To calculate the momentum of the stone just before it lands, we need to determine its velocity at that moment. We can use the principles of conservation of energy to find the velocity.
The potential energy of the stone at the initial height of 2m is given by: Potential Energy = m * g * h
where: m = mass of the stone = 2 kg g = acceleration due to gravity = 9.8 m/s^2 h = height = 2 m
Potential Energy = 2 kg * 9.8 m/s^2 * 2 m = 39.2 J
Since potential energy is converted to kinetic energy as the stone falls, we can equate the two:
Potential Energy = Kinetic Energy m * g * h = (1/2) * m * v^2
where v is the velocity of the stone just before it lands.
Simplifying the equation:
2 kg * 9.8 m/s^2 * 2 m = (1/2) * 2 kg * v^2 78.4 J = v^2
Taking the square root of both sides:
v = √(78.4 J) ≈ 8.85 m/s
The momentum of an object is given by the product of its mass and velocity:
Momentum = mass * velocity Momentum = 2 kg * 8.85 m/s = 17.7 kg·m/s
Therefore, the momentum of the stone just before it lands is 17.7 kg·m/s.
To calculate the impulse of the stone, we can use the relationship between impulse and momentum:
Impulse = Change in momentum
Since the stone starts from rest, its initial momentum is zero. The change in momentum is then equal to the final momentum:
Impulse = Final Momentum = 17.7 kg·m/s
Therefore, the impulse of the stone is 17.7 kg·m/s.