# If the momentum of an object is doubled, by what factor will its kinetic energy change?

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The momentum of an object with mass m and traveling at a velocity v is given by M = m*v and the kinetic energy is given by `KE = (1/2)*mv^2`

The momentum of the object can be made double by either increasing its mass to twice the initial mass with the velocity remaining the same or by increasing the velocity to twice the initial value.

Assuming the mass of the object is not altered and only the velocity is increased to twice the initial value, the kinetic energy of the object is increased to `(1/2)*m*(2*v)^2 = 4*(1/2)*m*v^2` , or it changes to four times the initial value.

**If the momentum of an object is doubled by increasing its velocity, its kinetic energy becomes 4 times the initial value.**

KE = (P²)/(2m)

KE = Kinetic energy

P = Momentium

m = mass

If the momentum of the object is doubled then...

KE is directly propotional to Square of the momentium

So...

(KE1)/(P²1) = (KE2)/(P²2)

KE1 = E

KE2 = x(?)

P²1 = P

P²2 = 4P

Therefore

E/P = x/4P

x = 4E

Thus KE increases by 4 times when momentum is doubled.

The relation between kinetic energy is proportional to the square of velocity. Momentum is directly proportional to velocity. If the momentum of an object is doubled, but its mass does not increase (so velocity remains well below the speed of light), then its velocity is doubled. If the velocity is doubled then the kinetic energy increases by the square of 2, or four time.

The relation between kinetic energy and linear momentum is given as:

K.E.=P*P/2M

Where p is linear momentum and

M is mass of the body.

in first case (K.E)1=P*P/2M.....................(1)

Now linear momentum gets doubled,keeping mass of body unchanged

in second case (K.E.)2=2P*2P/2M

or (K.E.)2=4P*P/2M..............(2)

Dividing eq.(2) by eq. (1)

(K.E.)2

_________=4

(K.E)1

Hence kinetic energy of the body will be **4 times** to its eirlier value.