Particle A of mass m1 moving with velocity 3i^+j^ ms−1 collides with another particle B of mass m2 which is at rest initially. Let V1→ and V2→ be the velocities of particles A and B after collision respectively. If m1=2m2 and after collision V1→=i^+3j^ms−1 , the angle between V1→ and V2→ is

Let V→2=Vxi^+Vyj^Applying momentum conservation along x and y direction,m1×3=m1+m2Vx ⇒Vx=23−1 And m1=3m1+m2Vy ⇒Vy=−23−1So V→2=23−1i^−23−1j^, which is making angle 45° with x axis as shownand V→1 =i^+3j^ , is making 600 with x axis as shown in figure So angle between V→1 and V→2 is 105°

Particle A of mass m1 moving with velocity 3i^+j^ ms−1 collides with another particle B of mass m2 which is at rest initially. Let V1→ and V2→ be the velocities of particles A and B after collision respectively. If m1=2m2 and after collision V1→=i^+3j^ms−1 , the angle between V1→ and V2→ is

Let V→2=Vxi^+Vyj^Applying momentum conservation along x and y direction,m1×3=m1+m2Vx ⇒Vx=23−1 And m1=3m1+m2Vy ⇒Vy=−23−1So V→2=23−1i^−23−1j^, which is making angle 45° with x axis as shownand V→1 =i^+3j^ , is making 600 with x axis as shown in figure So angle between V→1 and V→2 is 105°

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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 $\vec{V_{1}} a n d \vec{V_{2}}$ be the velocities of particles A and B after collision respectively. If $m_{1} = 2 m_{2}$ and after collision $\vec{V_{1}} = (\hat{i} + \sqrt{3} \hat{j}) m s^{- 1}$ , the angle between $\vec{V_{1}}$ and $\vec{V_{2}}$ is

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$15 °$

$- 45 °$

$60 °$

$105 °$

answer is D.

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Detailed Solution

Let ${\vec{V}}_{2} = V_{x} \hat{i} + V_{y} \hat{j}$
Applying momentum conservation along x and y direction,
$m_{1} \times \sqrt{3} = m_{1} + m_{2} V_{x} \Rightarrow V_{x} = 2 (\sqrt{3} - 1)$
And $m_{1} = \sqrt{3} m_{1} + m_{2} V_{y} \Rightarrow V_{y} = - 2 (\sqrt{3} - 1)$
So ${\vec{V}}_{2} = 2 (\sqrt{3} - 1) \hat{i} - 2 (\sqrt{3} - 1) \hat{j}$ , which is making angle 45° with x axis as shown