an alpha particle of mass m= 6.6 x 10^-27 kg has a speed of .81c, where the speed of light is c= 3 x 10^8 m/s. The alpha particle collides with a gold nucleus and rebounds with the same speed in the opposite direction. If the collision lasted for t= 4 x 10^-7s, what is the magnitude of the force that the gold nucleus exerted on the alpha particle?

Answer in units of N

2 answers

force=mv=2*restmass*dialation factor*time

the factor is 1/(sqrt(1-(v/c)^2)

check my thinking.
oops..

force*time=2*restmass*factor*velocity