Question

Particle A of mass \(m_1\) moving with velocity \((\sqrt3\,\hat i+\hat j)\ \mathrm{ms^{-1}}\) 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 \(m_1=2m_2\) and after collision \(\vec V_1=(\hat i+\sqrt3\,\hat j)\), the angle between \(\vec V_1\) and \(\vec V_2\) is :

• A. \(105^\circ\)
• B. \(15^\circ\)
• C. \(-45^\circ\)
• D. \(60^\circ\)

Hint:

Conserve linear momentum to get \(\vec V_2=2(\vec u-\vec V_1)\), then find its direction and compare with the direction of \(\vec V_1\) (60 degrees from the x-axis).

Answer 1

The Correct Option is A

Let \(\vec u=\sqrt3\hat i+\hat j\) and \(m_1=2m_2\). Momentum conservation: \(2m_2\vec u=2m_2\vec V_1+m_2\vec V_2\).
\(\vec V_2=2(\vec u-\vec V_1)=2\left[(\sqrt3-1)\hat i+(1-\sqrt3)\hat j\right]=2(\sqrt3-1)(\hat i-\hat j)\).
So \(\vec V_2\) makes \(-45^\circ\) with the x-axis (components \(+x,-y\)).
\(\vec V_1=\hat i+\sqrt3\hat j\) makes \(\tan^{-1}\sqrt3=60^\circ\) with the x-axis.
Angle between them \(=60^\circ+45^\circ=105^\circ\).
Check with dot product: \(\vec V_1\cdot\vec V_2=2(\sqrt3-1)(1-\sqrt3)=-2(\sqrt3-1)^2\); \(|\vec V_1|=2\), \(|\vec V_2|=2\sqrt2(\sqrt3-1)\); \(\cos\theta=\dfrac{-2(\sqrt3-1)^2}{4\sqrt2(\sqrt3-1)}=-\dfrac{\sqrt3-1}{2\sqrt2}=-0.259\Rightarrow\theta=105^\circ\).
Answer: (1) \(105^\circ\)