The potential energy would be zero. Only kinetic energy is present in this case. To find out what the answer is we do the equation: mv^2/2 soo...
KE =mv^2/2
KE= 1(2^2)/2 which the answer will come up by 2 Joules.
Explanation:
First, we will calculate the electric potential energy of two charges at a distance R as follows.
R = 2r
= 
= 0.2 m
where, R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.
U = 
= 
= 0.081 J
As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.
Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is,
.
Hence, K.E = 
= 
= 0.0405 J
Now, we will calculate the speed of balls as follows.
V = 
= 
= 0.142 m/s
Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.
In a displacement/time graph, the slope of the line is equal to the velocity
Answer:
The initial velocity of the ball is 28.714 m/s
Explanation:
Given;
time of flight of the ball, t = 2.93 s
acceleration due to gravity, g = 9.8 m/s²
initial velocity of the ball, u = ?
The initial velocity of the ball is given by;
v = u + (-g)t
where;
v is the final speed of the ball at the given time, = 0
g is negative because of upward motion
0 = u -gt
u = gt
u = (9.8 x 2.93)
u = 28.714 m/s
Therefore, the initial velocity of the ball is 28.714 m/s