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 +0.45 m/s . What is the inertia of the putty?

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• Let's let the standard cart be S, little i be "initial" and little f be "final." Conserve momentum for the initial collision:

mS*vSi + mA*vAi = mS*vSf + mA*vAf

1.0kg*0.40m/s + mAi*0m/s = 1.0kg*0.20m/s + mA*0.70m/s

solves to

mA = 0.2857 kg

Now conserve momentum for the collision of the putty (P) with cart A. Little c means "center" (collision), and the final velocities from the previous collision are the initial velocities here:

mA*vAi + mP*vPi = (mA + mP)*vAc

0.2857kg*-0.40m/s + mP*0m/s = (0.2857kg + mP)*vAc

0.2857kg + mP = -0.1143kg·m/s / vAc ← equation #1

And finally, conserve momentum for the final collision; P indicates the putty:

mS*vSc + (mA+mP)*vAc = mS*vSf + (mA+mP)*vAf

1.0kg*0.20m/s + (0.2857kg + mP)*vAc = 1.0kg*-0.20m/s + (0.2857kg + mP)*0.45m/s

substitute for (0.2857kg + mP) (equation #1) and do some multiplication:

0.20kg·m/s - (0.1143kg·m/s / vAc) * vAc = -0.20kg·m/s - (0.1143kg·m/s / vAc)*0.45m/s

0.0857 kg·m/s = -0.20kg·m/s - 0.051435kg·m²/s² / vAc)

0.2857 kg·m/s = -0.051435kg·m²/s² / vAc

vAc = -0.051435kg·m²/s² / 0.2857kg·m/s = -0.18 m/s

plug into equation #1

0.2857kg + mP = -0.1143kg·m/s / -0.18m/s = 0.6349 kg

and so

mP = 0.35 kg ◄

Not extremely difficult, just a nuisance keeping track of the three collisions.

Also note that one might have set up equation #1 to be

vAc = -0.1143kg·m/s / (0.2857kg + mP)

but I thought that would lead to slightly nastier math. It would, however, lead so a solution for mP without having to solve for vAc.

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