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Particle A of mass m1 moving with velocity (sqrt(3) i + j) m/s strikes a stationary particle B of mass m2. After impact, A moves with velocity v1 = (i + sqrt(3) j) m/s, while B moves with velocity v2. Given m1 = 2*m2, find the angle between v1 and v2.

1. 60 deg
2. 15 deg
3. -45 deg
4. 105 deg

Correct answer: 105 deg

Solution

Momentum conservation gives v2's components; v1 points at 60 deg and v2 points at -45 deg, so the angle between them is 105 deg.

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