Two billiard balls of equal mass are traveling straight toward each other with the same speed. they meet head-on in an elastic collision. what is the total momentum of the system containing the two balls before the collision? what about after the collision?

0 kg m/s before and after collision

Explanation:

Let m, v be the mass and speed of the 2 balls, respectively, before the collision. Since they have the same mass and same speed but in opposite direction, the total momentum of the system would be:

P = mv - mv = 0 kg m/s

As the collision is elastic. The total momentum after the collision is the same as the total momentum before the collision, which is 0.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge