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

m 1 u → 1 + m 2 u → 2 = m 1 v → 1 + m 2 v → 2 2 m 2 ( 3 i ^ + j ^ ) + m 2 × 0 = 2 m 2 ( i ^ + 3 j ^ ) + m 2 × v → 2 2 3 i ^ + 2 j ^ = 2 i ^ + 2 3 j ^ + v → 2 v → 2 = ( 3 - 1 ) i ^ - ( 3 - 1 ) j ^ v → 1 = i ^ + 3 j ^ Angle between v → 1 and v → 2 is 105 °

Start Free Mock LoginAlready registered? Login here

Physics - Center of Mass Question with Solution | TestHub

PhysicsCenter of MassCollisionHard2 minPYQ_2020

PhysicsHardsingle choice

Particle $A$ of mass $m_{1}$ moving with velocity $(\sqrt{3} \hat{i} + \hat{j}) {ms}^{- 1}$ 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 $m_{1} = 2 m_{2}$ and after collision ${\vec{v}}_{1} - (\hat{i} + \sqrt{3} \hat{j}) {ms}^{- 1}$ , the angle between ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ is :

Options:

Answer:

Solution:

$m_{1} {\vec{u}}_{1} + m_{2} {\vec{u}}_{2} = m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}$

$2 m_{2} (\sqrt{3} \hat{i} + \hat{j}) + m_{2} \times 0 = 2 m_{2} (\hat{i} + \sqrt{3} \hat{j}) + m_{2} \times {\vec{v}}_{2}$

$2 \sqrt{3} \hat{i} + 2 \hat{j} = 2 \hat{i} + 2 \sqrt{3} \hat{j} + {\vec{v}}_{2}$

${\vec{v}}_{2} = (\sqrt{3} - 1) \hat{i} - (\sqrt{3} - 1) \hat{j}$

${\vec{v}}_{1} = \hat{i} + \sqrt{3} \hat{j}$

Angle between ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ is $105^{°}$

Stream:JEESubject:PhysicsTopic:Center of MassSubtopic:Collision

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...