Particle A of mass m1 moving with velocity -3- ms-1 collides with another partice 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 V1-3-ms-1- the angle between V1 and V2 is -

Before collision: velocity of particle A, u_1=(√(3)î+ĵ) m/s Velocity of particle B, u_2=0 After collision: Velocity of particle A, v_1=(î+√(3)ĵ) m/s nbsp; ...

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Byju's Answer

Standard XII

Physics

Energy in Collision

Particle A of...

Question

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

15^{\circ}

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60^{\circ}

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- 45^{\circ}

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105^{\circ}

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Solution

The correct option is D $105^{\circ}$
Before collision:
velocity of particle $A, u_{1} = (\sqrt{3}^i +^j) m/s$

Velocity of particle $B, u_{2} = 0$

After collision:
Velocity of particle $A, v_{1} = (^i + \sqrt{3}^j) m/s$

Using conservation of linear momentum

$m_{1}_{1} + m_{2}_{2} = m_{1}_{1} + m_{2}_{2}$

$\Rightarrow 2 m_{2} (\sqrt{3^i} +^j) + (m_{2} \times 0) = 2 m_{2} (^i + \sqrt{3}^j) + (m_{2} \times_{2})$

$\Rightarrow 2 \sqrt{3}^i + 2^j = 2^i + 2 \sqrt{3}^j +_{2}$

$\Rightarrow_{2} = 2 (\sqrt{3} - 1)^i - 2 (\sqrt{3} - 1)^j$

And, $_{1} =^i + \sqrt{3}^j$

For angle between $_{1} and_{2},$

$cos θ = \frac{- \to v_{1} ._{2}}{|_{1} | |_{2} |} = \frac{2 (\sqrt{3} - 1) - 2 \sqrt{3} (\sqrt{3} - 1)}{2 \times 2 \sqrt{2} (\sqrt{3} - 1)}$

$cos θ = \frac{1 - \sqrt{3}}{2 \sqrt{2}}$

$\Rightarrow θ = 105^{\circ}$

So, the angle between $_{1}$ and $_{2}$ is $105^{\circ}$ .

Alternate solution:
Here, $_{1} =^i + \sqrt{3}^j$
$_{1}$ will make an agle $θ$ with $x -$ axis.

$∴ θ = {tan}^{- 1} (\sqrt{3} / 1) = 60^{\circ}$

Similarly, $_{2} = 2 (\sqrt{3} - 1)^i - 2 (\sqrt{3} - 1)^j$ will make angle $ϕ$ with $x -$ axis.

$∴ ϕ = {tan}^{- 1} (\frac{- 2 (\sqrt{3} - 1)}{2 (\sqrt{3} - 1)}) = - 45^{\circ}$

So, the angle between $_{1}$ and $_{2}$ will be $60^{\circ} + 45^{\circ} = 105^{\circ}$ .

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