Momentum and Collisions question. Help!?

Question

so my textbook says that after the collision of two objects,  the change in momentum of the first object is equal to and opposite the change in momentum of the second object so momentum is conserved. 
Can someone explain to me what happens after two objects both with initial momentum collide? So an action-reaction force pair occurs which determines the motion of each individual object and does the momentum simply flip flop? 
For example, if Object A has momentum of 60 p to the right and Object B has 40 p to the left. After the collision, would Object A have 40 p to the left and Object B would have 60 p to the right?

kuiperbelt2003 · Accepted Answer

momentum is conserved in all collisions, whether the initial momentum is zero or some other value

in your case, the initial momentum is +20 p (where we set to the right as the positive direction)

without knowing relative masses and other information, we cannot determine how much momentum each object will carry away after the collision; given the information you have given us, we can only say that the sum of momenta after the collision is + 20 p

Anonymous · Answer

Depends whether the collision is inelastic (they stick together) or bounce off each other conserving KE (elastic) Think about it inelastically and they stick together. 60+(-20) = 40 to the right. If totally elastic, you'd need their masses to find the speed and direction of the outcomes.

obee · Answer

it is probably no longer any help in any respect. a thank you to define, defining proper as useful (+), given the values: m1 = 3 kg u1 = 5 m/s v1 = m2 = 5 kg u2 = -3 m/s (a) Inelastic - kinetic power isn't conserved v * [m1 + m2] = m1u1 + m2u2 v * [ (3 kg) + (5 kg) ] = [ (3 kg) * (5 m/s) ] + [ (5 kg) * (-3 m/s) ] v * [ 8 kg ] = [ 15 kg-m/s ] + [ -15 kg-m/s ] v * [ 8 kg ] = [ 0.0 kg-m/s ] v = 0.0 kg-m/s (b) Elastic, kinetic power is conserved [0.5*m*u^2] + [0.5*m*u^2] = [0.5*m*v^2] + [0.5*m*v^2] [0.5 * (3 kg) * (5 m/s)^2] + [0.5 * (5 kg) * (-3 m/s)^2 ] = [0.5 * (3 kg) * v^2] + [0.5 * (5 kg) * v^2] [ (a million.5 kg) * (25 m^2/s^2) ] + [ (2.5 kg) * (9 m^2/s^2) ] = [ (a million.5 kg) * v^2 ] + [ (2.5 kg) * v^2 ] [ 37.5 J ] + [ 22.5 J ] = [ (a million.5 kg) * v^2 ] + [ (2.5 kg) * v^2 ] [ 60 J ] = [ (a million.5 kg) * (v1)^2 ] + [ (2.5 kg) * (v2)^2 ] [ (v1)^2 * (a million.5 kg) ] = [ 60 J ] - [ (2.5 kg) * (v2)^2 ] (v1)^2 = [ forty J ] - [ a million.sixty seven kg * (v2)^2 ] v1 = SQRT { [ forty J ] - [ a million.sixty seven kg * (v2)^2 ] }

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Momentum and Collisions question. Help!?

3 Answers