One-Dimensional Elastic Collisions with a Stationary Target

When we apply the conservation laws to this specific case, the general equations for final velocities simplify significantly:

$$v_1' = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) v_1$$ $$v_2' = \left(\frac{2m_1}{m_1 + m_2}\right) v_1$$

In these equations, $v_1$ is the initial velocity of the first mass, while $v_1'$ and $v_2'$ are the final velocities of the first and second masses, respectively.

Analyzing Mass Ratio Scenarios

The outcome of the collision—whether the first mass keeps moving forward, bounces backward, or stops entirely—depends heavily on the ratio of the two masses. We can examine three limiting cases to understand this behavior:

1. Heavy Projectile ($m_1 \gg m_2$)

When a very large mass hits a very small mass (like a bowling ball hitting a marble), the heavy mass continues almost unaffected, while the small mass is propelled forward at twice the initial speed:

• $v_1' \approx v_1$
• $v_2' \approx 2v_1$

2. Heavy Target ($m_1 \ll m_2$)

When a very small mass hits a very large, stationary mass (like a ball hitting a wall), the small mass bounces back with its original speed in the opposite direction, and the large mass remains nearly stationary:

• $v_1' \approx -v_1$
• $v_2' \approx 0$

3. Equal Masses ($m_1 = m_2$)

In the special case where both masses are identical, the moving mass comes to a complete stop, transferring all its momentum and energy to the target:

• $v_1' = 0$
• $v_2' = v_1$

Exam Tip: If you encounter a problem where a moving object hits an identical stationary object and the first object stops dead, you are looking at a one-dimensional elastic collision between equal masses.

Worked Example: The Ballistic Pendulum

To see these principles in action, consider the "Ballistic Pendulum" problem, which combines momentum conservation with energy conservation.

Problem: A bullet of mass $m = 0.010 \text{ kg}$ is fired into a wooden block of mass $M = 2.00 \text{ kg}$ suspended by a cord. The bullet embeds itself in the block, and the system swings upward to a maximum height of $h = 0.120 \text{ m}$. Find the initial speed of the bullet ($v_i$).

Step 1: Find the post-collision speed ($V$) using Energy Conservation. After the collision, the kinetic energy of the block-bullet system is converted into gravitational potential energy as it swings up:

$$\frac{1}{2}(m+M) V^{2} = (m+M) g h$$

Solving for $V$:

$$V = \sqrt{2gh} = \sqrt{2 \cdot 9.80 \cdot 0.120} \approx 1.534 \text{ m/s}$$

Step 2: Find the initial bullet speed ($v_i$) using Momentum Conservation. The collision itself is perfectly inelastic because the bullet sticks to the block. We use the conservation of momentum for the moment of impact:

$$m v_i = (m + M) V$$ $$v_i = \frac{m + M}{m} V = \frac{0.010 + 2.00}{0.010} \cdot 1.534 \approx 308 \text{ m/s}$$

This demonstrates how we can use different conservation laws at different stages of a physical process to solve for unknown variables.