Q.A ball of mass $m$, moving with a speed $2v_0$, collides inelastically ($e > 0$) with an identical ball at rest. Show that

Concept: Momentum Conservation and Coefficient of Restitution — In an inelastic collision, kinetic energy is not conserved, but momentum is. The coefficient of restitution $e$ (with $0 < e < 1$) relates relative speeds before and after impact.

1. Head-on collision Let masses be $m$, initial speeds: $u_1 = 2v_0$, $u_2 = 0$. After collision, let velocities be $v_1$, $v_2$ along the same line. By momentum conservation:

   $m(2v_0) = m v_1 + m v_2 \quad\Rightarrow\quad v_1 + v_2 = 2v_0$

   By definition of $e$:

   $e = \frac{v_2 - v_1}{2v_0 - 0} \quad\Rightarrow\quad v_2 - v_1 = 2e v_0$

   Solving:

   $v_1 = v_0(1 - e), \quad v_2 = v_0(1 + e)$

   Since $e > 0$, $v_1 > 0$ and $v_2 > 0$ — both balls move forward.
2. General (non-head-on) collision Take initial momentum vector $\mathbf{p}_0 = 2mv_0$ along x-axis. After collision, let momenta be $\mathbf{p}_1$ and $\mathbf{p}_2$ with angle $\theta$ between them. Momentum conservation: $\mathbf{p}_0 = \mathbf{p}_1 + \mathbf{p}_2$. Kinetic energy lost: initial KE $= \frac12 m (2v_0)^2 = 2mv_0^2$; final KE $= \frac{p_1^2}{2m} + \frac{p_2^2}{2m}$. Since $e < 1$, final KE < initial KE: …

Why This Problem Matters

This question tests your understanding of two fundamental ideas: the coefficient of restitution and the energy loss in inelastic collisions. The key insight is that $e>0$ means the collision is not perfectly inelastic — the balls separate after impact, but with less relative speed than they approached with. For identical masses, this has surprising consequences for both head-on and glancing collisions.

Let's work through each part.

Part (a): Head-on Collision — Both Balls Move Forward

1. Set up the head-on collision.

Take the moving ball (ball A) with initial velocity $2v_0$ to the right, and the stationary ball (ball B) at rest. After collision, let their velocities be $v_A$ and $v_B$, both along the same line. By momentum conservation:

which simplifies to:

2. Apply the coefficient of restitution.

For a head-on collision, the relative speed of separation equals $e$ times the relative speed of approach. The approach speed is $2v_0 - 0 = 2v_0$, so:

3. Solve for the velocities.

Add (1) and (2): $2v_B = 2v_0 + 2e v_0$, so:

Subtract (2) from (1): $2v_A = 2v_0 - 2e v_0$, so:

4. Interpret the result.

Since $e > 0$, we have $v_A = v_0(1 - e) > 0$ as long as $e < 1$. For a perfectly elastic collision ($e=1$), ball A stops ($v_A=0$) and ball B moves with $2v_0$. For any $e$ strictly between 0 and 1, $v_A$ is positive but less than $v_0$, and $v_B$ is greater than $v_0$. Both velocities are positive — both balls move forward.

Part (b): General Collision — Angle Between Velocities is Less Than $90^\circ$

1. Set up the general collision.

Now the collision is not head-on. Ball A comes in with velocity $2v_0$ along the x-axis. After collision, ball A moves at angle $\theta$ to the x-axis with speed $v_1$, and ball B moves at angle $\phi$ (on the other side of the x-axis) with speed $v_2$.

2. Write momentum conservation in components.

x-component: $m(2v_0) = m v_1 \cos\theta + m v_2 \cos\phi$

y-component: $0 = m v_1 \sin\theta - m v_2 \sin\phi$ (taking $\phi$ positive below the axis)

So:

3. Apply the coefficient of restitution along the line of impact.

For a general collision, the restitution relation applies only along the line joining the centers at the moment of impact (the line of impact). Let this line make some angle with the initial velocity. However, there's a more elegant approach using energy.

4. Use energy loss to bound the scattering angle.

Before collision, kinetic energy is:

After collision:

Since $e>0$ but $e<1$, we have $K_f < K_i$, so:

or:

5. Relate the angle to the velocities. …

Concept: Momentum Conservation + Coefficient of Restitution

Part (a): Head-on collision

Step 1: Set up momentum conservation

A ball of mass $m$ moving at $2v_0$ strikes an identical ball at rest. Let $v_1, v_2$ be their velocities after collision (along the same line):

Step 2: Apply the coefficient of restitution

Step 3: Solve the simultaneous equations

Adding and subtracting the two equations:

Step 4: Interpret

Since $0 < e \le 1$: $v_1 = v_0(1-e) \ge 0$ and $v_2 = v_0(1+e) > 0$. Both balls move forward (in the original direction of motion).

Part (b): General (non head-on) collision

Step 5: Momentum conservation, vector form

Step 6: Kinetic energy is NOT conserved (since $e<1$)

$K_f = \frac{p_1^2+p_2^2}{2m} < K_i = \frac{p_0^2}{2m} = 2mv_0^2 \implies p_1^2+p_2^2 < 4m^2v_0^2$ …