JEE Class main Answered
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Asked by kumarabhishekuprama | 10 Aug, 2021, 17:48: PM
Expert Answer
Let us consider that long piece paper is moving in the direction of +x-axis as shown in fig(1). Initially small block is moving in +y-axis direction.
Once small block is entered in the paper, it has x-component and y-component of motion. Trajectory of the block is shown in fig.(2) .
Fig.(3) gives the free body diagram of moving block . We see that friction force f = ( μ m g ) is opposing the movement of block.
Where μis friction coefficient, m is mass of block and g is acceleration due to gravity
Retardation a experienced by block = friction force / mass
Retardation a = ( μ m g ) / m = ( μ g )
Maximum velocity v of the block , if the block does not crosses the edge of paper is determined from the following equation of motion.
vf2 = v2 - ( 2 a L )
where vf = 0 is the final velocity , v is initial velocity , a is retardation and L is width of paper.
Hence , we get ,
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