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In a head-on (collinear) collision, a particle moving with initial speed v0 hits a stationary particle of equal mass. If the total kinetic energy after the collision is 50% more than the initial kinetic energy, what is the magnitude of the relative velocity of the two particles after the collision?

1. sqrt(2) v0
2. v0/2
3. v0/sqrt(2)
4. v0/4

Correct answer: sqrt(2) v0

Solution

Momentum gives v1+v2 = v0; energy gives v1²+v2² = 1.5 v0². Then (v1-v2)² = 2(1.5 v0²) - v0² = 2 v0², so |v1-v2| = sqrt(2) v0.

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