Answer:
A) 
denotes the resultant velocity of cart B post-collision.
B) 
C) 
D) 
E) 
F) Yes, kinetic energy remains conserved in this situation because both colliding bodies have identical mass.
G) Yes, momentum is conserved in every elastic collision.
Explanation:
Given:
- mass of car A,
- mass of car B,
- initial velocity of car A,
- final velocity of car A,
A)
The question mentions the cars experience an elastic collision:
By applying momentum conservation principles:
denotes the resulting velocity of cart B after collision.
B)
Initial kinetic energy of cart A:
C)
Initial kinetic energy of cart A:
D)
The final kinetic energy of cart A:
E)
The final kinetic energy of cart B:
F)
Yes, kinetic energy is conserved in this case due to both masses being identical in the collision.
G)
Indeed, momentum is consistently conserved in elastic collisions.
The answer is 10pi. I believe this will be helpful.
Result: -50.005 kJ
Details:
Provided Data
mass of the system = 10 kg
work done = 0.147 kJ/kg
Elevation change 
initial speed 
Final Speed 
Specific internal Energy 
according to the first Law of thermodynamics
where KE represents kinetic energy
PE indicates potential energy
U denotes internal Energy
Q = 1.47 + 3.375 - 4.850 - 50
Q = -50.005 kJ
This is somewhat misleading, and I encountered the same question in my homework. An electric field strength of 1*10^5 N/C is provided, along with a drag force of 7.25*10^-11 N, and the critical detail is that it maintains a constant velocity, indicating that the particle is in equilibrium and not accelerating.
<span>To solve, utilize F=(K*Q1*Q2)/r^2 </span>
<span>You'll want to equate F with the drag force, where the electric field strength translates to (K*Q2)/r^2; substituting the values results in </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
