We have:
Total Energy: KE + GPE
KE (Kinetic Energy) =
GPE (Gravitational Potential Energy) =
Data:
m (mass) = 2.0 Kg
v (speed) = 10 m/s
h (height) = 50 m
Use: g (gravity) = 10 m/s²
Formula:
Total Energy: KE + GPE
Solving:
Answer:
The neutron loses all of its kinetic energy to nucleus.
Explanation:
Given:
Mass of neutron is 'm' and mass of nucleus is 'm'.
The type of collision is elastic collision.
In elastic collision, there is no loss in kinetic energy of the system. So, total kinetic energy is conserved. Also, the total momentum of the system is conserved.
Here, the nucleus is still. So, its initial kinetic energy is 0. So, the total initial kinetic energy will be equal to kinetic energy of the neutron only.
Now, final kinetic energy of the system will be equal to the initial kinetic energy.
Now, as the nucleus was at rest initially, so the final kinetic energy of the nucleus will be equal to the initial kinetic energy of the neutron.
Thus, all the kinetic energy of the neutron will be transferred to the nucleus and the neutron will come to rest after collision.
Therefore, the neutron loses all of its kinetic energy to nucleus.
<span>Since there is no friction, conservation of energy gives change in energy is zero
Change in energy = 0
Change in KE + Change in PE = 0
1/2 x m x (vf^2 - vi^2) + m x g x (hf-hi) = 0
1/2 x (vf^2 - vi^2) + g x (hf-hi) = 0
(vf^2 - vi^2) = 2 x g x (hi - hf)
Since it starts from rest vi = 0
Vf = squareroot of (2 x g x (hi - hf))
For h1, no hf
Vf = squareroot of (2 x g x (hi - hf))
Vf = squareroot of (2 x 9.81 x 30)
Vf = squareroot of 588.6
Vf = 24.26
For h2
Vf = squareroot of (2 x 9.81 x (30 – 12))
Vf = squareroot of (9.81 x 36)
Vf = squareroot of 353.16
Vf = 18.79
For h3
Vf = squareroot of (2 x 9.81 x (30 – 20))
Vf = squareroot of (20 x 9.81)
Vf = 18.79</span>
<span>In the desert environment the chemical weathering of rocks is generally reduced because there is a lack of rain, in which most chemical weathering is caused by. No rain, no chemicals, no chemical weathering.</span>
