1) In the reference frame of one electron: 0.38c
To find the relative velocity of one electron with respect to the other, we must use the following formula:
where
u is the velocity of one electron
v is the velocity of the second electron
c is the speed of light
In this problem:
u = 0.2c
v = -0.2c (since the second electron is moving towards the first one, so in the opposite direction)
Substituting, we find:
2) In the reference frame of the laboratory: -0.2c and +0.2c
In this case, there is no calculation to be done. In fact, we are already given the speed of the two electrons; we are also told that they travel in opposite direction, so their velocities are
+0.2c
-0.2c
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The impulse required to decrease the speed of the boat is equal to the variation of momentum of the boat:
where
m=225 kg is the mass of the boat
is the variation of velocity of the boat
By substituting the numbers into the first equation, we find the impulse:
and the negative sign means the direction of the impulse is against the direction of motion of the boat.
when the ball hits the floor and bounces back the momentum of the ball changes.
the rate of change of momentum is the force exerted by the floor on it.
the equation for the force exerted is
f = rate of change of momentum
v is the final velocity which is - 3.85 m/s
u is initial velocity - 4.23 m/s
m = 0.622 kg
time is the impact time of the ball in contact with the floor - 0.0266 s
substituting the values
since the ball is going down, we take that as negative and ball going upwards as positive.
f = 189 N
the force exerted from the floor is 189 N
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
