Collision Setup: Elastic Head-on Impact

This problem describes a specific type of collision: a head-on elastic collision. Let's break down the initial conditions:

• Body A: Mass = $m$, Initial speed = $u$.
• Body B: Mass = $3m$, Initial speed = $0$ (at rest).

In an elastic collision, two fundamental conservation laws apply:

1. Conservation of Linear Momentum: The total momentum of the system before the collision equals the total momentum after the collision.
2. Conservation of Kinetic Energy: The total kinetic energy of the system before the collision equals the total kinetic energy after the collision.

Our goal is to find the fraction of energy lost specifically by Body A after this collision.

Conservation Laws Applied to the Collision

Let $v_A$ be the final velocity of Body A and $v_B$ be the final velocity of Body B after the collision.

1. Conservation of Momentum:

Initial momentum = Final momentum

$m \cdot u + (3m) \cdot 0 = m \cdot v_A + (3m) \cdot v_B$

Simplifying this equation by dividing by $m$ (since $m \neq 0$):

$u = v_A + 3v_B \quad \quad (1)$

2. Relative Velocity in Elastic Collisions:

For a head-on elastic collision, the relative speed of approach is equal to the relative speed of separation.

$u_{approach} = u_A - u_B = u - 0 = u$ $u_{separation} = v_B - v_A$

Equating them:

$u = v_B - v_A \quad \quad (2)$

Calculating Final Velocities Post-Collision

Now we have a system of two linear equations (1) and (2) with two unknowns ($v_A$ and $v_B$). Let's solve them:

• From equation (2), we can express $v_B$ as: $v_B = u + v_A$.
• Substitute this expression for $v_B$ into equation (1):
$u = v_A + 3(u + v_A)$ $u = v_A + 3u + 3v_A$ $u = 4v_A + 3u$
• Rearrange to solve for $v_A$:
$4v_A = u - 3u$ $4v_A = -2u$ $v_A = \frac{-2u}{4} = -\frac{u}{2}$
• Now, substitute the value of $v_A$ back into the expression for $v_B$:
$v_B = u + v_A = u + \left(-\frac{u}{2}\right) = \frac{u}{2}$

So, after the collision, Body A moves with velocity $v_A = -u/2$ (meaning it reverses direction) and Body B moves with velocity $v_B = u/2$.

Body A Energy Loss Calculation

We need to find the energy lost by Body A. First, let's calculate its initial and final kinetic energy ($KE$).

Initial Kinetic Energy of Body A ($KE_{A, initial}$):

$KE_{A, initial} = \frac{1}{2} m u^2$

Final Kinetic Energy of Body A ($KE_{A, final}$):

Using the final velocity $v_A = -u/2$:

$KE_{A, final} = \frac{1}{2} m (v_A)^2 = \frac{1}{2} m \left(-\frac{u}{2}\right)^2$ $KE_{A, final} = \frac{1}{2} m \left(\frac{u^2}{4}\right) = \frac{1}{8} m u^2$

Energy Lost by Body A:

Energy lost is the difference between initial and final kinetic energy:

$\Delta KE_A = KE_{A, initial} - KE_{A, final}$ $\Delta KE_A = \frac{1}{2} m u^2 - \frac{1}{8} m u^2$

To subtract, find a common denominator (which is 8):

$\Delta KE_A = \left(\frac{4}{8} - \frac{1}{8}\right) m u^2$ $\Delta KE_A = \frac{3}{8} m u^2$

Fraction of Energy Lost by Body A

The question asks for the fraction of energy lost by Body A. This is calculated by dividing the energy lost by Body A by its initial kinetic energy.

$ \text{Fraction Lost} = \frac{\Delta KE_A}{KE_{A, initial}} $ $ \text{Fraction Lost} = \frac{\frac{3}{8} m u^2}{\frac{1}{2} m u^2} $

The terms $m u^2$ cancel out:

$ \text{Fraction Lost} = \frac{3/8}{1/2} $

Dividing by a fraction is the same as multiplying by its reciprocal:

$ \text{Fraction Lost} = \frac{3}{8} \times \frac{2}{1} = \frac{6}{8} $

Simplifying the fraction:

$ \text{Fraction Lost} = \frac{3}{4} $

Therefore, the fraction of kinetic energy lost by colliding body A after the head-on elastic collision is $3/4$. This matches option 2.