Tricky collision problem but looks simple

[johns123]

Homework Statement

Mass1 has velocity1 = 10 m/s heading toward Mass2 of velocity2 = 5 m/s in the same direction. Mass1 collides with Mass2 in an elastic collision. What are their final velocities?

Homework Equations

Require solving the conservation of mass AND the conservation of Kinetic Energy together.

The Attempt at a Solution

I got the answer in the back of the book, but it sure ought to be simpler to do. That's my question. Is there a simpler way?

 

[haruspex]

You needed conservation of momentum too, right? Those are the laws to use. Whether there's a simpler way to use them to arrive at the answer I can't tell without seeing your work.

 

[gneill]

There's a nifty way to avoid using the conservation of KE equation in elastic collision problems, one that turns out to be equivalent to it but is simpler (mathematically) to manipulate when solving the two equations in two unknowns.

Suppose that the initial velocities are v1 and v2, and that the final velocities are u1 and u2. Then the closing speed of the two objects after collision is equal to the negative of their closing speed before collision. That is: (v1 - v2) = -(u1 - u2). See? No squares or square roots to spoil your day.

 

[johns123]

There's a nifty way to avoid using the conservation of KE equation in elastic collision problems, one that turns out to be equivalent to it but is simpler (mathematically) to manipulate when solving the two equations in two unknowns.

Suppose that the initial velocities are v1 and v2, and that the final velocities are u1 and u2. Then the closing speed of the two objects after collision is equal to the negative of their closing speed before collision. That is: (v1 - v2) = -(u1 - u2). See? No squares or square roots to spoil your day.

Yep. That was demonstrated in the chapter .. after a page and a half of subbing the momentum equation into the KE equation. But there, V2 initial = 0 to simplify the math. And this one has M2 moving :-( My "guess" is I could do a "relative" collision by setting V2 = 0 and solving for V1final and V2final, and tacking on the missing 5 m/s. Is that reasonable?

I used the magic equations ( from subbing momentum into KE ) V1final = ( m1 -m2)V1init/( m1 + m2 )
and
V2final = 2m1 x V1initial / ( m1 + m2 )

This thing is a stinker and makes no sense, but I got the right answer by tacking on the missing 5 m/s

 

[gneill]

Yep. That was demonstrated in the chapter .. after a page and a half of subbing the momentum equation into the KE equation. But there, V2 initial = 0 to simplify the math. And this one has M2 moving :-( My "guess" is I could do a "relative" collision by setting V2 = 0 and solving for V1final and V2final, and tacking on the missing 5 m/s. Is that reasonable?

You don't need to play any tricks with the velocities to force one to zero and "compensating" later; the method works as-is, regardless of the initial values of v1 and v2.

 

[haruspex]

There's a nifty way to avoid using the conservation of KE equation in elastic collision problems, one that turns out to be equivalent to it but is simpler (mathematically) to manipulate when solving the two equations in two unknowns.

Suppose that the initial velocities are v1 and v2, and that the final velocities are u1 and u2. Then the closing speed of the two objects after collision is equal to the negative of their closing speed before collision. That is: (v1 - v2) = -(u1 - u2). See? No squares or square roots to spoil your day.

True, but it is a consequence of the momentum and energy conservation laws. So you have to remember not to apply this unless you know energy is conserved. It is the perfectly elastic case of what is sometimes known as Newton's Experimental Law.

 

[johns123]

True, but it is a consequence of the momentum and energy conservation laws. So you have to remember not to apply this unless you know energy is conserved. It is the perfectly elastic case of what is sometimes known as Newton's Experimental Law.

How to you get both speeds after the collision? That crazy derivation gives both V1final and V2final .. assuming V2initial is = 0, and in terms of V1initial only. I made the assumption that I could simply ( minded ?? ) take 5 m/s off V2initial and V1initial giving V2initial = 0 and then solve for V1final and V2final .. and then tack 5 m/s back on ? Have no idea that I'm right.

Maybe I could try going through the derivation without V2initial = 0 ? I can just imagine what that's going to be like. In this problem, both energy and momentum are conserved.

 

[haruspex]

take 5 m/s off V2initial and V1initial giving V2initial = 0 and then solve for V1final and V2final .. and then tack 5 m/s back on ?

Yes, that's fine. Taking mass 2's initial trajectory as the reference frame, its initial velocity is zero. Use conservation of momentum and gneill's trick to figure out the two final velocities, then add V2initial back on.

 

[johns123]

Yes, that's fine. Taking mass 2's initial trajectory as the reference frame, its initial velocity is zero. Use conservation of momentum and gneill's trick to figure out the two final velocities, then add V2initial back on.

Thanks guys. If I'm learning something from this, it will surprise me :-)