. This is so much easier since at this level, you don't care about energy loss due to heat, sound, etc.

Zz.
 
  • #9
eddie said:
Thanks for your reply Mr.Kishki ,there will be a resultant force on the bullet and clay during the period of momentum transfer before they reach a common velocity. Momentum is only conserved when the resultant force on a system is zero.
The net resultant force on the system consisting of the clay plus bullet during the instant of the collision is gravity. If the collision is of negligible duration, then what can be said about the magnitude of the change in momentum resulting from this resultant force?
 
  • #10
ZapperZ said:
Why are you doing this using "forces" or even conservation of momentum? Why not solve this directly using conservation of energy? The bullet has an initial kinetic energy, and the bullet plus clay has a final potential energy since at the maximum height, it has no KE. This is so much easier since at this level, you don't care about energy loss due to heat, sound, etc.

Zz.
Thanks Zz for your reply ,the bullet loses energy in penetrating the clay.
 
  • #11
I hate questions that have different but valid answers depending on how much you know.
 
  • #12
ZapperZ said:
Why not solve this directly using conservation of energy?

Because the collision is inelastic. The clay is deformed. But you're right in that one should be thinking of the fastest way to solve this, which is to think about momentum conservation and not forces, or positions of the gun, or any other thing.
 

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