## Momentum, Friction, and Velocity, OH MY!

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

We are trying to predict the speed of all three blocks at the first instant they all have the same speed using only variables.

2. Relevant equations

fg = mg
fk = -μk(mg) = normal force
a = F/m
t = √(2Δx/a) derived from Δx = Vot + 1/2at^2
v = vo + at
p = mv = m(at)
p1 + p2 = pf (convserve momentum)

3. The attempt at a solution

This is my work and final solution of Vf at the bottom, i treated m1 and m2 as one object then used conservation of momentum to find the final velocity. it just does not seem right, i would have thought it would involve the kinetic friction of m3 somehow. i am assuming this is a perfectly inelastic "collision". But when i think about it more, i am thinking they are all 3 at the same speed the instant the impulse of the 2 moving blocks cause the 3rd block to have a momentum value, thus why i used the conservation of momentum to solve for final velocity. Can anyone guide me as to whether or not my thinking is on the right track?
 PhysOrg.com science news on PhysOrg.com >> Leading 3-D printer firms to merge in $403M deal (Update)>> LA to give every student an iPad;$30M order>> CIA faulted for choosing Amazon over IBM on cloud contract
 All the blocks will have the same speed after the string has become taut again, and when the string becomes taut, an impulsive tension acts on the block, and so momentum will not be conserved.
 okay i think i figured it out, or at least the answer seems a little more reasonable given that it involves block 3 now. here it is reworked with the answer at the bottom, would i still be able to use m1v1 + m2v2 = mfvf (final)? i'm sorry, we have not covered this in class yet, i read the book and am working through it on my own.

## Momentum, Friction, and Velocity, OH MY!

i just re-read your reply. you said momentum was not conserved? could you explain more, i thought momentum was always conserved, kinetic energy was not?
 Momentum is only conserved when there is no external force acting, in this case, there is friction as well as an impulse (A large force acting in a small time interval) by the string on the last block.
 if we were to assume the impulse approximation, wouldn't we just throw out the frictional force anyways since the impulse force would be much greater? i think that is how to use impulse approximation

 Tags force, friction, impulse, momentum, velocity

 Similar discussions for: Momentum, Friction, and Velocity, OH MY! Thread Forum Replies Introductory Physics Homework 2 Classical Physics 10 Classical Physics 7 General Physics 0 Introductory Physics Homework 1