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Question

# Particle A of mass m1 moving with velocity (√3^i+^j) 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=(^i+√3^j)ms−1, the angle between →V1 and →V2 is :

A
15
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B
60
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C
45
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D
105
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Solution

## The correct option is D 105∘Before collision: velocity of particle A, u1=(√3^i+^j) m/s Velocity of particle B, u2=0 After collision: Velocity of particle A, v1=(^i+√3^j) m/s Using conservation of linear momentum m1→u1+m2→u2=m1→v1+m2→v2 ⇒2m2(√3^i+^j)+(m2×0)=2m2(^i+√3^j)+(m2×→v2) ⇒2√3^i+2^j=2^i+2√3^j+→v2 ⇒→v2=2(√3−1)^i−2(√3−1)^j And, →v1=^i+√3^j For angle between →v1 and →v2, cosθ=−→v1.→v2|→v1||→v2|=2(√3−1)−2√3(√3−1)2×2√2(√3−1) cosθ=1−√32√2 ⇒θ=105∘ So, the angle between →v1 and →v2 is 105∘. Alternate solution: Here, →v1=^i+√3^j →v1 will make an agle θ with x− axis. ∴θ=tan−1(√3/1)=60∘ Similarly, →v2=2(√3−1)^i−2(√3−1)^j will make angle ϕ with x− axis. ∴ϕ=tan−1(−2(√3−1)2(√3−1))=−45∘ So, the angle between →v1 and →v2 will be 60∘+45∘=105∘.

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