A force of 15 N is acting on a particle of mass 3 kg towards north and a force of 20 N is acting on another particle of mass 2 kg towards east. The acceleration of the centre of mass of the two particle system is

7 m/s²

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5 m/s²

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17 m/s²

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55 m/s²

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Solution

The correct option is B
5 m/s²

Acceleration toward north, $a_{1} = \frac{F_{1}}{m_{1}} = \frac{15}{3} = 5 m / s^{2} (^i)$

Acceleration toward east, $a_{2} = \frac{F_{2}}{m_{2}} = \frac{20}{2} = 5 m / s^{2} (^j)$

$∴$ Acceleration of centre of mass of the two particle system
$a_{c m} = \frac{m_{1} {\to a}_{1} + m_{2} {\to a}_{2}}{m_{1} + m_{2}} = \frac{3 (5)^i + 2 (10)^j}{3 + 2}$

$\Rightarrow {\to a}_{c m} = \frac{15^i + 20^j}{5} = 3^i + 4^j$

$\Rightarrow {\to a}_{c m} = \sqrt{3^{2} + 4^{2}} = 5 m / s^{2}$

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A force of 15 N is acting on a particle of mass 3 kg towards north and a force of 20 N is acting on another particle of mass 2 kg towards east. The acceleration of the centre of mass of the two particle system is

The correct option is B 5 m/s2 Acceleration toward north, a1=F1m1=153=5m/s2(^i) Acceleration toward east, a2=F2m2=202=5m/s2(^j) ∴ Acceleration of centre of mass of the two particle system acm=m1→a1+m2→a2m1+m2=3(5)^i+2(10)^j3+2 ⇒→acm=15^i+20^j5=3^i+4^j ⇒→acm=√32+42=5m/s2