a) From the conservation of momentum,
m1v1+m2v2=m1v1′+m2v2′
v2′=m2m1(v1−v1′)+v2 Let the direction to the right be positive, the direction to the left be negative. So,
v2′=2.11.6(4−3)−2.5=−1.74sm b)From the conservation of energy,
0.5m1v12+0.5m2v22−0.5m1v1′2−0.5m2v2′2=0.5kx2
m1(v12−v1′2)+m2(v22−v2′2)=kx2
1.6(42−32)+2.1(2.52−1.742)=600x2
x=0.16 m
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