1. A base on Planet X fires a missile toward an oncoming space fighter. The miss

Question

1. A base on Planet X fires a missile toward an oncoming space fighter. The missile's speed according to the base is 0.80 c . The space fighter measures the missile's speed as 0.95 c

How fast is the space fighter traveling relative to Planet X?

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2. What are (a) the kinetic energy, (b) the rest energy, and (c) the total energy of a 1.60 g particle with a speed of 0.700 c ?

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3. A quarter-pound hamburger with all the fixings has a mass 210 g . The food energy of the hamburger is 2.20 MJ .

What is the energy equivalent of the mass of the hamburger?

By what factor does the energy equivalent exceed the food energy?

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Please answer these for me :)

Explanation / Answer

part(1)

speed of missile relative to the base planet x Vmx = 0.80c

speed of missile with respect to space fighter    Vmf = 0.95c

according to the given condition missile and space fighter both are traveling in opposite direction ,

so the speed of space fighter relative to the planet x(base)

Vmx = 0.8c (velocity of m with respect to x)

Vmf = 0.95c ( velocity of m with respect to fighter)

we know that Vmf = Vmx - Vfx ( as x is in rest)

so 0.95c = 0.80c - Vfx

Vfx = - 0.15c

answer ( there is -ve sign )

part(2)

given that

m = 1.60 g = 0.0016

v = 0.700 c = 0.700*3*10^8 m

(a)

kinetic energy = 1/2*m*v^2

KE = 1/2* 0.0016*(0.700*3*10^8)^2

KE = 0.003528*10^16

KE = 3.52*10^13 J

(b)

Rest energy = m*c^2

RE = 0.0016*(3*10^8)^2

RE = 0.0144*10^16

RE = 14.4*10^13 J

(c)

total energy = KE + RE

total energy = 3.52*10^13 + 14.4*10^13

total energy = 17.92*10^13 J

part(3)

given that

m = 210 g = 0.210 kg

food energy = 2.20 *10^6 J

energy equivalent of the mass of the hamburger

E = m*c^2

E = 0.210*(3*10^8)^2

E = 1.89*10^16 J

the energy equivalent exceed the food energy by the factor

F = E /food energy

F = 1.89*10^16 / 2.20*10^6

F = 0.85*10^10

F = 8.5*10^9

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