Let v1=0,v2=3.5m/s be the speeds of the balls before the collision, m1=2.5kg,m2=2m1=5kg the masses of the balls, v1′,v2′ the speeds of the balls after the collision.
Suppose the collision was head-on and ellastic. Thus, according to the momentum conservation law, have:
m1v1+m2v2=m1v1′+m2v2′ Since v1=0:
m2v2=m1v1′+m2v2′2m1v2=m1v1′+2m1v2′2v2=v1′+2v2′ According to the energy conservation law:
2m2v22=2m1v1′2+2m2v2′222m1v22=2m1v1′2+22m1v2′22v22=v1′2+2v2′2 Expressing v1′ from 2v2=v1′+2v2′ and substituting it into 2v22=v1′2+2v2′2, obtain:
3v2′2−4v2v2′+v22=0 Substituting the values and solving the quadratic equation, obtain:
v2′=1.166m/s or v2′=3.5m/s The last values in impossible, since it would mean that the first ball remains at rest after the collsion. Thus, subsituting v2′=1.166m/s into 2v2=v1′+2v2′, obtain:
v1′=2(v2−v2′)=4.66m/s And the kinetic energy is:
E1′=2m1v1′2=27.2J
Answer. 27.2J
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