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

MTotal = Mmatter + Mantimatter = 0.001+0.001 = 0.002kg

E= MTotal*c2 = 0.002*(3*10^8)^2 = 1.8*10^14 J

KE of car = E

KEcar = 1.8*10^14 J

By Newton’s 2nd law,

KEcar = Efriction = µk*mg*d

d= KEcar /(Efriction*µk*mg) = (1.8*10^14)/(0.75*1000*10) = 2.4*10^10 m

No of rounds = d/circumference = (2.4*10^10)/(4*10^7) = 600

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.