springs - elastic collision?

Question

I'm having trouble with this problem. I'm not sure what to do but I beilieve i would need to use momentum conservation.

when can assume when the springs velocity is 0 when the spring has reach it's max length .

the problem can be sloved by using the equation
m1v2 = m2v2

????

______________________________________

In a one dimensional air track two springs are collide head on in a elsatic collision. Spring 1 has mass of .081 kg and a velocity before contact with spring 2 of 5.3 m/s. Spring 2 has a mass of .053 kg and a velocity before contact with spring 1 of -3.6m/s. (the right side = +)

[1] what is the velocity of spring 1 and spring 2 when the two springs are at maximum compression?

[2] what is the velocity of spring 1 and spring 2 at the instant they touch each other?

Answer 1

1)u can conserve momentum.
2) considering two spring as a single system velocity of CM of system is
Vcm=.081*5.3-.053*3.6/.081+.053=1.78m/sec
let in collision x1 is displcement of spring 1 n x2 of spring 2
then x1+x2=total compression
diffrenciate wid rspt to time
v1+v2=dk/dt
at max compressiton dk/dt=0
v1=-v2
Vcm=m1v1+m2v2/m1+m2
since no external force is there so Vcm will remain constant ie=1.78
1.78=.053*V1-.081*v1/.053+.081
v1=-8.5m/sec
v2=+8.5m/sec

[2] for this jst conserve the momentum n kinetic energy u will get the answer

Answer 2

in case you learn the momenta of the blocks till ultimately now collision, the sum is 0 so the sum would desire to be 0 after the collision too although the only way that capability is likewise conserved (a call for of an elastic collision) is that if the blocks get extra applicable with their unique velocity yet indoors the countless direction so neither block useful properties or loses capability. you will probable be waiting to desire to unravel this mathematically by applying using those 2 equations M1V1 + M2V2 = M1V3 + M2V4 (conservation of momentum) and a million/2 M1V1^2 + a million/2M2V2^2 = a million/2M1V3^2 + a million/2M2V4^2 (conservation of capability) use the 1st equation to place V4 in terms of V3 then replace that into the 2nd equation and be certain for V3 you will discover that V3 = V1

Answer 3

[1] Yeah, momentum is conserved. In elastic collision, they just exchange momenta. They bounce in the opposite directions, so the velocities at the maximum comression gotta be zero. You can use the following argument. Momentum is conserved for the system, the system consists of 2 springs. m1v1 + m2v2 = m1u1 + m2u2
where u1 and u2 are the velocities after collision. The energy is also concerved: 1/2(m1V1^2) + 1/2(m2V2^2) = 1/2(m1u1^2) 1/2(m2u2^2). You can solve for u1 and u2.

[2] Strange question. V1 = 5.3 m/s, V2 = -3.6 m/s, didn't they just tell us so ?