The centers of two helium nuclei are held 2.5*10^-12 m apart, and are then released. Calculate the speed of each necleus when they are 0.75 m apart.
See Freda question below.
To calculate the speed of each helium nucleus when they are 0.75 m apart, we can make use of the principles of conservation of energy and apply the law of conservation of mechanical energy.
The total mechanical energy of the system (two helium nuclei) remains constant throughout the motion. At the initial position, all of the energy is in the form of potential energy (PE), given by the formula:
PE = k * q1 * q2 / r,
where k is the electrostatic constant (8.99 x 10^9 N m^2 / C^2), q1 and q2 are the charges on the nuclei (which are both equal and will cancel each other out), and r is the initial separation between the nuclei (2.5 x 10^-12 m).
At the final position, all of the potential energy will be converted into kinetic energy (KE), which is given by the formula:
KE = (1/2) * m * v^2,
where m is the mass of each helium nucleus (which is the same as the mass of a helium atom, roughly 6.64 x 10^-27 kg), and v is the velocity of each nucleus.
To find the speed, we need to equate the initial potential energy to the final kinetic energy:
PE = KE,
k * q1 * q2 / r = (1/2) * m * v^2.
Now, we rearrange the equation to solve for v:
v = √[2 * k * q1 * q2 / (m * r)].
Substituting the given values into the equation, we obtain:
v = √[2 * (8.99 x 10^9 N m^2 / C^2) * (1.6 x 10^-19 C)^2 / (6.64 x 10^-27 kg) * (0.75 m)].
Calculating this expression will give us the speed of each helium nucleus when they are 0.75 m apart.