Laws of Motion 3 Question 18

19. Two blocks A and B of equal masses are released from an inclined plane of inclination 45 at t=0. Both the blocks are initially at rest. The coefficient of kinetic

friction between the block A and the inclined plane is 0.2 while it is 0.3 for block B. Initially the block A is 2m behind the block B. When and where their front faces will come in a line?

(Take g=10m/s2 )

(2004, 5M)

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Answer:

Correct Answer: 19. sA=82m,2s

Solution:

  1. Acceleration of A down the plane,

aA=gsin45μAgcos45=(10)12(0.2)(10)12=42m/s2

Similarly acceleration of B down the plane,

aB=gsin45μBgcos45=(10)12(0.3)(10)12=3.52m/s2

The front face of A and B will come in a line when,

 or sA=sB+212aAt2=12aBt2+212×42×t2=12×3.52×t2+2

Solving this equation, we get t=2s

Further, sA=12aAt2=12×42×(2)2=82m

Hence, both the blocks will come in a line after A has travelled a distance 82m down the plane.



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