Question

A ball of mass $m_1$ moving with velocity $u_1$ makes a head-on collision with another ball moving with velocity $u_2$. Denoting the coefficient of restitution by the symbol $e$, deduce suitable expression for the loss of mechanical energy in the collision.

Answer 1

$\Delta KE = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (u_1 - u_2)^2 (1 - e^2)$

Answer 2

To deduce the expression for the loss of mechanical energy in a head-on collision, we will use the principles of conservation of linear momentum and the definition of the coefficient of restitution.

Let:

• $m_1$ and $m_2$ be the masses of the two balls.
• $u_1$ and $u_2$ be their initial velocities before collision.
• $v_1$ and $v_2$ be their final velocities after collision.
• $e$ be the coefficient of restitution.

1. Conservation of Linear Momentum:

For a collision, the total linear momentum of the system is conserved:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad (1)$

2. Coefficient of Restitution:

The coefficient of restitution $e$ is defined as the ratio of the relative speed of separation to the relative speed of approach:

$e = \frac{v_2 - v_1}{u_1 - u_2}$

This can be rewritten as:

$v_2 - v_1 = e(u_1 - u_2) \quad (2)$

3. Initial and Final Kinetic Energies:

The initial kinetic energy ($KE_i$) before collision is:

$KE_i = \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2$

The final kinetic energy ($KE_f$) after collision is:

$KE_f = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

4. Loss of Mechanical Energy ($\Delta KE$):

The loss of mechanical energy (kinetic energy) is the difference between the initial and final kinetic energies:

$\Delta KE = KE_i - KE_f = \left( \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 \right) - \left( \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \right)$

$\Delta KE = \frac{1}{2}m_1(u_1^2 - v_1^2) + \frac{1}{2}m_2(u_2^2 - v_2^2)$

$\Delta KE = \frac{1}{2}m_1(u_1 - v_1)(u_1 + v_1) + \frac{1}{2}m_2(u_2 - v_2)(u_2 + v_2) \quad (3)$

From equation (1), rearrange to get:

$m_1(u_1 - v_1) = m_2(v_2 - u_2) \quad (4)$

Let $X = m_1(u_1 - v_1) = m_2(v_2 - u_2)$.

So, $u_1 - v_1 = X/m_1$ and $u_2 - v_2 = -X/m_2$.

Substitute these into equation (3):

$\Delta KE = \frac{1}{2} \left( \frac{X}{m_1} \right) (u_1 + v_1) + \frac{1}{2} \left( \frac{-X}{m_2} \right) (u_2 + v_2)$

$\Delta KE = \frac{1}{2}X \left[ \frac{u_1 + v_1}{m_1} - \frac{u_2 + v_2}{m_2} \right]$

This approach can be complicated. Let's use an alternative manipulation from equation (3):

$\Delta KE = \frac{1}{2}m_1(u_1 - v_1)(u_1 + v_1) + \frac{1}{2}m_2(u_2 - v_2)(u_2 + v_2)$

Using $m_1(u_1 - v_1) = m_2(v_2 - u_2)$, we can write $m_2(u_2 - v_2) = -m_1(u_1 - v_1)$.

$\Delta KE = \frac{1}{2}m_1(u_1 - v_1)(u_1 + v_1) - \frac{1}{2}m_1(u_1 - v_1)(u_2 + v_2)$

$\Delta KE = \frac{1}{2}m_1(u_1 - v_1) [(u_1 + v_1) - (u_2 + v_2)]$

$\Delta KE = \frac{1}{2}m_1(u_1 - v_1) [(u_1 - u_2) - (v_2 - v_1)]$

Now, let $u_{rel} = u_1 - u_2$ (relative speed of approach) and $v_{rel} = v_2 - v_1$ (relative speed of separation).

From equation (2), $v_{rel} = e u_{rel}$.

So, $\Delta KE = \frac{1}{2}m_1(u_1 - v_1) [u_{rel} - e u_{rel}]$

$\Delta KE = \frac{1}{2}m_1(u_1 - v_1) u_{rel} (1 - e) \quad (5)$

Now we need to express $(u_1 - v_1)$ in terms of initial velocities and masses.

From equation (1): $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

From equation (2): $v_2 = v_1 + e(u_1 - u_2)$

Substitute $v_2$ into equation (1):

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 (v_1 + e(u_1 - u_2))$

$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v_1 + m_2 e(u_1 - u_2)$

$(m_1 + m_2) v_1 = m_1 u_1 + m_2 u_2 - m_2 e(u_1 - u_2)$

$v_1 = \frac{m_1 u_1 + m_2 u_2 - m_2 e(u_1 - u_2)}{m_1 + m_2}$

Now, find $(u_1 - v_1)$:

$u_1 - v_1 = u_1 - \frac{m_1 u_1 + m_2 u_2 - m_2 e(u_1 - u_2)}{m_1 + m_2}$

$u_1 - v_1 = \frac{u_1(m_1 + m_2) - (m_1 u_1 + m_2 u_2 - m_2 e(u_1 - u_2))}{m_1 + m_2}$

$u_1 - v_1 = \frac{m_1 u_1 + m_2 u_1 - m_1 u_1 - m_2 u_2 + m_2 e(u_1 - u_2)}{m_1 + m_2}$

$u_1 - v_1 = \frac{m_2 u_1 - m_2 u_2 + m_2 e(u_1 - u_2)}{m_1 + m_2}$

$u_1 - v_1 = \frac{m_2(u_1 - u_2) + m_2 e(u_1 - u_2)}{m_1 + m_2}$

$u_1 - v_1 = \frac{m_2(u_1 - u_2)(1 + e)}{m_1 + m_2}$

$u_1 - v_1 = \frac{m_2 u_{rel} (1 + e)}{m_1 + m_2}$

Substitute this expression for $(u_1 - v_1)$ back into equation (5):

$\Delta KE = \frac{1}{2} \left( \frac{m_2 u_{rel} (1 + e)}{m_1 + m_2} \right) u_{rel} (1 - e)$

$\Delta KE = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (1 + e)(1 - e) u_{rel}^2$

$\Delta KE = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (1 - e^2) (u_1 - u_2)^2$

This is the suitable expression for the loss of mechanical energy in the collision.

The second part of the question "of the same mass at rest on a frictionless horizontal." seems incomplete or a remnant from another question and is not considered for the general deduction.

The final answer is $\boxed{\frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (u_1 - u_2)^2 (1 - e^2)}$.

Explanation of the solution:

1. Apply the principle of conservation of linear momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.
2. Use the definition of the coefficient of restitution: $e = \frac{v_2 - v_1}{u_1 - u_2}$.
3. The loss of mechanical energy is the difference between initial and final kinetic energies: $\Delta KE = (\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2) - (\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2)$.
4. Algebraically manipulate these equations to express $\Delta KE$ in terms of initial parameters ($m_1, m_2, u_1, u_2$) and the coefficient of restitution ($e$). The derivation involves expressing final velocities in terms of initial velocities and $e$, then substituting into the energy loss equation, or by a more elegant method of factoring and substituting relative velocities. The result is the standard formula for kinetic energy loss in a collision.