To calculate the momentum of the stone just before it lands, we need to consider its velocity. Assuming there is no air resistance, we can use the equation for the velocity of an object in free fall:
v = √(2gh)
where: v is the velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height (2 m in this case).
Plugging in the values, we get:
v = √(2 * 9.8 * 2) ≈ √39.2 ≈ 6.26 m/s
The momentum (p) of an object is given by the equation:
p = mv
where: m is the mass of the stone (2 kg in this case), and v is the velocity (6.26 m/s).
Plugging in the values, we get:
p = 2 kg * 6.26 m/s ≈ 12.52 kg·m/s
Therefore, the momentum of the stone just before it lands is approximately 12.52 kg·m/s.
Now, let's calculate the impulse of the stone. Impulse (J) is defined as the change in momentum and can be calculated using the equation:
J = Δp = p_final - p_initial
Since the stone starts from rest (initial velocity = 0 m/s), the initial momentum (p_initial) is 0 kg·m/s. The final momentum (p_final) is the momentum just before it lands, which we calculated to be 12.52 kg·m/s.
Therefore, the impulse of the stone is:
J = p_final - p_initial = 12.52 kg·m/s - 0 kg·m/s = 12.52 kg·m/s