Two particles of equal masses have velocity v→$$1$$ $$=$$ $$2i^$$ $$m/s$$ and $$v1$$→ $$=$$ $$2j^$$ $$m/s$$. The first particle has an acceleration $$a1$$→ $$=$$ (3i^+3j^) $$m/s2$$ while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a:

Question

Two particles of equal masses have velocity v→$$1$$ $$=$$ $$2i^$$ $$m/s$$ and $$v1$$→ $$=$$ $$2j^$$ $$m/s$$. The first particle has an acceleration $$a1$$→ $$=$$ (3i^+3j^) $$m/s2$$ while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a:

Infinity Learn · Accepted Answer

vcm→ $$=$$ $$m2i^+m2j^m+m$$ $$=$$ $$i^+j^$$ $$m/sacm$$→  $$=$$ $$m(3i^+3j^)+m$$×$$0m+m$$ $$=$$ $$32i^+32j^Since$$ acceleration and velocity are in same direction, so path should be straight line.

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Two particles of equal masses have velocity v→1 = 2i^ m/s and v1→ = 2j^ m/s. The first particle has an acceleration a1→ = (3i^+3j^) m/s2 while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a:

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