Solving for Velocity and Mass in an Elastic Collision

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SUMMARY

The discussion focuses on solving for the velocity and mass of a target ball involved in an elastic collision with a softball of mass 0.220 kg moving at 6.7 m/s. After the collision, the softball bounces backward at 4.2 m/s. The conservation of momentum and kinetic energy principles are applied to derive the equations necessary for solving the problem. The initial momentum equation is set as m1v1 + m2v2 = m1v1(final) + m2v2(final), and the kinetic energy equation is .5m1v1^2 + .5m2v2^2 = .5m1v1^2(final) + .5m2v2^2(final).

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  • Understanding of elastic collisions in physics
  • Knowledge of conservation of momentum
  • Familiarity with conservation of kinetic energy
  • Basic algebra for solving equations
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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?
 

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