Two certain small objects A and B, each have mass m. A is suspended using a lightweight, inelastic string, and B is placed on a rough horizontal floor in a position where it is in contact with the vertically suspended A. Next, A is raised to height h above its original position while the string is kept taut, and then is gently released. A collides elastically with B, which then slides along the floor distance d before coming to a rest
(mA)(vA1)+(mB)(vB1) = (mA)(vA2)+(mA)(vA2)
K1 + U1 + W other = K2 + U2
The Attempt at a Solution
To my understanding, in collision problems the momentum is always conserved. However, energy is not necessarily conserved.
So I used the momentum equation and and inserted the variables. Since object B is at rest before the collision, and object A is at rest after the collision; I got rid of (mB)(vB1) and (mA)(vA2).
(mA)(vA1) = (mB)(vB2) since both masses are equal I got that both velocities are equal.
(vA1) = (vB2)
Then I use the energy equation where
K1 = 0
U1 = mgh
W other = (Ff)(d) = (μk)(n) = (μk)(-mg)
K2 = 1/2mv^2
U2 = 0
mgh - (μk)(mg)(d) = 1/2mv^2
gh - (μk)(g)(d) = 1/2v^2
h - (μk)(d) = (v^2)/(2g)
(μk) = (2gh - v^2)/(2g)(d)
I don' really know where I went wrong. But I think my energy equation is wrong.
The answer is supposed to be h/d.