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

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 that force (F) is equal to the change in momentum (Δp) divided by the change in time (Δt).

The momentum (p) of an object is given by the product of its mass (m) and velocity (v): p = m * v.

In this case, the alpha particle collides with the gold nucleus and rebounds with the same speed in the opposite direction. Since the speed remains the same, we know that the magnitude of the change in momentum is equal to two times the initial momentum.

Therefore, the magnitude of the change in momentum (Δp) is given by: Δp = 2 * m * v.

Now, we can calculate the magnitude of the force (F) using the formula F = Δp / Δt.
Plugging in the values, we get:

Δp = 2 * (6.6 x 10^-27 kg) * (.81c)
Δp = 2 * (6.6 x 10^-27 kg) * (.81 * 3 x 10^8 m/s) (converting c to m/s)
Δp = 2 * (6.6 x 10^-27 kg) * (2.43 x 10^8 m/s)
Δp ≈ 3.1344 x 10^-18 kg m/s

Now, we can substitute this value of Δp along with the given time (Δt) into the equation for force:

F = Δp / Δt
F = (3.1344 x 10^-18 kg m/s) / (4 x 10^-7 s)
F ≈ 7.836 x 10^-12 N

Therefore, the magnitude of the force exerted by the gold nucleus on the alpha particle is approximately 7.836 x 10^-12 N.