Two particles of masses m1 and m2 have equal linear momenta. The ratio of their kinetic energy

Let's assume that the linear momenta of the two particles are equal, denoted as p1 and p2.

The linear momentum of a particle is given by the product of its mass and velocity, so we can write:

p1 = m1 * v1 p2 = m2 * v2

Since the linear momenta are equal, we have:

p1 = p2

Substituting the expressions for p1 and p2, we get:

m1 * v1 = m2 * v2

Now, let's consider the kinetic energy of each particle. The kinetic energy of a particle is given by the formula:

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

For the first particle, we have:

K.E.1 = (1/2) * m1 * v1^2

For the second particle, we have:

K.E.2 = (1/2) * m2 * v2^2

To find the ratio of their kinetic energies, we can divide K.E.1 by K.E.2:

(K.E.1) / (K.E.2) = [(1/2) * m1 * v1^2] / [(1/2) * m2 * v2^2]

Simplifying the expression, we get:

(K.E.1) / (K.E.2) = (m1 * v1^2) / (m2 * v2^2)

Since we know that m1 * v1 = m2 * v2, we can substitute this into the equation:

(K.E.1) / (K.E.2) = (m1 * v1^2) / (m2 * v2^2) = (m1 * (m1 * v1)^2) / (m2 * (m2 * v2)^2)

Simplifying further, we get:

(K.E.1) / (K.E.2) = (m1^3 * v1^2) / (m2^3 * v2^2)

Therefore, the ratio of their kinetic energies is (m1^3 * v1^2) / (m2^3 * v2^2).