Inelastic collision (finding e, don't understand concept)

In summary, the conversation discusses two different approaches to solving a problem involving equations and velocities. The first approach uses NEL and substitution, while the second approach uses a formula to determine the velocity. The student also questions their teacher's method and his reasoning behind it. They also discuss the concept of positive and negative velocities and how it affects the solutions. Ultimately, it is concluded that the teacher's method was incorrect and the correct solution is obtained by measuring all velocities in the same direction.
  • #1
coconut62
161
1
I have problems with a(ii). (please view the first image.)

Now I have found that w = (13-8e)u/7 and v = (20e+13)u/7

What I did is like this:

(Using NEL)

v = 4ue
Where the LHS is actually v-0 because I assume Q = 0.

(20e+13)u/7 = 4ue

e = 33/28 >1 ∴ e cannot exist.

1. Is this way correct?

2. My teacher used another way. (please view second image.)

I don't understand the reason behind his workings.

Actually he like to use the formula like this :
V1- V2 = -e(U1-U2) ,even when V2 is faster than V1. Because he said in his brain, 2 always comes after 1, so he doesn't want to use V2-V1. I don't follow him, I follow the book, using speed of separation = fast - slow. But I still don't understand why he used V-W = -e(4u) to assume Q is brought to rest. (or did he even make that assumption at all?)

My interpretation is like this: Since he uses "slower-faster = negative e (speed of approach)"
Then in his workings, W should be assumed faster than V. And in the situation, both P and Q are moving to the left after collision, which makes their velocities negative. So if you want to assume Q is brought to rest, then W = 0. Since 0 is larger than negative, then his assumption is valid.

This is just simply a hypothesis. I only have a vague picture of the concept but I cannot fully grasp it. Someone please explain it to me.

3. The second part of a(ii), determining the least possible value of w. I just substituted e=1 into v-w = 4ue and got the answer. But after thinking about it, I suddenly don't understand why it should be e=1 which is substituted inside. I tried e=0.9, and the resulting w is smaller than the answer when e=1. So why is the least value obtained when e=1?
 

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  • #2
coconut62 said:
(20e+13)u/7 = 4ue
e = 33/28 >1 ∴ e cannot exist.
You went a bit astray in there. You should have got e = 13/8. But your method is fine.
2. My teacher used another way. (please view second image.)
I'm afraid your teacher has blundered. He has the sign wrong on the RHS. His method should have given e = e.
he like to use the formula like this :
V1- V2 = -e(U1-U2) ,even when V2 is faster than V1.
That's fine. In this case, that gives v-w = -e(-u - 3u), so v - w = 4eu.
3. The second part of a(ii), determining the least possible value of w. I just substituted e=1 into v-w = 4ue and got the answer. But after thinking about it, I suddenly don't understand why it should be e=1 which is substituted inside.
w = (13-8e)u/7, so as e increases w decreases. Therefore the smallest w goes with the largest e, and the largest possible e is 1.
 
  • #3
haruspex said:
I'm afraid your teacher has blundered. He has the sign wrong on the RHS. His method should have given e = e.

Why is it that when e=e, there is no solution? Isn't e equal to e? (err...)

haruspex said:
That's fine. In this case, that gives v-w = -e(-u - 3u), so v - w = 4eu.
I thought the right hand side should be -e (speed of approach) ? Why is it (-u-3u)? The particles are moving in opposite directions, which makes their speeds add up to become the total speed, no?
 
  • #4
coconut62 said:
Why is it that when e=e, there is no solution? Isn't e equal to e? (err...)
Arriving at e = e does not mean there is no solution. What it means is that the method your teacher used was ineffective. This can happen quite often when manipulating equations. If you use the same equation twice you can find everything cancels out and you end up with 0 = 0.
I thought the right hand side should be -e (speed of approach) ? Why is it (-u-3u)? The particles are moving in opposite directions, which makes their speeds add up to become the total speed, no?
You must measure all the velocities in the same direction. The given u and v are in opposite directions, so you have to flip the sign of one of them.
 
  • #5
haruspex said:
You must measure all the velocities in the same direction. The given u and v are in opposite directions, so you have to flip the sign of one of them.

This (-u-4u) thing only applies when using my teacher's version right? If I use W-V instead of V-W for speed of separation, then it would be just (u+4u)?
 

1. What is an inelastic collision?

An inelastic collision is a type of collision where kinetic energy is not conserved. This means that after the collision, the objects involved stick together and move with a common velocity.

2. How is the coefficient of restitution (e) calculated in an inelastic collision?

The coefficient of restitution (e) is calculated by taking the ratio of the relative velocity of separation to the relative velocity of approach. In other words, it is the ratio of the final velocity to the initial velocity of the objects involved in the collision.

3. What does a coefficient of restitution (e) of 0 mean in an inelastic collision?

A coefficient of restitution (e) of 0 means that the objects involved in the collision stick together and move with a common velocity after the collision. This type of collision is known as a perfectly inelastic collision.

4. How is kinetic energy affected in an inelastic collision?

In an inelastic collision, kinetic energy is not conserved and is typically reduced after the collision due to the loss of energy in the form of heat, sound, or deformation of the objects involved.

5. Can two objects with the same mass have different coefficients of restitution (e) in an inelastic collision?

Yes, two objects with the same mass can have different coefficients of restitution (e) in an inelastic collision. The coefficient of restitution depends on the materials and surfaces of the objects involved, as well as the angle and speed at which they collide.

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