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a 2.0kg block slides along a frictionless surface at 1.0 m/s. A second block, sliding at...

a 2.0kg block slides along a frictionless surface at 1.0 m/s. A second block, sliding at a faster 4.0 m/s, collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0 m/s. What was the mass of the second block?
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Answer #1
By conservation of momentum,
2*1 + m*4 = (2+m)*2
2m = 2
m = 1kg
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Answer #2

By conservation of momentum,
m1v1+m2v2 = m1v1' + m2v2'
here

v1' = v2'

given m1 = 2

v1 = 1 m /s

m2 = m (let)

v2 = 4 m/s

2*1 + m*4 = (2+m)*2
2m = 2
m = 1kg

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