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Solving Collision Problems

Solving collision problems involves applying the principle of momentum conservation—and sometimes kinetic energy conservation—depending on the type of collision.

General Strategy

Example 1: Perfectly Inelastic Collision

Problem: A 1.5 kg cart moving at 2 m/s collides with a 0.5 kg cart at rest. They stick together. What is their final velocity?

Step 1: Identify the type

Perfectly inelastic – objects stick together, momentum conserved, KE not.

Step 2: Apply conservation of momentum
\[ v_f = \frac{m_1v_{1,i} + m_2v_{2,i}}{m_1 + m_2} \]
Step 3: Plug in values
\[ v_f = \frac{1.5(2) + 0.5(0)}{1.5 + 0.5} = \frac{3}{2} = 1.5 \, \text{m/s} \]

Answer: They move together at 1.5 m/s.

Example 2: Elastic Collision (Equal Masses)

Problem: A 0.6 kg ball moving at 3 m/s hits a stationary 0.6 kg ball. What are the final velocities?

Step 1: Elastic collision with equal masses

In 1D, identical masses simply swap velocities in an elastic collision.

Step 2: Final velocities
\[ v_{1,f} = 0 \quad v_{2,f} = 3 \, \text{m/s} \]

Answer: First ball stops; second ball moves at 3 m/s.

Example 3: Elastic Collision (Unequal Masses)

Problem: A 2 kg cart moving at 4 m/s hits a 1 kg stationary cart. What are the final velocities if the collision is elastic?

Step 1: Use momentum and kinetic energy conservation
\[ v_{1,f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1,i} + \frac{2m_2}{m_1 + m_2} v_{2,i} \] \[ v_{2,f} = \frac{2m_1}{m_1 + m_2} v_{1,i} + \frac{m_2 - m_1}{m_1 + m_2} v_{2,i} \]
Step 2: Plug in values
\[ v_{1,f} = \frac{2 - 1}{3} \cdot 4 = \frac{1}{3} \cdot 4 = 1.33 \, \text{m/s} \] \[ v_{2,f} = \frac{2 \cdot 2}{3} \cdot 4 = \frac{4}{3} \cdot 4 = 5.33 \, \text{m/s} \]

Answer: Cart 1: 1.33 m/s, Cart 2: 5.33 m/s

Example 4: Inelastic but Not Perfectly

Problem: A 3 kg object moving at 6 m/s hits a 2 kg object moving at 2 m/s. After colliding, they bounce off with different velocities. Final velocity of object 2 is 4 m/s. What is object 1’s final velocity?

Step 1: Use momentum conservation
\[ m_1v_{1,i} + m_2v_{2,i} = m_1v_{1,f} + m_2v_{2,f} \]
Step 2: Plug in knowns
\[ 3(6) + 2(2) = 3v_{1,f} + 2(4) \Rightarrow 18 + 4 = 3v_{1,f} + 8 \Rightarrow 22 - 8 = 3v_{1,f} \Rightarrow v_{1,f} = \frac{14}{3} \approx 4.67 \, \text{m/s} \]

Answer: Object 1’s final velocity is approximately 4.67 m/s

Example 5: Kinetic Energy Loss

Problem: A 2 kg object moving at 3 m/s collides with a 2 kg object at rest. They stick together. How much kinetic energy is lost?

Step 1: Initial Kinetic Energy
\[ KE_i = \frac{1}{2} m v^2 = \frac{1}{2}(2)(3)^2 = 9 \text{ J} \]
Step 2: Final Velocity
\[ v_f = \frac{2(3) + 2(0)}{4} = 1.5 \text{ m/s} \]
Step 3: Final Kinetic Energy
\[ KE_f = \frac{1}{2}(4)(1.5)^2 = \frac{1}{2}(4)(2.25) = 4.5 \text{ J} \]
Step 4: Energy Lost
\[ \Delta KE = KE_f - KE_i = 4.5 - 9 = -4.5 \text{ J} \]

Answer: 4.5 J of kinetic energy is lost.

Example 6: Percent KE Loss

Problem: In the previous example, what percent of the initial kinetic energy is lost?

\[ \% \text{ Loss} = \frac{|KE_f - KE_i|}{KE_i} \times 100 = \frac{4.5}{9} \times 100 = 50\% \]

Answer: 50% of the initial kinetic energy was lost in the collision.

Example 7: Ballistic Pendulum

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Problem: A 0.02 kg bullet traveling at 300 m/s embeds into a 1 kg wooden block hanging from a string. How high does the block-bullet system rise after the collision?

Step 1: Use conservation of momentum to find velocity after collision
\[ v_f = \frac{m_1v_{1,i}}{m_1 + m_2} = \frac{(0.02)(300)}{1.02} \approx 5.88 \text{ m/s} \]
Step 2: Use conservation of energy to find height
\[ KE = PE \Rightarrow \frac{1}{2}(1.02)(5.88)^2 = (1.02)(9.8)h \Rightarrow h \approx \frac{17.6}{9.996} \approx 1.76 \text{ m} \]

Answer: The system rises approximately 1.76 meters.

Key Tips for Solving

Summary

Collision problems are highly solvable with momentum conservation. The collision type tells you whether you can also use kinetic energy conservation. Practice identifying types and setting up equations systematically!