Impulse imparted to A by B during collision

[tooter555]

Homework Statement

Small object A (mass: [m][/A]) is traveling 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.

Correct answer is 1.

Homework Equations

Impulse = mdv
v - 0 = [v][/B] - [v][/A]
and maybe [m][/A]v = [m][/B][v][/B] + [m][/A][v][/A]?

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.

 

[tooter555]

Homework Statement

Small object A (mass: [m][/A]) is traveling 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.

http://imgur.com/BICAM14

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.

http://imgur.com/H8LOF8x

Correct answer is 1.

Homework Equations

Impulse = mdv
v - 0 = [v][/B] - [v][/A]
and maybe [m][/A]v = [m][/B][v][/B] + [m][/A][v][/A]?

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.

 

[jbriggs444]

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.

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

 

[haruspex]

Since collision was elastic, change in velocity be v - 0 = v.

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.

 

[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.

 

[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.

 

[haruspex]

v - 0 = [v][/B] - [v][/A]

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.

total Kinetic Energy after the collision equals the total Kinetic Energy before the collision.

Yes, but using momentum conservation and NEL is equivalent, and avoids quadratics.

 

[scottdave]

Yes. I know this as (a special case of) Newton's Experimental Law. It can be deduced from conservation of momentum+conservation of energy...using momentum conservation and NEL is equivalent, and avoids quadratics.

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.