First slide

Motion of centre of mass

Question

A wedge shaped block 'A' of mass M is at rest on a smooth horizontal surface. A small block 'B' of mass 'm' placed at the top edge of inclined plane of length 'L' as shown in the figure. By the time, the block 'B' reaches the bottom end, the wedge A moves a distance of

Difficult

A

$\frac{{mL}}{{M\cos \theta }}$
B

$\,\frac{{mL\cos \theta }}{{m + M}}$
C

$\frac{{mL}}{{m + M}}$
D

zero

Solution

Since net horizontal force on the system is zero, its horizontal momentum must remain conserved. If V be the velocity of wedge relative to ground and u be the velocity of block relative to the wedge, we can write,
MV + m(V - ucosθ) = M x 0 + m x 0
$\, \Rightarrow V = \frac{{m\cos \theta }}{{(m + M)}}.u$
$\, \Rightarrow \frac{{dx}}{{dt}} = \frac{{m\cos \theta }}{{(m + M)}}.\frac{{dl}}{{dt}}$
$\Rightarrow \int\limits_0^x {dx = \frac{{m\cos \theta }}{{(M + m)}}.\int\limits_0^L {dl} }$ (where x = Displacement of wedge and l= displacement of block with respect to wedge)

$\Rightarrow x = \frac{{mL\cos \theta }}{{(M + m)}}$

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