# HELP!! been stuck on this question for so long it pains me

An alpha particle 6.7×10 ^−27kg is fired at the nucleus in a gold atom with a speed of 3.5×10^6ms−1. It bounces off at the same speed in the opposite direction. If the collision takes 10^−19s, what is the magnitude of the average force? Give your answer to 2 significant figures.

i got 0.2345 but for some reason its wrong, this was the help given "The alpha particle does not come to rest, so you need to use its final velocity to work out the change in momentum." Isaac physics needs a better way of helping students hahaha
Heya, personally I have no idea aha. But I gave this question a quick google and this is what I found: F delta(t) = m delta(v)
F ( 1E-19 s) = 6.7E-27 kg (3.5E6 m/s - (-3.5E-6 m/s)) = 4.69E-27
F = 4.69E-13 N or 4.7E-13 N to 2 sig figs

Hope that maybe helped?! :0
Original post by Swroop12
An alpha particle 6.7×10 ^−27kg is fired at the nucleus in a gold atom with a speed of 3.5×10^6ms−1. It bounces off at the same speed in the opposite direction. If the collision takes 10^−19s, what is the magnitude of the average force? Give your answer to 2 significant figures.

i got 0.2345 but for some reason its wrong, this was the help given "The alpha particle does not come to rest, so you need to use its final velocity to work out the change in momentum." Isaac physics needs a better way of helping students hahaha

final velocity is the same so its just the same momentum again. Just need to add the two momentums to get the magnitude.
force is rate of change of momentum, so divide the change of momentum by the time taken.