Two balls with masses `m_(1)=3` and `m_(2)=5`kg have initial velocities `v_(1)=v_(2)=5m//s` in the directions shown in figure. They collide at the origin. a. find the velocioty of the CM `3s` before the collision. b. Find the position of the CM `2s` after the collision.
Text Solution
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a. Given time is of no consequence since `v_(CM)` is fixed for all times `v_(CM)(x)=(m_(1)v_(1x)+m_(2)v_(2x))/(m_(1)+m_(2))` `=((3)(-5cos37^(@))+(5)(0))/(8kg)=-1.5m//s` `v_(CM)(y)=(m_(1)v_(1)y+m_(2)y_(2))/(m_(1)+m_(2))` `=(3)(-5sin37^(@))+(5xx5)/(8kg)=+2m//s` `:. vecv_(CM)=1.5hati+2hatjm//s` b. Since the collision occurs at the origin `(vecr_(1)=0)`, the position of the centre of mass `2s` later is `vecr_(CM)=vecr_(1)+vecv_(CM^(t))` `=vecv_(CM^(t))=-3hati+4hatjm`
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