# At what speed v will the mass of an object be 10% greater than its rest mass?

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### 2 Answers

The mass-energy equivalence relation `E = mc^2` gives the energy released when an object with mass m is completely converted to energy. Alternately, as the energy of an object is increased there is a corresponding increase in its mass.

If an object with rest mass `M_o` moves at v m/s, the mass of the object is increased to `M = M_o/(sqrt(1 - v^2/c^2))` , where c is the speed of light.

Let the mass of an object be 10% greater than the rest mass at speed v,

`1.1*M_o = M_o/(sqrt(1 - v^2/c^2))`

=> `sqrt(1 - v^2/c^2) = 1/1.1`

=> `1 - v^2/c^2 = 1/1.21`

=> `v^2/c^2 = 21/121`

=> `v^2 = (21/121)*c^2`

=> `v = (sqrt(21)/11)*c`

=> `v ~~ 0.4166*c`

**The mass of an object is 10% greater than the rest mass when its speed is 0.4166*c**

**Sources:**

In the theory of special relativity, if the object accelerates to the speed close to the speed of light, it becomes harder and harder to accelerate. The closer the speed to the speed of light, `3*10^8 ` m/s in the vacuum, the more energy it would take to increase the speed of the object further. It is therefore impossible for an object with non-zero rest mass to reach the speed equal to the speed of light.

Mass is the measure of inertia of objects; that is, it reflects how much force is required to give an object a given acceleration. This means that an object will become more massive as its speed approaches the speed of light. The mass of the object is determined by

`m = gamma*m_0` , where `m_0` is the rest mass and

`gamma = 1/sqrt(1-v^2/c^2)`

If the mass *m *is 10% greater than the rest mass, then `m = m_0 + 0.1m_0 = 1.1m_0` , and

`gamma = 1.1`

`1/sqrt(1 - v^2/c^2) = 1.1`

Squaring this and getting rid of the fraction results in

`1-v^2/c^2 = 0.826`

`v^2/c^2 =0.174`

`v=0.417*c = 1.25*10^8 m/s`

**The mass of an object will be 10% greater than its rest mass when the object has a speed of 125 million meters per second or 0.417 of the speed of light in the vacuum.**

**Sources:**