Question

the momentum of particle in numerical double the kinetic energy what is the velocity of the particle

Answer 1

Answer:

I take the phrase numerically equal to mean that the mismatch of units is to be disregarded.

Set 12mv2=mv.

Divide both sides by m. But before dividing by a variable, we must consider what happens if it is zero. In this case the left side is the wrong expression for energy and the right is wrong for momentum. We know that the velocity in this case is c. If units are chosen so that the momentum of a massless particle is numerically equal to its energy then c is a valid solution, so we include c in the solution set. Now divide both sides by m.

12v2=v.

Divide both sides by v. As before, we must check what happens if v is zero. Zero happens to be a valid solution, so we shall add it to the solution set. Now divide both sides by v.

12v=1, or

v=2.

The solutions are 0, 2 and possibly c, where c is expressed in units that make energy numerically equal to momentum. I leave it as an exercise for the reader to determine whether such units exist and, if so, what the numerical value of c would be.

Answer 2

Answer:

V = 2 m / s

Explanation:

Linear momentum is, P=mv and the kinetic energy is KE= 1/2 mv²

Mv = 1/2 mv²

V²-2v = 0 (Cross multiplied here)

or v(v−2)=0.

So, either v=0 m/s i.e. a trivial case or v=2m/s.

So, v=2m/s is the solution.

Hope it helps and please mark me as brainliest.