1. The problem statement, all variables and given/known data 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). 2. Relevant 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) 3. 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?