Question

The momentum of a body is increased by 20%.The percentage increase in its kinetic energy is

Answer 1

momentum = mass * velocity

kinetic energy = 1/2 m*v^2

Therefore

You need to find a mathematical connection between the momentum (p) and the kinetic energy (K) and study what happens if you increase K by 20%.

We know that

K=(1/2)*m*v^2

p=m*v

so you can see the relation by computing this two.

Let us massage equation 2 a bit, for convenience:

p=m*v <=> v=p/m

Now we can plug this into our Kinetic energy equation, by replacing v for p/m, which yields:

K=(1/2)*m*(p/m)^2

which, after manipulating a little will shorten to:

K=(1/2m)*p^2

And now we have a relation between K and p. Let us now see what happens when you increase p by 20%.

An increase by 20% in p can be expressed as 1.2*p . So if you replace, in the last equation, p by 1.2*p, you will end up with

K=(1.44/2m)*p^2

So, if you increase the momentum by 20%(1.2), the Kinetic energy will increase by a factor of 1.44 or, in percentage, by 44%.