An electron is projected, with an initial speed of v= 2.18e+05 m/sec, directly towards a proton that is essentially at rest. If the electron is initially a great distance from the proton, at what distance from the proton is its speed instantaneously equal to twice its initial value? Express your answer in meters.
See my answer to a similar question below under related questions.
The initial speed is different but the same equation can be used.
Thanks
To find the distance from the proton where the speed of the electron is twice its initial value, we can use conservation of energy.
The initial kinetic energy of the electron is given by:
KE_initial = (1/2) * m * v_initial^2
Where m is the mass of the electron and v_initial is its initial speed.
The final kinetic energy of the electron when its speed is twice its initial value is given by:
KE_final = (1/2) * m * (2 * v_initial)^2
We can equate these two expressions for kinetic energy to find the distance from the proton:
(1/2) * m * v_initial^2 = (1/2) * m * (2 * v_initial)^2
Canceling out the factors and simplifying:
v_initial^2 = 4 * v_initial^2
Dividing both sides by v_initial^2:
1 = 4
This equation is not possible; it implies that 1 is equal to 4, which is not true. Therefore, there is no distance from the proton where the speed of the electron is instantaneously equal to twice its initial value.