A rigid body of mass m is held at a height on two smooth wedges are th...
A rigid body of mass m is held at a height on two smooth wedges are th...
**Explanation:**
**1. Initial conditions:**
- The rigid body of mass m is held at a height H from the ground.
- The two smooth wedges are at rest on a horizontal frictionless floor.
**2. Release of the rigid body:**
- When the rigid body is released, it moves down due to gravity.
- As it moves down, it pushes aside the wedges.
**3. Conservation of momentum:**
- The total momentum of the system (rigid body + wedges) is conserved.
- Initially, the wedges are at rest, so their momentum is zero.
- Therefore, the total momentum of the system is initially zero.
**4. Motion of the rigid body:**
- As the rigid body moves down, it gains momentum.
- The momentum gained by the rigid body is equal and opposite to the momentum gained by the wedges.
- This is due to the conservation of momentum.
- The velocity gained by the rigid body can be calculated using the equation:
m * v = m * g * t
where v is the velocity gained by the rigid body, g is the acceleration due to gravity, and t is the time taken to reach the ground.
**5. Motion of the wedges:**
- The wedges move apart from each other as the rigid body pushes them aside.
- The velocity of recession of the wedges can be calculated using the equation:
v_w = (m * v) / (M1 + M2)
where v_w is the velocity of recession of the wedges, m is the mass of the rigid body, M1 and M2 are the masses of the wedges.
**6. Conclusion:**
- The velocity of recession of the wedges from each other when the rigid body is at a height H from the ground can be calculated using the above equation.
- The velocity of recession depends on the mass of the rigid body and the masses of the wedges.
- As the mass of the rigid body or the masses of the wedges increase, the velocity of recession will decrease.
- Similarly, as the height H from the ground increases, the velocity of recession will also increase due to the increased potential energy of the rigid body.
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