When equal amounts of matter and antimatter collide, energy is released accordin

Question

When equal amounts of matter and antimatter collide, energy is released according to Einstein's famous E = mc^2 equation, where m is the sum of the matter antimatter masses and c = 3 times 10^8 m/s is the speed of light. Suppose 1 g of antimatter collides with 1 g of matter and all of that mass energy is delivered to a 1000 kg car as kinetic energy. The startled driver immediately slams on the breaks and the car eventually skids to a halt, with a coefficient of friction between tires and road of mu_k = 3/4. How far does the car travel before coming to rest? (Use g = 10 m/s^2 as needed) The circumference of Earth is 4 times 10^7 m. How many times around Earth does the driver travel?

Explanation / Answer

let m = 1 + 1 = 2 grams = 0.002 kg

M = 1000 kg

let u is the speed of the car when the matter is converted into energy

apply, (1/2)*M*u^2 = m*c^2

u = sqrt(2*m*c^2/M)

= sqrt(2*0.002*(3*10^8)^2/1000)

= 6*10^5 m/s

acceleration of the car when breaks are applied, a = -g*mue_k

= -10*(3/4)

= -7.5 m/s^2

let d is the distance travelled before coming to rest.

use, v^2 - u^2 = 2*a*d

==> d = (v^2 - u^2)/(2*a)

= (0^2 - (6*10^5)^2/(2*(-7.5))

= 2.4*10^10 m <<<<<<<<<<<---------------Answer

no of revolutions made around the earth, N = d/circumference of the earth

= 2.4*10^10/(4*10^7)

= 600   <<<<<<<<<<<---------------Answer

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