#### nomadreid

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**1. The problem statement, all variables and given/known data**

Three bodies A,B,C on frictionless surface

masses= 1 kg each,

Positions at time 0: A is at x=0,. B is at x=1, C is at x=2 (unit is 1 meter).

velocities at time 0: A : 1m/s (to the right), B = 0 m/s, C = -1 m/s

Assume sizes are uniform, or just ignore the sizes in calculations

Collision A-B is elastic. Collision C-B is inelastic, and C sticks to B.

Question: what direction is the motion of the BC combination after collision, and why?

The relevant choices:

(a) to the right (i.e., positive velocity), due to the conservation of momentum

(d) to the left (i.e., negative velocity), due to difference of time force is applied by the respective bodies

**2. Relevant equations**

conservation of momentum, conservation of force , and (with

**v, F, p**being vectors):

**p**=m

**v**,

**F**=Δ

**p**/Δt (possibly KE=mv

^{2/2}, E= KE + friction, but I doubt it)

**3. The attempt at a solution**

My choice was (a) , due to the following: total momentum before and after, zero, so if there are two parts, they must be opposite signs. Since you cannot have A passing through BC, A then goes to the left (has a negative velocity), and BC has a positive. Hence BC goes to the right.

However, the answer is given as (d).

Analyzing it in terms of forces and time

The forces must also sum up to zero, that is, if after the collision the velocities are

**v**

_{A}and

**v**

_{BC}, then [m

**v-**m

**v**_{A}]/Δt

_{A}= [-(-M

**v**-2M

**v**

_{BC})]/Δt

_{BC}, which has two variables, so assuming WLOG that the loss of energy due to friction is negligible, and that the velocities after the collision are both negative (as D would imply), I do not end up with a total zero force.

Is my answer correct, or is the answer given correct?

Thank you.