JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Conservation of Momentum

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 centre of mass of the particles is:

(A) $\frac{a}{5} (\hat{i} - \hat{j})$ ✓ Correct
(B)Zero
(C) $\frac{a}{5} (\hat{i} + \hat{j})$
(D) $a (\hat{i} + \hat{j})$

Answer: (A)

Solution

The acceleration of the centre of mass is the mass-weighted average

a_{c m} = \frac{\sum m _{i} a _{i}}{\sum m _{i}}

. Reading the directions from the figure,

a_{A} = - a \hat{i}

a_{B} = a \hat{j}

a_{C} = a \hat{i}

and

a_{D} = - a \hat{j}

, with masses

m, 2 m, 3 m, 4 m

and total

10 m

. The weighted sum is

m (- a \hat{i}) + 2 m (a \hat{j}) + 3 m (a \hat{i}) + 4 m (- a \hat{j}) = 2 ma \hat{i} - 2 ma \hat{j}

, so

a_{c m} = \frac{2 ma i ^ - 2 ma j ^}{10 m} = \frac{a}{5} (\hat{i} - \hat{j})

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Conservation of Momentum

Solution

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