A 2—kg ball B travelling at 22 m.s–¹ overtakes a 4—kg ball travelling in the same direction at 10 m.s—¹. Find the velocity of each ball after collision
Answer
Initial momentum is given
"P_i=mu+m'u'\\\\=2\\times22+4\\times10\\\\=84kg-m\/sec"
So velocity after collision
For first particle B
"v=\\frac{P_i+m'(u'-u) }{m+m'}\\\\=\\frac{84+4(10-22) }{2+4}\\\\=6m\/sec"
Now A
"v'=\\frac{P_i+m(u-u') }{m+m'}\\\\=\\frac{84+2(22-10) }{2+4}\\\\=18m\/sec"
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