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# JEE Class main Answered

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Asked by kumarabhishekuprama | 10 Aug, 2021, 05:48: PM
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 ,
Answered by Thiyagarajan K | 11 Aug, 2021, 09:19: AM
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