Question

Two men of same mass m, hold the two ends of a rope and start pulling each other on a frictionless plane. Find the position where they meet. If masses are changed to m and 4m. What is the new location​

Answer 1

Answer:

assuming pulley is rotating clockwise with an acceleration a.

also a

mg

=a

mr

+a

rg

, so acceleration of m

1

is a+1.2 and acceleration of m

2

is (−a+2) assuming both in upward direction.

writing equation of motion

T−600=60(−a+2)

and T−400=(a+1.2)

solving above equations we get a=2.72 in clockwise direction.

also tension in cable T=556.8N

now net acceleration of both the man relative to ground is m

1

=3.84m/s

2

upwards and m

2

=0.72m/s

2

downwards

acceleration of m

2

wrt m

1

is 4.56m/s

2

upwards and distance between both is 5m.

using equation S=ut+

2

1

at

2

where u=0,a=4.56 we get T=1.48sec.

so best possible options are B,C.

solution

Explanation:

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