Determine the velocity after collision

[goldfish9776]

Homework Statement

three small spheres, A , B and C with mass of 3kg , 4kg , 7kg respectively are arranged as shown in the figure. Initially , sphere B is placed in static condition,while the sphere A is moving with a velocity of 4u towards B and collides . Then , sphere C move to right with velocity u , the elastic coefficient between A and B is 0.75 , and B and C is 0.5, determine

the velocity of sphere A and B after the first collision . How to do this ?

3(4u) +4(0) = 3V1 +4V2

12u = 3V1 +4V2

V2 - V1 / ( U1 -U2 ) = 0.75

since U2 = 0 ,

V2 - V1 = 0.75 u

12u - 3( V2- 0.75u ) +4V2

14.25u = 7V2 , V2 = 2.035 u , V1 = 1.29u ,

is my ans correct ? i don't have the ans btw

 

[theodoros.mihos]

I think that elastic coefficient refers to kinetic energy loss ratio.

 

[goldfish9776]

Homework Statement

three small spheres, A , B and C with mass of 3kg , 4kg , 7kg respectively are arranged as shown in the figure. Initially , sphere B is placed in static condition,while the sphere A is moving with a velocity of 4u towards B and collides . Then , sphere C move to right with velocity u , the elastic coefficient between A and B is 0.75 , and B and C is 0.5, determine

the velocity of sphere A and B after the first collision . How to do this ?

3(4u) +4(0) = 3V1 +4V2

12u = 3V1 +4V2

V2 - V1 / ( U1 -U2 ) = 0.75

since U2 = 0 ,

V2 - V1 = 0.75 u

12u - 3( V2- 0.75u ) +4V2

14.25u = 7V2 , V2 = 2.035 u , V1 = 1.29u ,

is my ans correct ? i don't have the ans btw

what will occur after that ? sphere A and B sticked together and collide with C or ?

 

[theodoros.mihos]

Take all together in one step with 3 bodies.

 

[goldfish9776]

Take all together in one step with 3 bodies.

can you show how to do that ?

 

[theodoros.mihos]

Sorry, I'm wrong. IP ask two steps, 1st A collides B and 2nd by B and C are collides.
Take momentum conservation for every step combined with the correct energy relation.

 

[haruspex]

I think that elastic coefficient refers to kinetic energy loss ratio.

I would think it refers to the coefficient of restitution. This relates to velocity ratios. The fraction of KE retained will be the square of this.

 

[theodoros.mihos]

I would think it refers to the coefficient of restitution. This relates to velocity ratios. The fraction of KE retained will be the square of this.

For this reason, if we divide velocities after with the same factor, the momentum conservation is not correct.