Cart/putty collision question (conservation of momentum)

[mnphys]

Homework Statement

A 1.0-kg standard cart collides on a low-friction track with cart A . The standard cart has an initial x component of velocity of +0.40 m/s , and cart A is initially at rest. After the collision the x component of velocity of the standard cart is +0.20 m/s and the x component of velocity of cart A is +0.70 m/s . After the collision, cart A continues to the end of the track and rebounds with its speed unchanged. Before the carts collide again, you drop a lump of putty onto cart A , where it sticks. After the second collision, the x component of velocity of the standard cart is -0.20 m/s and the x component of velocity of cart A is +0.55 m/s .

What is the inertia of the putty? (Express your answer with the appropriate units).

Homework Equations

pi = pf (conservation of momentum) [did not use this]
p = mv (equation for momentum) [did not use this]
m1/m2 = -Δv2/Δv1 mass to velocity ratio for momentum)

The Attempt at a Solution

I have tried this question several different ways, and gone over each method at least twice. Here's the one I found the most promising, and yet I still cannot get the correct answer. Note that m1 = mass of cart 1, and m2 = mass of cart 2.

- For the first collision, the mass of the first cart is 1kg and the mass of the second cart is unknown. The change in velocity for the first cart is -.2 m/s (.2 - .4) and the change in velocity for the second cart is .7 m/s (.7 - 0). Substituting known values using m1/m2 = -Δv2/Δv1: 1/m2 = -.7/-.2. This resolves to 1/m2 = 3.5, or m2 = 1/3.5 = approximately .285714... kg. Makes enough sense; it should be a lot lighter than the first cart.
- For the second collision, use the same equation. m1 = 1. change in velocity for the first cart is (-.2 - .2 = -.4). change in velocity for the second cart is (.55 - -.7 = 1.25) (note that I got -.7 for the initial speed because the second cart rebounds from .7 m/s and does not change speed, hence -.7 m/s. I did not take acceleration from the putty drop into account; I'm not sure how one would do that because we haven't learned anything involving application of forces or the like yet). This leads to 1/m2 = -1.25 / -.4 =
3.125. If 1/m2 = 3.125 then m2 = 1/3.125 = .32 kg.
- If the cart 2 was .2857... kg before the putty and .32 kg after the putty, then the putty weighs .32 - .2857 = approximately .034 kg... but the program tells me this is nowhere near the correct result.

Where did I go wrong?

 

[haruspex]

You do need to calculate how adding the putty affected the speed of cart A. Assume the putty was dropped vertically, and use conservation of momentum.

 

[mnphys]

Thanks! This was one of those questions you spend hours on and then when you finally realize what you're looking for it's absurdly simple. I assumed that the velocity of the second cart did not change after dropping the putty on it - wrong, of course. But without knowing how much the putty weighs, how do I know what the new velocity is? I tried backward substitution, all sorts of crazy algebra, and nothing worked...

And then I realized: the momentum before the putty drop is (mass of second cart * -.7 m/s). Adding the putty does not change that figure, only the mass and velocity themselves! With that in mind it's very easy to find the momentum of the system after the rebound, and with the momentum of the system in mind + the final velocity of the cart available, there is only one variable left: the mass of the second cart + the putty.

So I ended up solving the whole thing without ever knowing what the velocity of the second cart was after the putty drop. I feel kinda silly but... hey. As long as I got there eventually, right? ;)

 

[haruspex]

Thanks! This was one of those questions you spend hours on and then when you finally realize what you're looking for it's absurdly simple. I assumed that the velocity of the second cart did not change after dropping the putty on it - wrong, of course. But without knowing how much the putty weighs, how do I know what the new velocity is? I tried backward substitution, all sorts of crazy algebra, and nothing worked...

And then I realized: the momentum before the putty drop is (mass of second cart * -.7 m/s). Adding the putty does not change that figure, only the mass and velocity themselves! With that in mind it's very easy to find the momentum of the system after the rebound, and with the momentum of the system in mind + the final velocity of the cart available, there is only one variable left: the mass of the second cart + the putty.

So I ended up solving the whole thing without ever knowing what the velocity of the second cart was after the putty drop. I feel kinda silly but... hey. As long as I got there eventually, right? ;)

Good, that is probably the easiest way. You could have put the putty mass in as an unknown, found the velocity in terms of it, and later solved simultaneous equations, but as you say, whatever works.

 

[SDewan]

You do need to calculate how adding the putty affected the speed of cart A. Assume the putty was dropped vertically, and use conservation of momentum.

Just one thing to ask, when I put the putty onto the cart, will the cart not get an impulse? I mean, in the question nowhere it is mentioned that the lump is "gently" kept on the cart. Since it is "dropped" onto the cart, the cart is bound to get an impulse. Please give me a reasoning for that.

 

[haruspex]

Just one thing to ask, when I put the putty onto the cart, will the cart not get an impulse? I mean, in the question nowhere it is mentioned that the lump is "gently" kept on the cart. Since it is "dropped" onto the cart, the cart is bound to get an impulse. Please give me a reasoning for that.

Vertical impulse doesn't matter since that will be balanced by an equal and opposite normal impulse from the ground. The horizontal impulse results from the putty arriving vertically onto a moving cart. Since it "immediately" acquires a horizontal velocity, there is a limit to how gentle that can be.

 

[SDewan]

Got it.