Find the velocity of each object after the collision.

[AG1189]

Question: A 10.0 g object moving to the right at 19.0 cm/s makes an elastic head-on collision with a 15.0 g object moving in the opposite direction at 36.0 cm/s. Find the velocity of each object after the collision.

-I am little lost with this problem, I tried using the conservation of momentum equation (M1V1i+M2V2i=M1V1f+M2V2f), but since there are two unknowns I don't know what other equation to use. The answer for both problems need to be in cm/s. Can anyone help me with this?

 

[cristo]

Since the collision is perfectly elastic, kinetic energy will be conserved also. Try using this.

 

[chromium blade]

The collision is not perfectly elastic, it is "elastic". So energy is not conserved. And besides, perfect elasticity is an unrealistic assumption, especially where two "objects" are concerned. You need more information.

 

[PhanthomJay]

The collision is not perfectly elastic, it is "elastic". So energy is not conserved. And besides, perfect elasticity is an unrealistic assumption, especially where two "objects" are concerned. You need more information.

Nothing in life is perfect, and nothing in physics is perfect either. But I would venture to guess that the problem implies that this is a perfectly elastic collision, and therefore energy, as well as momentum, is conserved. I didn't write the problem though, so my guess is as good as yours, isn't it?

 

[Kurdt]

To be honest if its in an introductory physics forum its pretty safe to bet on the fact that the question implies perfectly elastic collisons.

 

[cristo]

The collision is not perfectly elastic, it is "elastic". So energy is not conserved. And besides, perfect elasticity is an unrealistic assumption, especially where two "objects" are concerned. You need more information.

If you take this stance, then more or less every introductory physics question can be answered with "I need more information" or "this is an unrealistic assumption."

In this sort of question, when the collision is said to be elastic, it usually means perfectly elastic.

 

[PhanthomJay]

If you take this stance, then more or less every introductory physics question can be answered with "I need more information" or "this is an unrealistic assumption."

In this sort of question, when the collision is said to be elastic, it usually means perfectly elastic.

Yes, and if not perfectly elastic, the problem is usually worded as 'inelastic' or, if the masses stick together, 'totally inelastic'. The word 'elastic' or 'near elastic' implies 'perfectly elastic' when doing the calculations, unless specifically stated otherwise.

 

[chromium blade]

okay if you're making the assumption that it's a perfectly elastic collision, note that this implies that the coefficient of restitution between the particles would equal 1. Hence the particles will have the same final speed as their initial, with their direction of motion reversed. However, If you solve a rather awkward simultaneous equation using energy instead, you'll find this is true. So you can say what the answer is without doing any work !

 

[cristo]

okay if you're making the assumption that it's a perfectly elastic collision, note that this implies that the coefficient of restitution between the particles would equal 1. Hence the particles will have the same final speed as their initial, with their direction of motion reversed. However, If you solve a rather awkward simultaneous equation using energy instead, you'll find this is true. So you can say what the answer is without doing any work !

I tell you what, Chromium; instead of trying to criticise, and giving out incorrect advice, how about you solve the "awkward simultaneous equations" and then come back to me with the results.

To AG1189; if you're still reading this thread, set up two equations, one for conservation of momentum, and one for conservation of energy, then solve for v₁ and v₂.

 

[chromium blade]

Cristo I'm not criticising anybody's work here; I made no reference to your method being incorrect in any way. If you don't understand collisions then consult someone who does, please don't say my work is "incorrect" when it isn't, you're just making it difficult for students who need help. And besides, isn't it in violation of forum terms and conditions to give out solutions, as you kindly let me know. This shall be my last post on the topic as it seems the "homework helpers" need help with their own understanding of collisions (and I have my doubts about other areas too).

 

[cristo]

Cristo I'm not criticising anybody's work here; I made no reference to your method being incorrect in any way. If you don't understand collisions then consult someone who does, please don't say my work is "incorrect" when it isn't, you're just making it difficult for students who need help.

Your work is incorrect. What makes you think that after an elastic collision, the velocities are the same but in the opposite direction? Furthermore, how does the fact that the coefficient of restitution is equal to one imply this? The only thing I can think of is that you've interpreted e=\frac{v_{2f}-v_{1f}}{v_1-v_2}=1 as implying that v_{2f}=-v_2 and v_{1f}=-v_1.

And besides, isn't it in violation of forum terms and conditions to give out solutions, as you kindly let me know.

Kindly point out where I broke the forum rules. I advised the OP how to go about answering the question, and may have told him more than usual, since he was bound to be confused by contradicting advice!

This shall be my last post on the topic as it seems the "homework helpers" need help with their own understanding of collisions (and I have my doubts about other areas too).

This can be discussed via PM.

 

[weatherhead]

chromium_blade: ... the masses are not equal... that is all.

 

[cristo]

Thanks for pointing that out, weatherhead.. chromium blade must have thought that the equations for equal masses also applies to the situtation in the question. However, this is not true!

AG1189; again, to solve the problem, you will need to consider conservation of energy and conservation of momentum.