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Laws of Motion Part 1

Section: Physics

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Question : 80 of 100

Marks: +1, -0

Two fixed frictionless inclined planes making an angle

30^{\circ}

60^{\circ}

with the vertical are shown in the figure. Two blocks

A

B

are placed on the two planes. What is the relative vertical acceleration of

A

with respect to

B

$4.9 \text{ ms}^{-2}$ in horizontal direction
$9.8 \text{ ms}^{-2}$ in vertical direction
Zero
$4.9 \text{ ms}^{-2}$ in vertical direction

Solution:

Along inclined plane the equation of motion of the body

m g \sin \theta = m a \;\; \therefore a = g \sin \theta

where

a

is along the inclined plane.

\therefore

vertical component of acceleration is

(g \sin \theta) \sin \theta = g \sin^2 \theta

(Along vertical)

For block A,

a_{A(\text{ along vertical })} = g \sin^2 60^{\circ}

For block B,

a_{B(\text{ along vertical })} = g \sin^2 30^{\circ}

\therefore

relative vertical acceleration of

A

with respect to

B

is

\; g \sin^2 60^{\circ} - g \sin^2 30^{\circ}

\; = g ( \sin^2 60 - \sin^2 30 ]

\; = 4.9 \text{ m/s}^2 \text{ in vertical direction }

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