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Standard XII

Physics

Velocity of COM

Two particles...

Question

Two particles $A$ and $B$ are connected by a rigid rod $A B$ . The rod slides on perpendicular rails as shown in figure. The velocity of $A$ to the left is $10 m / s$ . What is velocity of $B$ when $α = 60^{o}$ ? (in m/s)

4.3

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17.3

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Solution

The correct option is D $17.3$
Given : $V_{A} = 10 m / s$
Both the particles move can move parallel to the surface but not perpendicular to the surface.
From figure,
$tan 60^{o} = \frac{V_{B}}{V_{A}}$

$⟹ V_{B} = V_{A} tan 60^{o}$
$V_{B} = 10 \times \sqrt{3}$
$V_{B} = 17.3 m / s$

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Two particles A and B are connected by a rigid rod AB. The rod slides on perpendicular rails as shown in figure. The velocity of A to the left is 10m/s. What is velocity of B when α=60o? (in m/s)

The correct option is D 17.3Given : VA=10m/sBoth the particles move can move parallel to the surface but not perpendicular to the surface.From figure, tan60o=VBVA⟹VB=VAtan60o VB=10×√3 VB=17.3 m/s

Two particles $A$ and $B$ are connected by a rigid rod $A B$ . The rod slides on perpendicular rails as shown in figure. The velocity of $A$ to the left is $10 m / s$ . What is velocity of $B$ when $α = 60^{o}$ ? (in m/s)

The correct option is D $17.3$
Given : $V_{A} = 10 m / s$
Both the particles move can move parallel to the surface but not perpendicular to the surface.
From figure,
$tan 60^{o} = \frac{V_{B}}{V_{A}}$

$⟹ V_{B} = V_{A} tan 60^{o}$
$V_{B} = 10 \times \sqrt{3}$
$V_{B} = 17.3 m / s$