Physics
posted by NADIA on .
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

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