Answer:
When the body reaches the floor its velocity V1 is obtained
from the conservation of energy:
mgH = (1/2)mV1^2 which gives V1 = (2gH)^1/2 where H =8m.
After the bounce its velocity V2 is obtained in the same way.
mgh = (1/2)mV2^2 . So V2 = (2gh)^1/2 where h = 2m.
Obtain the change of momentum
Dp =mDV = m(V2-V1) = - m[(2gH)^1/2 - (2gh)^1/2] =- 0.63kg(m/s)
From the law FDt = mDV obtain the force exerted on the body
F = Dp/Dt = - m[(2gH)^1/2 - (mgh)^1/2]/Dt , with Dt = 0.1s
F = -6.3N .
Answer:
[tex] \sf{p \approx 1.253 \, \text{kg} \cdot \text{m/s}} [/tex]
Explanation:
1. Velocity when the ball hits the ground (v1):
We'll use the conservation of energy principle where the potential energy at the height of 8m will be converted to kinetic energy just before impact. The formula for potential energy (PE) is [tex] \rm{PE = mg} [/tex] , and the formula for kinetic energy (KE) is [tex] \rm{KE = \dfrac{1}{2}mv^2} [/tex] . Equating PE and KE and solving for v gives us the velocity just before impact.
[tex] \rm{v = \sqrt{2gh}} [/tex]
[tex] \rm{v = \sqrt{2 \times 9.8 \times 8}} [/tex]
[tex] \rm{v = \sqrt{156.8}} [/tex]
[tex] \boxed{\rm{v \approx 12.53 \, \text{m/s}}} [/tex]
2. Velocity with which the ball leaves the ground (v2):
Similarly, we can calculate the velocity with which the ball leaves the ground after bouncing to a height of 2m.
[tex] \rm{v' = \sqrt{2gh'}} [/tex]
[tex] \rm{v' = \sqrt{2 \times 9.8 \times 2}} [/tex]
[tex] \rm{v' = \sqrt{39.2}} [/tex]
[tex] \rm{v' \approx 6.26 \, \text{m/s}} [/tex]
3. Momentum when the ball hits the ground (p):
Momentum is calculated using the formula p = mv .
[tex] \rm{p = m \times v} [/tex]
[tex] \rm{p = 0.1 \times 12.53} [/tex]
[tex] \boxed{\bf{p \approx 1.253 \, \text{kg} \cdot \text{m/s}}} [/tex]
These calculations assume that air resistance is negligible and the collision is perfectly elastic. In real-world scenarios, some energy would be lost due to air resistance and inelastic deformation of the ball and floor during the collision.
[tex] \rule{180pt}{3pt} [/tex]
