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Step-by-Step Solution
Step 1: Understand the Given Information
• Particle A has mass $m_{1} = 2m_{2}$ and is initially moving with velocity $\left(\sqrt{3}\,\hat{i} + \hat{j}\right)\,\text{m s}^{-1}$.
• Particle B has mass $m_{2}$ and is initially at rest.
• After collision, the velocity of A becomes $\vec{V}_{1} = \hat i + \sqrt{3} \,\hat j$.
• We need to find the angle between $\vec{V}_{1}$ and $\vec{V}_{2}$ (the velocity of B after collision).
Step 2: Assign Masses
Let $m_{2} = m$. Then $m_{1} = 2m$.
Step 3: Apply Conservation of Momentum
Initially, only particle A has momentum. Particle B is at rest. So the total initial momentum $\vec{p}_{i}$ is
$\vec{p}_{i} = m_{1} \left(\sqrt{3}\,\hat{i} + \hat{j}\right) = 2m \left(\sqrt{3}\,\hat{i} + \hat{j}\right).$
Let the final velocity of B be $\vec{V}_{2}$. After collision, the final total momentum $\vec{p}_{f}$ is
$\vec{p}_{f} = m_{1}\,\vec{V}_{1} + m_{2}\,\vec{V}_{2} = 2m\left(\hat i + \sqrt{3}\,\hat j\right) + m\,\vec{V}_{2}.$
By conservation of momentum, $\vec{p}_{i} = \vec{p}_{f}$. Hence,
$2m \left(\sqrt{3}\,\hat{i} + \hat{j}\right) = 2m\left(\hat i + \sqrt{3}\,\hat j\right) + m\,\vec{V}_{2}.$
Divide through by $m$ to simplify:
$2\left(\sqrt{3}\,\hat{i} + \hat{j}\right) = 2\left(\hat i + \sqrt{3}\,\hat j\right) + \vec{V}_{2}.$
Rearrange to solve for $\vec{V}_{2}:$
$\vec{V}_{2} = 2\left(\sqrt{3}\,\hat{i} + \hat{j}\right) - 2\left(\hat i + \sqrt{3}\,\hat j\right).$
Step 4: Simplify the Expression for $\vec{V}_{2}$
$\begin{aligned}
\vec{V}_{2}
&= 2\sqrt{3}\,\hat{i} + 2\,\hat{j} \;-\; 2\hat i - 2\sqrt{3}\,\hat j \\
&= \left(2\sqrt{3} - 2\right)\hat{i} + \left(2 - 2\sqrt{3}\right)\hat{j} \\
&= 2\left(\sqrt{3} - 1\right)\hat{i} - 2\left(\sqrt{3} - 1\right)\hat{j} \\
&= 2\left(\sqrt{3} - 1\right)\left(\hat{i} - \hat{j}\right).
\end{aligned}$
Step 5: Use the Dot Product to Find the Angle
We want the angle $\theta$ between $\vec{V}_{1} = \hat i + \sqrt{3}\,\hat j$ and $\vec{V}_{2} = 2(\sqrt{3}-1)(\hat{i} - \hat{j})$.
The cosine of the angle is given by
$\cos \theta = \frac{\vec{V}_{1} \cdot \vec{V}_{2}}{\left|\vec{V}_{1}\right|\,\left|\vec{V}_{2}\right|}.$
Step 5a: Compute the Dot Product $\vec{V}_{1} \cdot \vec{V}_{2}$
$\begin{aligned}
\vec{V}_{1} \cdot \vec{V}_{2}
&= \left(\hat i + \sqrt{3}\,\hat j\right) \cdot 2\left(\sqrt{3}-1\right)\left(\hat{i} - \hat{j}\right) \\
&= 2\left(\sqrt{3}-1\right)\Big[(1)\cdot(1) + (\sqrt{3})\cdot(-1)\Big] \\
&= 2\left(\sqrt{3}-1\right)\left(1 - \sqrt{3}\right).
\end{aligned}$
Step 5b: Compute the Magnitudes
1. $\left|\vec{V}_{1}\right| = \sqrt{(1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2.$
2. $\left|\vec{V}_{2}\right| = \left| 2 (\sqrt{3}-1)(\hat{i} - \hat{j}) \right|.$
Since $\hat{i} - \hat{j}$ has magnitude $\sqrt{(1)^2 + (-1)^2} = \sqrt{2}$, we get
$\left|\vec{V}_{2}\right| = 2(\sqrt{3}-1)\times \sqrt{2}.$
Step 5c: Plug Values into the Cosine Formula
$\begin{aligned}
\cos \theta
&= \frac{2\left(\sqrt{3}-1\right)\left(1 - \sqrt{3}\right)}{\big(2\big)\big(2(\sqrt{3}-1)\sqrt{2}\big)} \\
&= \frac{2\left(\sqrt{3}-1\right)\left(1 - \sqrt{3}\right)}{4\sqrt{2}\left(\sqrt{3}-1\right)}.
\end{aligned}$
Notice that $\left(\sqrt{3} - 1\right)$ appears both in numerator and denominator, so it cancels out (provided $\sqrt{3}-1 \neq 0$). Thus,
$\cos \theta = \frac{1 - \sqrt{3}}{2\sqrt{2}}.$
Step 6: Determine the Angle
We recognize that $\cos \theta = \frac{1 - \sqrt{3}}{2\sqrt{2}}$ is negative (because $\sqrt{3} \approx 1.732$, making $1 - \sqrt{3}$ a negative number).
By evaluating or comparing with known cosine values, we get
$\theta = 105^\circ.$
Answer: 105°
