Impule and Momentum. Second Collision of a box and spring

[FordRanger4x4]

Homework Statement

a 10-kg block A is released from rest 2m above the 5kg plate P, which can slide freely along the smooth vertical guides BC and DE determine the velocity of the block and plate just after impact. the coefficient of restitution between the block and the plate is e=.75 also find the maximum compression of the spring due to impact. the spring has and unstretched length of 600m. k=1500N/m

V of the block is 2.61m/s
V of the spring is 7.308m/s
Smax is .32m

Ive solved for all of the questions its asking for here, answers have been verified, my teacher wants us to find the result of the secondary collision between the objects in the problem.

Homework Equations

Coefficient of Restitution
e=((Vb)2-(Va)2)/((Va)1-(Vb)1)
Conservation of Momentum
mb(Vb)1+ma(Va)1=mb(Vb)2+ma(Va)2
Conservation of Energy
To+Vo=T1+V1

The Attempt at a Solution

I thought id start out trying to find out how fast the block was moving when it started back up from the spring. Then with that answer use kinematics to find out how far up it went, then solve for how fast it would move back down and hit the block. I used T1+(Sum)U1-2=T2

0+1/2(1500)(.32)^2-15(9.8)(.32)=1/2(15)(V_2)^2
T1=0 because my starting point is at rest, the energies are from the spring and gravity multiplied by the displacement .32, equal to the KE of the block and plate moving back up.
my V_2 is 2.46m/s

My issue is, I don't know how to solve for the velocity of just the box once it leaves the plate. I tried using the conservation of momentum, finding an expression for the velocity of the plate and the box, and then using the coefficient of restitution formula to find a different formula for the plate and box and then solving, but because my initial velocity is 0 it negates the coefficient of restitution formula.

any help would be fantastic, thanks.

 

[KataKoniK]

Thank you for your post. You have made good progress in solving the initial problem and determining the velocity and maximum compression of the spring. To solve for the secondary collision between the objects, you can use the conservation of momentum and energy equations again. Since the objects are now moving in opposite directions, you will need to use negative signs for the velocities of the objects.

First, you can use the conservation of momentum equation to find the velocity of the block after the secondary collision. This can be written as:

mb(Vb)2+ma(Va)2=mb(Vb)3+ma(Va)3

Where (Vb)3 is the final velocity of the block after the collision and (Va)3 is the final velocity of the plate after the collision. You can use the velocity of the block from the initial collision and the mass of the plate to solve for (Va)3.

Then, you can use the conservation of energy equation to find the velocity of the plate after the secondary collision. This can be written as:

To+Vo=T3+V3

Where To is the initial kinetic energy of the plate and block, and T3 is the final kinetic energy of the plate and block after the collision. You can use the velocity of the block from the initial collision and the mass of the plate to solve for V3.

Once you have both (Va)3 and V3, you can use the coefficient of restitution formula to find the final velocity of the block after the secondary collision. This can be written as:

e=((Vb)3-(Va)3)/((Va)2-(Vb)2)

Solving for (Vb)3 will give you the final velocity of the block after the secondary collision. I hope this helps. Let me know if you need any further clarification. Good luck with your research!