Posted by Shane on Wednesday, December 9, 2009 at 6:21pm.
A softball of mass 0.220 kg that is moving with a speed of 6.5 m/s (in the positive direction) collides headon and elastically with another ball initially at rest. Afterward it is found that the incoming ball has bounced backward with a speed of 4.8 m/s.
Calculate the mass of the target ball when the final velocity of the target ball is 1.7 m/s

physics  drwls, Wednesday, December 9, 2009 at 7:15pm
Conservation of momentum tells you that
0.22*6.5 = 0.22*4.8 + m v
mv = 2.486 kg m/s (positive direction)
Conservation of kinetic energy tells you that
0.22[(6.5)^2  (4.8)^2] = (1/2) m v^2
mv^2 = 8.452 kg m^2/s^2
m = (mv)^2/mv^2 = 0.714 kg