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

To find the magnitude of the force exerted by the gold nucleus on the alpha particle, we can use Newton's second law of motion, which states: force = mass × acceleration.

First, we need to find the acceleration of the alpha particle during the collision. The acceleration can be calculated using the equation: acceleration = (change in velocity) / time.

Since the alpha particle rebounds with the same speed in the opposite direction, the change in velocity is equal to twice its initial speed: Δv = 2 × (0.81c).

Now, we need to convert the speed of light (c) to meters per second to be consistent with the units in the problem. c = 3 × 10^8 m/s.

Calculating the change in velocity:
Δv = 2 × (0.81c) = 2 × (0.81 × 3 × 10^8 m/s) = 4.86 × 10^8 m/s.

Next, we can substitute the values of mass (m), change in velocity (Δv), and time (t) into the equation for acceleration:
acceleration = (Δv) / t = (4.86 × 10^8 m/s) / (4 × 10^(-7) s).

Now that we have the acceleration, we can calculate the force exerted on the alpha particle. Using Newton's second law, force = mass × acceleration. In this case, the mass is given as m = 6.6 × 10^(-27) kg.

Substituting the values into the equation:
force = mass × acceleration = (6.6 × 10^(-27) kg) × [(4.86 × 10^8 m/s) / (4 × 10^(-7) s)].

Now, we can perform the calculation to find the magnitude of the force exerted by the gold nucleus on the alpha particle:

force = (6.6 × 10^(-27) kg) × [(4.86 × 10^8 m/s) / (4 × 10^(-7) s)].

force = 3.795 × 10^(-19) Newtons (N).

Therefore, the magnitude of the force exerted by the gold nucleus on the alpha particle is approximately 3.795 × 10^(-19) Newtons (N).