In the figure shown C is a fixed wedge on horizontal surface. Blocks A and B are of masses m and $$2$$ m respectively are kept as shown in figure. They can slide along the inclined plane smoothly. The pulley and string are massless. Take θ$$=30o$$ and g $$=$$ $$10$$ $$m/s2$$. The inclined planes are very Long. A and B are released from rest. $$2$$ seconds after the release, B is caught for a moment and released again. Find out the speed of A just before the instant when the string becomes tight again.

Question

In the figure shown C is a fixed wedge on horizontal surface. Blocks A and B are of masses m and $$2$$ m respectively are kept as shown in figure. They can slide along the inclined plane smoothly. The pulley and string are massless. Take θ$$=30o$$ and g $$=$$ $$10$$ $$m/s2$$. The inclined planes are very Long. A and B are released from rest. $$2$$ seconds after the release, B is caught for a moment and released again. Find out the speed of A just before the instant when the string becomes tight again.

Infinity Learn · Accepted Answer

Initially acceleration of B and A is same along the string which is given by $$a=2mgsin$$⁡$$30$$∘−mgsin⁡$$30$$∘$$3m=g6After$$ $$2$$ seconds, speed of both A and B $$isvB=vA=g6$$×$$2=103m/sNow$$, when B is caught for a moment and released again, speed of B becomes zero, while A is still having a speed $$103m/s$$ up the inclined due to which string will become slack. But A will decelerate and B will accelerate. Because of this the string will become tight again after A and B travel the same distance. Let this time interval be $$t103$$×t−$$12gsin$$⁡$$30$$∘×$$t2=12gsin$$⁡$$30$$∘×$$t2$$⇒ $$103=g2t$$ as t≠$$0t=23secAt$$ this time speed of A is given $$byv=103$$−$$g2$$×$$t=103$$−$$102$$×$$23=0$$

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