Consider a ball of mass m1 moving with a velocity u1 collides elastically with a ball of mass m2 at rest.Show that the maximum energy transfer takes place when m1=m2.

To show that the maximum energy transfer occurs when m1 = m2, we can use the conservation of momentum and the principle of conservation of kinetic energy in an elastic collision.

Let's denote the initial velocities of the balls as u1 and u2 for m1 and m2, respectively, and the final velocities as v1 and v2. Since the second ball is at rest initially, we have u2 = 0.

Conservation of momentum:

Before the collision: m1u1 + m2u2 = m1v1 + m2v2

Since m2u2 = 0, the equation simplifies to: m1u1 = m1v1 + m2v2

Conservation of kinetic energy:

Before the collision: (1/2)m1u1^2 + (1/2)m2u2^2 = (1/2)m1v1^2 + (1/2)m2v2^2

Since u2 = 0, the equation simplifies to: (1/2)m1u1^2 = (1/2)m1v1^2 + (1/2)m2v2^2

Now, to find the maximum energy transfer, we need to differentiate the expression (1/2)m1v1^2 + (1/2)m2v2^2 with respect to m1 and equate it to zero.

Differentiating with respect to m1, we have:
dv1/dm1 = -dv2/dm1

Now, substituting this result into the equation m1u1 = m1v1 + m2v2:

m1u1 = m1v1 - m2v1

Re-arranging the terms:

(m1 - m2)v1 = m1u1

Since we want to find the maximum energy transfer, we need to maximize (1/2)m1v1^2 + (1/2)m2v2^2.

Substituting v1 from the equation above, we have:

(1/2)m1((m1 - m2)v1)^2 + (1/2)m2v2^2

Expanding and simplifying:

(1/2)(m1^2 - 2m1m2 + m2^2)v1^2 + (1/2)m2v2^2

Since v1^2 and v2^2 are always positive, to maximize the energy transfer, we need to maximize the term (m1^2 - 2m1m2 + m2^2).

To maximize this term, we can differentiate it with respect to m1 and equate it to zero:

d((m1^2 - 2m1m2 + m2^2))/dm1 = 0

Simplifying this expression, we get: 2m1 - 2m2 = 0

This result implies that m1 = m2.

Therefore, we have shown that the maximum energy transfer occurs when m1 = m2 in an elastic collision.