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I've been working on this problem, and I can't seem to get it. Please, can someone help me out...

Two particles, of masses m and 4m , are moving toward each other along the x-axis with the same initial speeds of 6.53 m/s. Mass m is traveling to the left, and mass 4m is traveling to the right. They undergo a head-on elastic collision, and each rebounds along the same line as it approached.

I need to find the final speed of the heavier particle.
Any help would be very much appreciated, Thanks in advance.

please, i cant get it!

Answer 1

Momentum = P_total
= 4*m*u - m*u
= 3*m*u
Kinetic energy = KE_total
= 4*m*u^2/2 + m*u^2/2
= 5*m*u^2/2
where u = 6.53

After collision,
Momentum = P_total
= 4*m*v1 + m*v2
= m*(4*v1 + v2)
KE = 4*m*v1^2/2 + m*v2^2/2
= (m/2)*(4v1^2 + v2^2)
So:
3*m*u = P_total = m*(4*v1 + v2)
3*u = 4*v1 + v2
=> v2 = 3*u - 4*v1
and:
(5/2)*m*u^2 = KE = (1/2)*m*(4*v1^2 + v2^2)
5*u^2 = 4*v1^2 + v2^2
= 4*v1^2 + (3*u - 4*v1)^2
= 4*v1^2 + 9*u^2 +16*v1^2 - 24*u*v1
= 20*v1^2 + 9*u^2 - 24*u*v1
0 = 20*v1^2 + 4*u^2 -24*u*v1
0 = 5*v1^2 + u^2 - 6*u*v1
0 = 5*x^2 -6*x + 1 , where x = v1/u
Using the quadratic formula:
x = 1 or 0.2
If x = 1, v1 = u, v2 = - u: this takes us back to the original situation! So this is not the right answer.
So x = 0.2, v1 = u/5, v2 = (3 - 0.8)*u = 2.2*u

So the speed of the heavier particle is v1 = u/5.

Notice something: v2 - v1 = 2.2*u - 0.2*u = 2*u = u - (-u)
In other words, the magnitude of the relative velocity of the two particles did NOT change. THIS IS GENERALLY TRUE FOR ELASTIC COLLISIONS. If you had used this fact from the beginning, you could have solved the problem by using conservation of momentum (no quadratic formula!).

Answer 2

In a collision, momentum is always conserved, and the center of mass maintains the same momentum. Velocity is a vector quantity
so
4*m*6.53-m*6.53 =4*m*v1+m*v2

Also, an elastic collision implies that kinetic energy is conserved

.5*4*m*6.53^2+.5*m*6.53^2=
.5*4*m*v1^2+.5*m*v2^2

solve for v1

I find the solution that works is
that the large mass continues with a positive velocity of 1.31 m/s and the small mass reflects with a velocity of 14.4 m/s. Which means they are both moving in the same direction post collision. This solution satisfies all of the equations.

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