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Three particles A, B and C of equal mass...

Three particles` A, B` and `C` of equal mass move with equal speed `V = 5 ms^(-1)` along the medians of an equilateral triangle as shown in the figure. They collide at the centroid `G` of the triangle. After the collision, `A` comes to rest, `B` retraces its path with the speed `V = 5 ms^(-1)`. What is the speed of `C`?

Text Solution

Verified by Experts

The correct Answer is:
5
Initial momentum
`vecp_(i)=mvhatj-mv((sqrt(3))/2hati+1/2hatj)+mv((sqrt(3))/2hati+1/2hatj)`
Final momentum
`vecp_(f)=-mv((sqrt(3))/2hati+1/2hatj)+mvecv_(c)`
Now `vecp_(f)=vecp_(i)`
`implies vecv_(c)=-vhatj+v(sqrt3hati+hatj)+v(-sqrt(3)/2hati+1/2hatj)`
`implies vec_(c)=v((sqrt(3))/2hati+1/2hatj)impliesv_(C)=5ms^(-1)`
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