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Three interacting particles of masses 100 g, 200 g and 400 g each move at 20 m/s, directed along the positive x-axis, y-axis and z-axis respectively. After interaction the third particle (400 g) comes to rest, and the second particle (200 g) moves with velocity (10 j + 5 k) m/s. Find the velocity of the first particle (100 g). (Use conservation of momentum.)

1. 20 i + 20 j + 70 k
2. 10 i + 20 j + 8 k
3. 30 i + 10 j + 7 k
4. 15 i + 5 j + 60 k

Correct answer: 20 i + 20 j + 70 k

Solution

Conserving momentum, 0.1 v1 = p_initial - p2 - p3, giving v1 = 20 i + 20 j + 70 k m/s.

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