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A smooth rope of mas m and length L lies...

A smooth rope of mas `m` and length `L` lies in a heap on a smooth horizontal floor, with one end attached to a block of mass `M`. The block is given a sudden kick and instantaneously acquires a horizontal velocity of magnitude `V_(0)` as shown in figure 1. As the block moves to right pulling the rope from heap, the rope being smooth, the heap remains at rest. At the instant block is at a distance `x` from point `P` as shown in figure `-2( P` is a point on the rope which has just started to move at the given instant `)`, choose correct options for next three question.

The tension in rope at point `P` is

A

`(mM^(2))/(L)(V_(0)^(2))/((M+(m)/(L)x)^(2))`

B

`(m^(2)M)/(L)(V_(0)^(2))/((M+(m)/(L)x)^(2))`

C

`(m^(3))/(L)(V_(0)^(2))/((M+(m)/(L)x)^(2))`

D

`(M^(3))/(L)(V_(0)^(2))/((M+(m)/(L)x)^(2))`

Text Solution

Verified by Experts

The correct Answer is:
A

The mass of moving material is `M+(m)/(L)x`
From conservation of momentum `MV_(0)=(M+(m)/(L)x)V`
`:. `Velocity of moving block and moving rope is
`V=(MV_(0))/(M+(m)/(L)x)`
`(A)` The tension at point `P` is what give momentume to next tiny piece `(` to left of `P )` that starts moving . The speed of this piece increases from 0 to `V` in time `dt.`
`rArr dp =dmV`
`or F=(dP)/(dt)=(dm)/(dt)V=((m)/(L)dx)/(dt)V=(m)/(L)V^(2)`
`:. F_(P)=(m)/(L)(M^(2)V_(0)^(2))/((M+(m)/(L)x)^(2))`
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