A block of mass m starts from rest and slides | vedclass

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Question

(A) $1$. Velocity of the block just before hitting the putty (by conservation of mechanical energy):$K.E._i + P.E._i = K.E._f + P.E._f$$0 + mgh = \fra

Vedclass · Accepted Answer

$h/4$

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(A) $1$. Velocity of the block just before hitting the putty (by conservation of mechanical energy):$K.E._i + P.E._i = K.E._f + P.E._f$$0 + mgh = \frac{1}{2}mv^2$$v = \sqrt{2gh}$$2$. Collision (conservation of linear momentum):Let the common velocity of the block and putty after sticking together be $v'$.$mv = (m + m)v'$$m\sqrt{2gh} = 2mv'$$v' = \frac{\sqrt{2gh}}{2} = \sqrt{\frac{gh}{2}}$$3$. Maximum height reached by the system (by conservation of mechanical energy):Let the maximum height reached be $h'$.$\frac{1}{2}(2m)(v')^2 = (2m)gh'$$\frac{1}{2} \cdot \frac{gh}{2} = gh'$$h' = h/4$

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