Final answer:
The momentum of a stone with a mass of 200 g dropped from a height of 2 m just before it hits the floor is approximately 0.886 kg·m/s, calculated by first determining its velocity through conservation of energy and then applying the momentum formula.
Step-by-step explanation:
Calculating Momentum Before Impact
The question pertains to finding the momentum of a stone with a mass of 200 g (0.2 kg) just before it strikes the floor after being dropped from a height of 2 m. To solve for momentum, we apply the principles of physics, particularly the equation for gravitational potential energy (PE) at the height it was dropped and kinetic energy (KE) just before impact.
First, we calculate the potential energy at the height of 2 m using the equation PE = mgh, where m is mass in kg, g is the acceleration due to gravity (9.8 m/s²), and h is the height in meters. The potential energy is entirely converted to kinetic energy just before the stone hits the ground as per the principle of conservation of energy.
To find the velocity (v) just before impact, we use the equation for kinetic energy, KE = 0.5 * m * v². Since potential energy at the start is equal to the kinetic energy just before impact, we can set them equal and solve for v.
Next, momentum (p) is given by the equation p = mv. By substituting the calculated velocity into this equation, we determine the momentum of the stone just before it hits the floor.
Let's perform the calculations:
- Calculate the velocity using potential energy:
PE = KE
mgh = 0.5 * m * v²
0.2 kg * 9.8 m/s² * 2 m = 0.5 * 0.2 kg * v²
v² = (0.2 kg * 9.8 m/s² * 2 m) / (0.5 * 0.2 kg)
v = sqrt(19.6 m²/s²)
v ≈ 4.43 m/s - Calculate the momentum:
p = mv
p = 0.2 kg * 4.43 m/s
p ≈ 0.886 kg·m/s
Therefore, the momentum of the stone just before it hits the floor is approximately 0.886 kg·m/s.