Question

Particle A of mass $m _{1}$ moving with velocity $(\sqrt{3} \hat{ i }+\hat{ j }) ms ^{-1}$ collides with another particle $B$ of mass $m _{2}$ which is at rest initially. Let $\vec{ V_{1} }$, and $\vec{ V }_{2}$ be the velocities of particles $A$ and $B$ after collision respectively. If $m _{1}=2 m _{2}$ and after collision $\vec{ V }_{1}=(\hat{ i }+\sqrt{3} \hat{ j }) \,ms ^{-1}$, the angle between $\vec{ V }_{1}$ and $\vec{ V }_{2}$ is :

• A. $60^{\circ}$
• B. $15^{\circ}$
• C. $-45^{\circ}$
• D. $105^{\circ}$

Answer 1

Answer: (D)

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