Three particles of masses 8 kg, 4 kg and 4 kg situated at (4,1), (–2,2) and (1,–3) are acted upon by external forces 6j¯ N, –6i¯ N and 14i¯ N. The acceleration of centre of mass of the system is

m1 = 8kg, m2 = 4kg, m3 = 4kg F¯1=6j^, F¯2=-6i^, F¯3=14i^a¯cm=F¯1+F¯2+F¯3m1+m2+m3=6j^-6i^+14i^16=8i^+16j^16acm=(8)2+(6)2(16)2=100256=1016=0.625 m/s2

Three particles of masses 8 kg, 4 kg and 4 kg situated at (4,1), (–2,2) and (1,–3) are acted upon by external forces 6j¯ N, –6i¯ N and 14i¯ N. The acceleration of centre of mass of the system is

m1 = 8kg, m2 = 4kg, m3 = 4kg F¯1=6j^, F¯2=-6i^, F¯3=14i^a¯cm=F¯1+F¯2+F¯3m1+m2+m3=6j^-6i^+14i^16=8i^+16j^16acm=(8)2+(6)2(16)2=100256=1016=0.625 m/s2

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Three particles of masses 8 kg, 4 kg and 4 kg situated at (4,1), (–2,2) and (1,–3) are acted upon by external forces 6 $\bar{j}$ N, –6 $\bar{i}$ N and 14 $\bar{i}$ N. The acceleration of centre of mass of the system is

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0.625 ms^-2

6.25 ms^-2

2.2 ms^-2

22 ms^-2

answer is A.

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Detailed Solution

m₁ = 8kg, m₂ = 4kg, m₃ = 4kg
${\bar{F}}_{1} = 6 \hat{j}, {\bar{F}}_{2} = - 6 \hat{i}, {\bar{F}}_{3} = 14 \hat{i}$

${\bar{a}}_{cm} = \frac{{\bar{F}}_{1} + {\bar{F}}_{2} + {\bar{F}}_{3}}{m_{1} + m_{2} + m_{3}} = \frac{6 \hat{j} - 6 \hat{i} + 14 \hat{i}}{16} = \frac{8 \hat{i} + 16 \hat{j}}{16}$

$a_{cm} = \sqrt{\frac{{(8)}^{2} + {(6)}^{2}}{{(16)}^{2}}} = \sqrt{\frac{100}{256}} = \frac{10}{16} = 0.625 m / s^{2}$

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