Collisions between 3 particles (perfectly elastic)

[Taniaz]

Homework Statement

Three perfectly elastic particles A, B, C with masses 4 kg, 2 kg, 3 kg respectively, lie at rest in a straight line on a smooth horizontal table. Particle A is projected towards B with speed 15 m/s and after A has collided with B, B collides with C. Find the velocities of the particles after the second collision and state whether there will be a third collision.

Homework Equations

v1-v2= -e (u1-u2)
Momentum before collision and momentum after collision equations using the principle of conservation of momentum.[/B]

The Attempt at a Solution

Well I formed the equations using the equations stated above. Should I include the momentum of all 3 before collision together and do the same for after or should I do it separately between A and B first and then B and C? If I do it for all 3 together then how will this equation v1-v2= -e (u1-u2) change? e=1 for a perfectly elastic collision.[/B]

 

[haruspex]

should I do it separately between A and B first and then B and C?

Why do anything more complicated?

 

[Taniaz]

This is what I've done so far.
Let u1= velocity of A before collision, v1=velocity of A after collision
u2= velocity of B before collision, v2=velocity of B after collision
u3= velocity of C before collision, v3=velocity of C after collision

u1= 15 and u2 = 0
v1=? and v2=?

Momentum before collision = 4(15) + 0(2) = 60
Momentum after collision = 4v1 + 2v2 = 60 since momentum before collision = momentum after collision

AND v1-v2 = -e(u1-u2) where e=1 for a perfectly elastic collision.
v1-v2 = -1(15-0) so v1-v2 = -15 so v1= -15+v2

Plugging it into 4v1+2v2=60 gives us 4(-15+v2) + 2v2 = 60 so v2 = 60 m/s and therefore v1 = 45 m/s

Now I know the velocity with which B hits C so u2 = 45 and u3=0
And v2-v3= -1(u2-u3) and the same procedure...

 

[haruspex]

v2 = 60 m/s and therefore v1 = 45 m/s

Anything strike you as surprising about those answers?

 

[Taniaz]

They're huge compared to the initial velocities

It's because of the coefficient of restitution

 

[Taniaz]

Btw, I'm still not sure of part g of the gravitational potential question

 

[haruspex]

They're huge compared to the initial velocities

It's because of the coefficient of restitution

No matter how good the restitution, the KE cannot increase,

 

[Taniaz]

So I'm getting all the velocities as positive which means they're all traveling in the same direction after collision so a third collision is unlikely?

 

[haruspex]

So I'm getting all the velocities as positive which means they're all traveling in the same direction after collision so a third collision is unlikely?

Too early to say. The velocities you quote are clearly wrong. Can you spot your mistake?

 

[Taniaz]

Is it the equations?

 

[Taniaz]

Ok I see it 6 v2=120 so v2=20

 

[Taniaz]

So now I get v1 = 5m/s, v2= 20 m/s and v3=16m/s

 

[Vibhor]

So now I get v1 = 5m/s, v2= 20 m/s and v3=16m/s

V2 = 20m/s is velocity of B after first collision ? What about velocity of B after second collision ? Will there be a third collision ?

 

[Taniaz]

Velocity of B after second collision is -4 m/s

 

[Taniaz]

So there is a possibility of a collision between A and B

 

[Vibhor]

Velocity of B after second collision is -4 m/s

Right

So there is a possibility of a collision between A and B

Yes . But why use the word "possibility" ? They surely will collide .