Mastering Momentum and Collisions for Your Physics Exam

What the examiner is testing

The examiner assesses your ability to apply the principle of conservation of momentum to solve problems involving collisions and explosions. Marks are awarded for correctly identifying the initial and final states of the system, setting up the conservation equation, and accurately calculating unknown velocities or masses.

The method

1. Identify the system: Determine which objects are involved in the collision or explosion and will be included in your momentum calculation.
2. Define positive direction: Choose a direction (e.g., right or left) as positive. All velocities in that direction will be positive, and all velocities in the opposite direction will be negative.
3. Calculate initial momentum: For each object, calculate its initial momentum using the formula \(p = mv\). Sum these individual momenta to find the total initial momentum of the system.
4. Calculate final momentum: For each object, calculate its final momentum using the formula \(p = mv\). Sum these individual momenta to find the total final momentum of the system.
5. Apply conservation of momentum: Set the total initial momentum equal to the total final momentum.
   $$ \Sigma (mv)_{\text{initial}} = \Sigma (mv)_{\text{final}} $$
6. Solve for the unknown: Rearrange the equation to solve for the required quantity (e.g., final velocity, mass).

Worked example

A 2 kg trolley moving at \(3 \text{ m/s}\) collides with a stationary 4 kg trolley. After the collision, the two trolleys stick together. Calculate their combined velocity after the collision.

• Step 1: Identify the system. The system consists of the two trolleys.
• Step 2: Define positive direction. Let the initial direction of the 2 kg trolley be positive.
• Step 3: Calculate initial momentum.
  • Momentum of 2 kg trolley: \(p_1 = m_1 v_1 = 2 \text{ kg} \times 3 \text{ m/s} = 6 \text{ kg m/s}\)
  • Momentum of 4 kg trolley: \(p_2 = m_2 v_2 = 4 \text{ kg} \times 0 \text{ m/s} = 0 \text{ kg m/s}\)
  • Total initial momentum: \(P_{\text{initial}} = p_1 + p_2 = 6 \text{ kg m/s} + 0 \text{ kg m/s} = 6 \text{ kg m/s}\)
• Step 4: Calculate final momentum.
  • Since the trolleys stick together, they move as a single mass \(M_{\text{final}} = m_1 + m_2 = 2 \text{ kg} + 4 \text{ kg} = 6 \text{ kg}\).
  • Let their combined final velocity be \(V_{\text{final}}\).
  • Total final momentum: \(P_{\text{final}} = M_{\text{final}} V_{\text{final}} = 6 \text{ kg} \times V_{\text{final}}\)
• Step 5: Apply conservation of momentum.
  $$ P_{\text{initial}} = P_{\text{final}} $$
  $$ 6 \text{ kg m/s} = 6 \text{ kg} \times V_{\text{final}} $$
• Step 6: Solve for the unknown.
  $$ V_{\text{final}} = \frac{6 \text{ kg m/s}}{6 \text{ kg}} $$
  $$ V_{\text{final}} = 1 \text{ m/s} $$
  • Sanity check: The final velocity is less than the initial velocity of the moving trolley, and the combined mass is greater. This makes sense as momentum is conserved, so a larger mass must move slower. The direction is also positive, consistent with the initial direction.

Worked example: a harder one

A student fires a \(0.01 \text{ kg}\) pellet from a \(1.5 \text{ kg}\) toy gun. The pellet leaves the gun at \(120 \text{ m/s}\). Calculate the recoil velocity of the gun.

• Step 1: Identify the system. The system is the gun and the pellet.
• Step 2: Define positive direction. Let the direction the pellet travels be positive.
• Step 3: Calculate initial momentum.
  • Before firing, both the gun and the pellet are stationary.
  • Momentum of gun: \(m_g v_g = 1.5 \text{ kg} \times 0 \text{ m/s} = 0 \text{ kg m/s}\)
  • Momentum of pellet: \(m_p v_p = 0.01 \text{ kg} \times 0 \text{ m/s} = 0 \text{ kg m/s}\)
  • Total initial momentum: \(P_{\text{initial}} = 0 \text{ kg m/s} + 0 \text{ kg m/s} = 0 \text{ kg m/s}\)
• Step 4: Calculate final momentum.
  • Momentum of pellet: \(m_p v_p = 0.01 \text{ kg} \times 120 \text{ m/s} = 1.2 \text{ kg m/s}\)
  • Let the recoil velocity of the gun be \(V_g\).
  • Momentum of gun: \(m_g V_g = 1.5 \text{ kg} \times V_g\)
  • Total final momentum: \(P_{\text{final}} = 1.2 \text{ kg m/s} + (1.5 \text{ kg} \times V_g)\)
• Step 5: Apply conservation of momentum.
  $$ P_{\text{initial}} = P_{\text{final}} $$
  $$ 0 \text{ kg m/s} = 1.2 \text{ kg m/s} + (1.5 \text{ kg} \times V_g) $$
• Step 6: Solve for the unknown.
  $$ -1.2 \text{ kg m/s} = 1.5 \text{ kg} \times V_g $$
  $$ V_g = \frac{-1.2 \text{ kg m/s}}{1.5 \text{ kg}} $$
  $$ V_g = -0.8 \text{ m/s} $$
  • Why the obvious first move fails (or why students get this wrong): Students often forget that the initial momentum of the entire system (gun + pellet) is zero. They might only consider the pellet's momentum or incorrectly assume the gun's initial momentum is negligible. The negative sign for the gun's velocity is crucial; it indicates recoil in the opposite direction to the pellet.

Practice

1. A \(50 \text{ kg}\) person running at \(4 \text{ m/s}\) jumps onto a stationary \(10 \text{ kg}\) skateboard. What is the combined velocity of the person and skateboard immediately after the person lands on it?
2. A \(0.2 \text{ kg}\) billiard ball moving at \(2 \text{ m/s}\) collides head-on with a stationary \(0.2 \text{ kg}\) billiard ball. After the collision, the first ball stops dead. What is the velocity of the second ball?
3. A \(1000 \text{ kg}\) car travelling at \(15 \text{ m/s}\) crashes into the back of a \(1500 \text{ kg}\) lorry travelling in the same direction at \(10 \text{ m/s}\). After the collision, the two vehicles stick together. Calculate their common velocity immediately after the collision.
4. A \(75 \text{ kg}\) astronaut is floating motionless in space. They throw a \(0.5 \text{ kg}\) wrench away from them at a speed of \(10 \text{ m/s}\). What is the recoil speed of the astronaut?

Answers:

1. \(3.33 \text{ m/s}\)
2. \(2 \text{ m/s}\)
3. \(12 \text{ m/s}\)
4. Working for question 4:
   • Initial momentum of astronaut + wrench = \(0 \text{ kg m/s}\) (both motionless).
   • Final momentum of wrench = \(m_w v_w = 0.5 \text{ kg} \times 10 \text{ m/s} = 5 \text{ kg m/s}\).
   • Let the recoil speed of the astronaut be \(v_a\). Final momentum of astronaut = \(m_a v_a = 75 \text{ kg} \times v_a\).
   • By conservation of momentum:
     $$ 0 = 5 \text{ kg m/s} + (75 \text{ kg} \times v_a) $$
     $$ -5 \text{ kg m/s} = 75 \text{ kg} \times v_a $$
     $$ v_a = \frac{-5 \text{ kg m/s}}{75 \text{ kg}} $$
     $$ v_a = -0.067 \text{ m/s} \text{ (to 2 s.f.)} $$
   • The recoil speed is \(0.067 \text{ m/s}\). The negative sign indicates the direction is opposite to the wrench.

The three mistakes that lose marks

1. Forgetting to define a positive direction and use negative signs for opposing velocities: This leads to incorrect sums of momentum. For example, if a car \(m_1\) moving right at \(v_1\) collides with a car \(m_2\) moving left at \(v_2\), and you calculate initial momentum as \(m_1 v_1 + m_2 v_2\) (both positive), you will get an inflated total initial momentum. The correct calculation requires \(m_1 v_1 + m_2 (-v_2)\).
2. Mixing up initial and final states or only considering one object's momentum: This often happens in explosion problems. Forgetting that the initial momentum of a stationary system (like a gun and bullet before firing) is zero, and only calculating the bullet's final momentum, would lead to the incorrect conclusion that the gun has no recoil. The total initial momentum must equal the total final momentum for all objects in the system.
3. Incorrectly combining masses for objects that stick together: While the total mass is indeed the sum, students sometimes forget to apply this combined mass to the final velocity in the momentum equation. For example, if two objects \(m_1\) and \(m_2\) stick together with final velocity \(V\), their combined final momentum is \((m_1 + m_2)V\), not \(m_1 V + m_2 V\). This is algebraically the same, but conceptually, treating them as separate masses with the same final velocity can lead to errors if the student then tries to apply individual masses to other parts of the calculation.

30-second recap

Momentum is conserved in collisions and explosions, meaning the total momentum before equals the total momentum after. Remember to define a positive direction and use negative signs for velocities in the opposite direction. For objects that stick together, their masses combine for the final momentum calculation.