A block of mass M moving on a frictionless ho | vedclass

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(C) According to the law of conservation of mechanical energy,the initial kinetic energy of the block is converted into the potential energy of the sp

Vedclass · Accepted Answer

$L \sqrt{MK}$

Vedclass · Answer

(C) According to the law of conservation of mechanical energy,the initial kinetic energy of the block is converted into the potential energy of the spring at maximum compression.Initial kinetic energy,$KE = \frac{1}{2} Mv^2$Final potential energy of the spring,$PE = \frac{1}{2} KL^2$Equating the two:$\frac{1}{2} Mv^2 = \frac{1}{2} KL^2$$Mv^2 = KL^2$$v^2 = \frac{K}{M} L^2$$v = L \sqrt{\frac{K}{M}}$The momentum of the block is given by $p = Mv$.Substituting the value of $v$:$p = M \left( L \sqrt{\frac{K}{M}} \right)$$p = L \sqrt{M^2 \cdot \frac{K}{M}}$$p = L \sqrt{MK}$

$A$ block of mass $M$ moving on a frictionless horizontal surface collides with a spring of spring constant $K$,as shown in the figure. If the spring compresses by a length $L$,then the maximum momentum of the block is:

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