# Successive Collisions

• Feb 9th 2008, 02:44 PM
free_to_fly
[SOLVED] Successive Collisions
I'm stuck on yet another mechanics question

Three perfectly elastic particles A, B and C with masses 3kg, 2kg and 1kg respectively lie at rest in a straight line on a smooth horizontal table with B between A and C. Particle A is projected directly towards B with speed 5m/s and after A has collided with B, B collides with C. Find the speed of each particle after the second impact.

I considered the coefficient of resitution between A and B, and between B and C, as well as conservation of momentum before and after the two impacts, and came up with 5 unknown and only 4 equations:
Let v be the speed of A after 1st impact, w be the speed of B after 1st imacpt, x be the speed of B after the 2nd impact and y be the speed of C after the 2nd impact:
15=3v+2w (1)
coefficient of restitution between A and B e=(w-v)/5 (2)
2w-2x+y (3)
coefficient of resitution between B and c e=(y-x)/w (4)

I need a bit more information to continue or have I missed something in the question? Any help would be greatly appreciated.
• Feb 9th 2008, 04:23 PM
TwistedOne151
Perfectly elastic
By perfectly elastic particles, it means there is no loss of kinetic energy in the collision (see here, for example), so coefficients of restitution are not needed (they are all one). You just have that the total momentum is the same before and after each collision, and that the total kinetic energy is the same before and after each collision.

--Kevin C.
• Feb 11th 2008, 01:46 AM
free_to_fly
So I've come up with 4 equations to solve:
v is the speed of A after first collision,
w is the speed of B after first collision,
x is the speed of B after the second collision,
y is the spped of C after the second collision

Considering the conservation of momentum:
15=3v+2w (1)
2w=2x+y (2)
Then considering the conservation of KE:
75=3v^2+2w^2 (3)
2w^2=2x^2+y^4 (4)
When I solved these I found:
v=3m/s
w=3m/s
x=1m/s
y=4m/s
but according to my book the speeds v=1m/s, x=4m/s and y=8m/s. Where did I go wrong? Help please?
• Feb 11th 2008, 12:03 PM
TwistedOne151
Solving
Your equations look correct, but you must have made an error in solving them, as your solutions don't fit when substituted back in. For example, you have v=w=3 m/s. But if you plug these into your equation (3), you see $3v^2+2w^2$ is 45 for these values, not the needed 75.

To solve for v and w, use substitution:
$3v+2w=15$
$w=\frac{15-3v}{2}$
$3v^2+2w^2=75$
$3v^2+2(\frac{15-3v}{2})^2=75$
$3v^2+\frac{(15-3v)^2}{2}=75$
$6v^2+(15-3v)^2=150$
$6v^2+9v^2-90v+225=150$
$15v^2-90v+75=0$
$v^2-6v+5=0$
Solving this quadratic will give you v. (Note that one of the two solutions will be the initial value for the velocity, which is 5 m/s. You will want the other solution). From v, you can obtain w, and then with those values, you can use (2) and (4) to solve for x and y in the same manner

--Kevin C.
• Feb 11th 2008, 02:09 PM
free_to_fly
I've redone my calculations and obtained the following results:
v=1m/s
w=6m/s
x=2m/s
y=4m/s

Unfortunately this is still incorrect, because according to my text book, v=1m/s, x=4m/s and y=8m/s. Do you also get these solutions?
• Feb 11th 2008, 04:39 PM
TwistedOne151
I don't see how the book answer can work: it gives a final momentum of
3*1+2*4+1*8=19 kg*m/s, but the initial momentum is only 15 kg*m/s

The v and the w are correct.
This gives us $2x^2+y^2=2*6^2=72$ and $2x+y=2*6=12$

Solving the second for y, $y=12-2x$, so:
$2x^2+(12-2x)^2=72$
$2x^2+4x^2-48x+144=72$
$6x^2-48x+72=0$
$2x^2+4x^2-48x+144=72$
$x^2-8x+12=0$
$(x-6)(x-2)=0$
The x=6 (and y=0) solution is the pre-impact condition, so the post-impact condition is x=2 m/s, y=8 m/s

We see that this result has the right total momentum (15 kg*m/s) and energy.

--Kevin C.