Solving for Velocity and Mass in an Elastic Collision

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In an elastic collision involving a 0.220 kg softball moving at 6.7 m/s colliding with a stationary ball, both momentum and kinetic energy are conserved. The momentum equation is set up as m1v1 + m2v2 = m1v1(final) + m2v2(final), where the second ball's initial velocity is zero. The kinetic energy equation is .5m1v1^2 + .5m2v2^2 = .5m1v1^2(final) + .5m2v2^2(final), with the initial energy of the target ball also being zero. To solve for the target ball's velocity and mass, these two equations can be combined and manipulated to isolate the unknowns. The discussion emphasizes the need to correctly apply conservation laws to find the solutions.
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A softball of mass 0.220 kg that is moving with a speed of 6.7 m/s collides head-on and elastically with another ball initially at rest. Afterward it is found that the incoming ball has bounced backward with a speed of 4.2 m/s. (Assume the positive direction is forward.)

(a) Calculate the velocity of the target ball after the collision.

(b) Calculate the mass of the target ball.

I know I have to use conservation of momentum along with conservation of energy, but I don't know how to set them up or combine the 2 equations.
 
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You are right, momentum and kinetic energy will be conserved. Can you write an equation to show that?
 
momentum equation:
initial momentum = final momentum
m1v1 + m2v2 = m1v1(final) + m2v2(final)

I know that m2v2 (initial) = 0 because the target ball is at rest.

Kinetic energy equation:
initial energy = final energy
.5m1v1^2 + .5m2v2^2 = .5m1v1^2(final) + .5m2v2^2(final)

.5m2v2^2 (initial) = 0 because target ball is at rest.

How would I solve these 2 equations?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

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