Total momentum a total energy of the system is conserved in the elastic collisions. We can write following:
m1v1+m2v2=m1v1′+m2v2′
m1v12+m2v22=m1v1′2+m2v2′2 (here we have already divided full equation by 2)
m1(v1−v1′)=m2(v2′−v2) (1)
m1(v1+v1′)(v1−v1′)=m2(v2′+v2)(v2′−v2) (2)
dividing (1) by (2):
v1+v1′=v2′+v2 (3)
After solving (1)+(2) and taking into account (3) we have:
v1′=−v2+2m1+m2m1v1+m2v2=−2+2710+10=3.714 m/s.
v2′=−v1+2m1+m2m1v1+m2v2=−5+2710+10=0.714 m/s.
After the collision the balls are moving into opposite directions, so the speed of separation is
v1′+v2′=3.714+0.714=4.428 m/s.
Answer: 4.428 m/s.
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