# Impulse imparted to A by B during collision

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1. Jun 11, 2017

### tooter555

1. The problem statement, all variables and given/known data
Small object A (mass: [m][/A]) is travelling in a linear path on a horizontal surface and collides elastically with small object B ([m][/B]), at rest on the same line, with a speed of v. Friction between the objects and the horizontal surface is negligible.

What is the impulse imparted to A by B during the collision? The positive direction of impulse is defined as the direction of A's velocity before collision.

2. Relevant equations
Impulse = mdv
v - 0 = [v][/B] - [v][/A]
and maybe [m][/A]v = [m][/B][v][/B] + [m][/A][v][/A]?
3. The attempt at a solution
Isn't the impulse simply -[m][/A]v? Since collision was elastic, change in velocity be v - 0 = v. The direction would be towards the negative because [m][/A] would be moving in the opposite direction after collision.

2. Jun 11, 2017

### tooter555

3. Jun 11, 2017

### jbriggs444

What does "elastic" mean?
Which of the four relevant velocities is given by "v"?

4. Jun 11, 2017

### haruspex

You seem to be saying that A is stationary after the collision. That will only be true (for an elastic collision) if the masses are equal.

5. Jun 12, 2017

### tooter555

v is the initial speed of mass A. I think v - 0 = [v][/B] - [v][/A] is correct since the the collision is elastic. And subbing v = [v][/B] - [v][/A] into [m][/A]v = [m][/B][v][/B] + [m][/A][v][/A], I was able to get the answer. Thanks for the advices.

6. Jun 12, 2017

### scottdave

If I remember correctly, a perfectly elastic collision means that the total Kinetic Energy after the collision equals the total Kinetic Energy before the collision. You can use this information to create an equation, to help you solve for the unknowns.

7. Jun 12, 2017

### haruspex

Yes. I know this as (a special case of) Newton's Experimental Law. It can be deduced from conservation of momentum+conservation of energy.
The general form involves coefficient of restiturion, for imperfect elasticity.
Yes, but using momentum conservation and NEL is equivalent, and avoids quadratics.

8. Jun 12, 2017

### scottdave

Cool. Like I said, it'd been awhile. So I've forgotten about NEL, but I don't think I'll ever forget Conservation of Momentum, and Energy. Yes I do now recall about using the coefficient of restitution, when collision is inelastic.