m _{1} દળ અને (sqrt{3} hat{i}+hat{j}), m | vedclass

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Question

$\overrightarrow{ v }_{01}=(\sqrt{3 \hat{ i }}+\hat{ j }) m / s$

$\overrightarrow{ v }_{02}=\overrightarrow{0}$

$m _{1}=2 m _{2}$

After colli

Vedclass · Accepted Answer

$105$

Vedclass · Answer

$\overrightarrow{ v }_{01}=(\sqrt{3 \hat{ i }}+\hat{ j }) m / s$

$\overrightarrow{ v }_{02}=\overrightarrow{0}$

$m _{1}=2 m _{2}$

After collision, $\overrightarrow{ v }_{1}=(\hat{ i }+\sqrt{3} \hat{ j }) m / s$

$\overrightarrow{ v }_{2}=?$

Applying conservation of linear momentum,

$m _{1} \overrightarrow{ v }_{01}+ m _{2} \overrightarrow{ v }_{0 2}= m _{1} \overrightarrow{ v }_{1}+ m _{2} \overrightarrow{ v }_{2}$

$2 m_{2}(\sqrt{3 \hat{i}}+\hat{j})+0=2 m_{2}(\hat{i}+\sqrt{3} \hat{j})+m_{2} \vec{v}_{2}$

$\overrightarrow{ v }_{2}=2(\sqrt{3 \hat{ i }}+\hat{ j })-2(\hat{ i }+\sqrt{3} \hat{ j })$

$=2(\sqrt{3 \hat{1}}-\hat{j})+2(\hat{i}-\sqrt{3} \hat{j})$

$\overrightarrow{ v }_{2}=2(\sqrt{3}-1)(\hat{ i }-\hat{ j })$

for angle between $\overrightarrow{ v }_{1} \& \overrightarrow{ v }_{2}$

$\cos 	heta=\frac{\overrightarrow{ v }_{1} \cdot \overrightarrow{ v }_{2}}{\overrightarrow{ v }_{1} \overrightarrow{ v }_{2}}=\frac{2(\sqrt{3}-1)(1-\sqrt{3})}{2 	imes 2 \sqrt{2}(\sqrt{3}-1)}$

$\cos 	heta=\frac{1-\sqrt{3}}{2 \sqrt{2}} \Rightarrow 	heta=105^{\circ}$

કોલમ $-I$	કોલમ $-II$
$(1)$ ન્યૂટનનો ગતિનો ત્રીજો નિયમ	$(a)$ ${\vec F _{12}}\, = \, - \,{\vec F _{21}}$
$(2)$ વેગમાનના સંરક્ષણનો નિયમ	$(b)$ $\Delta \,\vec p \, = \,0$
	$(c)$ ${\vec F _{12}}\, = \,{\vec F _{21}}$

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