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

Answer 1

Depends...P = MV; so did you double the mass (M) or the velocity (V)? Or did you do a little bit of both?

KE = 1/2 M V^2; where KE is kinetic energy

So if you doubled the mass, KE doubles (i.e., KE new = 2 (KE old) = 2 [1/2 MV^2]). If you doubled the velocity, KE quadruples (i.e., KE new = 4 (KE old) = 4 [1/2 MV^2]).

The lesson to learn here is that velocity has far more impact on changing kinetic energy than mass does. This results from the square in V^2.

Answer 2

As we know momentum P of a body having mass m and moving at velocity v is defined by equation
P=m*v (1)
On the other hand the kinetic energy is
K = (1/2) * m *v^2 (2)

Solving (1) for v we get
v = P/m
Substituting in (2) we have
K = (1/2) * m * (P/m)^2 ->
K = P^2/(2*m)
So we see that kinetic energy is proportional to momentum squared.
So if P'=2*P then K'=4*K

All this stands in cases of stable mass!
If the mass does change (as for example in a case of a truck moving under heavy rain where m increases, or a track carrying sand which falls on the road through small holes, in which case m decreases) as well as the velocity, then we need more input to answer the question.

Answer 3

KE = 1/2 mv^2 = p^2/2m. So if p is doubled, KE is increased by a factor of 4.