Two particles of masses 1kg and 2kg are moving with constant velocities 2m/s(i^) and 5m/s(i^) respectively and crosses the y-axis simultaneously at t = 0sec and are moving on a smooth horizontal xy-plane. The separation between the two particles is 10 meter at t = 0. The angular momentum of 2kg particle with respect to 1kg particle at t = 5sec is
Question
Two particles of masses 1kg and 2kg are moving with constant velocities 2m/s(i^) and 5m/s(i^) respectively and crosses the y-axis simultaneously at t = 0sec and are moving on a smooth horizontal xy-plane. The separation between the two particles is 10 meter at t = 0. The angular momentum of 2kg particle with respect to 1kg particle at t = 5sec is
Solution
To solve this problem, we need to understand that the angular momentum (L) of a particle is given by the cross product of the position vector (r) and the linear momentum (p) of the particle. The formula is L = r x p.
Step 1: Determine the position vector (r) At t = 0, the separation between the two particles is 10m. Since they are moving with constant velocities, the 1kg particle will have moved 2m/s * 5s = 10m and the 2kg particle will have moved 5m/s * 5s = 25m at t = 5s. Therefore, the separation between the two particles at t = 5s is 25m - 10m = 15m. This is the magnitude of the position vector (r) of the 2kg particle with respect to the 1kg particle.
Step 2: Determine the linear momentum (p) The linear momentum (p) of the 2kg particle is its mass times its velocity, which is 2kg * 5m/s = 10kg*m/s.
Step 3: Calculate the angular momentum (L) The angular momentum (L) is the cross product of the position vector (r) and the linear momentum (p). Since both vectors are in the same direction (along the x-axis), their cross product is zero. Therefore, the angular momentum of the 2kg particle with respect to the 1kg particle at t = 5s is 0.
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