Question

A .50kg ball is rolling on a frictionless surface at a speed of .75m/s. it collides with a second ball with a mass of 1kg which is also moving in the same direction as the first ball with a speed of .38m/s. After the collision the firt ball continues at a reduced speed of .35m/s. what is the new speed of the second ball after the collision?
The new speed of the second ball is whatever conserves linear momentum.
Call it V2f

M1*V1i + M2*V2f = M1*V1f + M2*V2f
0.50*0.75 + 1.00*0.38 = 0.50*0.35 + 1.00*V2f

V2f = 0.58 m/s

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