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

E=mc2

M= 1g+1g =2g=2*10-3kg

E= (2* 10-3 )* (3*108)2

E=18*1013J

This energy is used as K.E. to a car of 1000kg.

i.e. ½ mv2=18*1013J

Hence v= 6*105m/s

Since µk = F/N

i.e. ma/mg = ¾

    a/g = ¾

( v/x )/g = ¾

X= 4v/3g

X= 4 * (6*105)/3 * 10

X= 8 *104m

X=80 km

Hence car continues to travel for 80km.

Since circumference is 4 * 10 7 m the the driver can travel (4 * 10 7 )/(8 *104) = 500 time around the earth.

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