JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Conservation of Momentum

Particle

A

of mass

m_{1}

moving with velocity

(3 \hat{i} + \hat{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 aftr collision

V_{1} = (\hat{i} + 3 \hat{j}) ms^{- 1}

, the angle between

V_{1}

and

V_{2}

is:

Answer: (D)

Solution

Since

B

starts at rest, momentum conservation reads

m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}

. Putting

m_{1} = 2 m_{2}

u_{1} = 3 \hat{i} + \hat{j}

and

v_{1} = \hat{i} + 3 \hat{j}

, we get

v_{2} = 2 [(3 \hat{i} + \hat{j}) - (\hat{i} + 3 \hat{j})] = 2 (3 - 1) (\hat{i} - \hat{j})

. The angle between

v_{1}

and

v_{2}

follows from

cos θ = \frac{v _{1} \cdot v _{2}}{∣ v _{1} ∣ ∣ v _{2} ∣} = \frac{1 - 3}{2 2}

, which corresponds to

θ = 10 5^{\circ}

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option