Question

Particle A of mass $m_1$ moving with velocity $(\sqrt{3}\hat{i}+\hat{j})m{s}^{-1}$ collides with another particle $B$ of mass $m_2$ which is at rest initially. Let $\overrightarrow{V_1}$, and ${\vec{V}}_2$ be the velocities of particles $A$ and $B$ after collision respectively. If $m_1=2m_2$ and after collision ${\vec{V}}_1=(\hat{i}+\sqrt{3}\hat{j})m{s}^{-1}$, 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,
$m_1{\vec{v}}_{01}+m_2{\vec{v}}_{02}=m_1{\vec{v}}_1+m_2{\vec{v}}_2$
$2m_2(\sqrt{3}\hat{i}+\hat{j})+0=2m_2(\hat{i}+\sqrt{3}\hat{j})+m_2{\vec{v}}_2$
${\vec{v}}_2=2(\sqrt{3}\hat{i}+\hat{j})-2(\hat{i}+\sqrt{3}\hat{j})$
$=2(\sqrt{3}\hat{i}-\hat{j})+2(\hat{i}-\sqrt{3}\hat{j})$
${\vec{v}}_2=2(\sqrt{3}-1)(\hat{i}-\hat{j})$
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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