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# If the kinetic energy of a body changes by 20% then its momentum would change by – (A) 20% (B) 24% (C) 40% (D) 44%

Arun
25763 Points
one year ago
Momentum is proportional to speed, so when momentum is increased by 20%, or multiplied by a factor 1.2 , then then speed must have increased by 20%. Kinetic energy is proportional to the square of speed, so with speed increasing by a factor 1.2 , Kinetic Energy wil increase by the square of the factor 1.2, which is 1.44 or 44%
Vikas TU
14149 Points
one year ago

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%.

Kind regards,