In a perfectly elastic collision the momentum conservation law holds. Thus, the total momentum before the collision:
p=m1v1is equal to the total momentum after the collision:
p′=m1v1′+m2v2′where m1=40kg,m2=29kg are the masses of the objects, v1=25m/s is the speed of the first object before the collision, and v1′,v2′ are their speeds after the collsion. Thus, obtain:
m1v1=m1v1′+m2v2′
Futhermore, the energy conservation law also holds for an elastic collision:
2m1v12=2m1v1′2+2m2v2′2 Expressing v1′ from the first equation and substituting it into the last one, after doing some algebra obtain:
v2′2(m1m22+m2)−2m2v1v2′=0v2′2(m1m2+1)−2v1v2′=0v2′(v2′(m1m2+1)−2v1)=0 One solution is v2′=0, which is not possible. Another one is:
v2′(m1m2+1)−2v1=0v2′=m1m2+12v1=m1+m22m1v1 Substituting the numbers, obtain:
v2′=40+292⋅40⋅25≈28.99 m/s
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