A small body of mass m is placed over a larger mass M whose surface is horizontal near the small body and gradually curves to become vertical. The small body is pushed on the larger mass at a speed v and the system is

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Quantrex Academy · Accepted Answer

Correct answer: C. (Mv^2)/(2g(M + m))

Let the initial velocity of the small body of mass

m

be

v and the larger mass

M

be at rest.<br><br>Since there is no external force acting on the system in the horizontal direction, the horizontal component of momentum is conserved.<br><br>At the maximum height

h

, the small body is at rest relative to the larger mass. Thus, both masses move with the same horizontal velocity

V

.<br><br>Applying conservation of linear momentum in the horizontal direction:<br><br>

mv = (M + m)V

<br><br>

V = \dfrac{mv}{M + m}

<br><br>Applying conservation of mechanical energy for the system:<br><br>

\dfrac{1}{2}mv^2 = \dfrac{1}{2}(M + m)V^2 + mgh

<br><br>Substituting the value of

V

:<br><br>

\dfrac{1}{2}mv^2 = \dfrac{1}{2}(M + m)\left(\dfrac{mv}{M + m}\right)^2 + mgh

<br><br>

\dfrac{1}{2}mv^2 = \dfrac{m^2v^2}{2(M + m)} + mgh

<br><br>

mgh = \dfrac{1}{2}mv^2 - \dfrac{m^2v^2}{2(M + m)}

<br><br>

mgh = \dfrac{mv^2}{2} \left(1 -

\dfrac{m}{M + m}\right)<br><br>

mgh =

\dfrac{mv^2}{2}\left(\dfrac{M + m - m}{M + m}\right)

mgh = \dfrac{mMv^2}{2(M + m)}

h = \dfrac{Mv^2}{2g(M + m)}

A small body of mass m is placed over a larger mass M whose surface...

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