A small block of mass m3=2kg of negligible size is placed on a plank of mass m2=4kg and length 4m which is further placed on a frictionless surface as shown in figure. If m1=8kg and system starts from rest,
A
Time after which m3 falls from m2 is ∞
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B
Time after which m3 falls from m2 is 2 sec
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C
Velocity of m1 at 2 sec is 12 m/s
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D
Velocity of m2 at t = 2 sec is 11.4 m/s
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Solution
The correct option is C Velocity of m1 at 2 sec is 12 m/s Assume all the blocks move together, then a=m1gm1+m2+m3 =808+4+2=8014=5.7m
Friction acting on m3=m3a=2×5.7 =11.4N
But maximum friction on m3 can only be, f=m3gμ=20×0.4=8N
Therefore, m3 sliding on m2, and f=8N
a=m1g−fm1+m2=80−812=6m/s2
acceleration (a0) of m3 is a0=fm3=82=4
Relative acceleration of m3 with respect to m2 is 6−4=2m/s2
The distance it has to cover is 4m. ∴s=ut+12at2 4=12×2×t2 ⇒t=2sec
Velocity of m1 and m2 at t = 2 sec is u+at =0+6×2=12m/s