A 75.0 kg person bends his knees and then jumps straight up. After his feet leave the floor, his motion is unaffected by air resistance and his center of mass rises by a maximum of 13.5 cm. Model the floor as completely solid and motionless. With what momentum does the person leave the floor ?
To find the momentum with which the person leaves the floor, we can apply the principle of conservation of momentum. According to this principle, the total momentum before and after the jump remains the same, assuming no external forces are acting.
The initial momentum of the person can be calculated using the concept of gravitational potential energy. When the person jumps, the gravitational potential energy is converted into kinetic energy, which is then shared as momentum.
First, let's calculate the change in potential energy when the person's center of mass rises by 13.5 cm. The change in potential energy (ΔPE) is given by the formula:
ΔPE = m * g * h
where m is the mass of the person (75.0 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the vertical displacement (13.5 cm = 0.135 m).
ΔPE = 75.0 kg * 9.8 m/s² * 0.135 m
ΔPE = 98.35 J
This change in potential energy is equal to the change in kinetic energy and momentum. So, we can write:
ΔPE = ΔKE = Δp
Now, let's solve for the momentum (p):
p = Δp
p = ΔPE
p = 98.35 kg·m²/s
Therefore, the person leaves the floor with a momentum of 98.35 kg·m²/s.