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Conservation of Momentum

Four particle...

Question

Four particles $A, B, C$ and $D$ with masses $m_{A} = m$ , $m_{B} = 2 m$ , $m_{C} = 3 m$ and $m_{D} = 4 m$ are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the center of mass of the particles is

\frac{a}{5} (^i +^j)

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\frac{a}{5} (^i -^j)

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zero

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a (^i +^j)

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Solution

The correct option is B $\frac{a}{5} (^i -^j)$

The acceleration of $A, {\to a}_{A} = - a^i$

The acceleration of $B, {\to a}_{B} = a^j$

The acceleration of $C, {\to a}_{C} = a^i$

The acceleration of $D, {\to a}_{D} = - a^j$

The acceleration of $C M, {\to a}_{c m}$

$= \frac{m_{a} {\to a}_{a} + m_{b} {\to a}_{b} + m_{c} {\to a}_{c} + m_{d} {\to a}_{d}}{m_{a} + m_{b} + m_{c} + m_{d}}$

${\to a}_{c m} = \frac{- m a^i + 2 m a^j + 3 m a^i - 4 m a^j}{10 m}$

$= \frac{2 m a^i - 2 m a^j}{10 m} = \frac{a}{5}^i - \frac{a}{5}^j$

$= \frac{a}{5} (^i -^j)$

Final answer: (a)

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Conservation of Momentum

Standard X Physics

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Four particles A,B,C and D with masses mA=m, mB=2m, mC=3m and mD=4m are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the center of mass of the particles is

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