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Question

Physics Question on System of Particles & Rotational Motion

Find the velocity of centre of mass of the system shown in the figure?

A

(2+233)i^−23j^\left( \frac{2+2\sqrt{3}}{3} \right)\hat{i}-\frac{2}{3}\hat{j}

B

4i^4\hat{i}

C

(2−233)i^−13j^\left( \frac{2-2\sqrt{3}}{3} \right)\hat{i}-\frac{1}{3}\hat{j}

D

None of the above

Answer

(2+233)i^−23j^\left( \frac{2+2\sqrt{3}}{3} \right)\hat{i}-\frac{2}{3}\hat{j}

Explanation

Solution

Here, m1=1 kg,v1→=2i^m_{1}=1 \,kg , \overrightarrow{ v _{1}}=2 \hat{ i } m2=2kgv⃗2→=2cos⁡30i^−2sin⁡30j^m_{2} =2 kg \overrightarrow{\vec{v}_{2}}=2 \cos 30 \hat{ i }-2 \sin 30 \hat{ j } v⃗cm=m1v→1+m2v2→m1+m2\vec{v}_{ cm } =\frac{m_{1} \overrightarrow{ v }_{1}+m_{2} \overrightarrow{ v _{2}}}{m_{1}+m_{2}} =1×2i^+2(2cos⁡30i^−2sin⁡30i^1+2=\frac{1 \times 2 \hat{i}+2(2 \cos 30 \hat{i}-2 \sin 30 \hat{i}}{1+2} =2i^+23i^−2j^3=\frac{2 \hat{i}+2 \sqrt{3} \hat{i}-2 \hat{j}}{3} =(2+233)i^−23j^=\left(\frac{2+2 \sqrt{3}}{3}\right) \hat{i}-\frac{2}{3} \hat{j}