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A particle of mass 5 g undergoes projectile motion. It is launched from point A with an initial speed of 5*sqrt(2) m/s at an angle of 45 deg to the horizontal, and air resistance is negligible. Point B is the point where the particle lands back at the same horizontal level. What is the magnitude of the change in momentum of the particle between A and B, expressed as x*10⁻² kg m/s? Give x to the nearest integer.

1. 5
2. 10
3. 3
4. 7

Correct answer: 5

Solution

The horizontal momentum is conserved, so the change in momentum equals the change in vertical momentum. By symmetry the vertical velocity reverses from +5 to -5 m/s, giving |Delta p| = m*2*vy = 0.005*10 = 0.05 kg m/s = 5*10⁻² kg m/s.

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