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Figure 10 shows two railway trucks A and B travelling towards each other on the same railway line which is straight and horizontal - AQA - A-Level Physics - Question 6 - 2018 - Paper 1

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Question 6

Figure-10-shows-two-railway-trucks-A-and-B-travelling-towards-each-other-on-the-same-railway-line-which-is-straight-and-horizontal-AQA-A-Level Physics-Question 6-2018-Paper 1.png

Figure 10 shows two railway trucks A and B travelling towards each other on the same railway line which is straight and horizontal. The trucks move a distance of 15... show full transcript

Worked Solution & Example Answer:Figure 10 shows two railway trucks A and B travelling towards each other on the same railway line which is straight and horizontal - AQA - A-Level Physics - Question 6 - 2018 - Paper 1

Step 1

6.1 State the quantity that is not conserved in an inelastic collision.

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Answer

The quantity that is not conserved in an inelastic collision is the total kinetic energy.

Step 2

6.2 Show that the speed of the joined trucks immediately after the collision is about 0.3 m s⁻¹.

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Answer

To find the final speed of the joined trucks, we can apply conservation of momentum. The total momentum before the collision is given by:

egin{align*} \text{Total momentum before} & = m_A v_A + m_B v_B
& = (16,000 ext{ kg} \times 2.8 ext{ m s}^{-1}) + (12,000 ext{ kg} \times 3.1 ext{ m s}^{-1})
& = 44,800 ext{ kg m s}^{-1} + 37,200 ext{ kg m s}^{-1}
& = 82,000 ext{ kg m s}^{-1}. \end{align*}

After they collide, let their combined mass be ( m_C = m_A + m_B = 16,000 ext{ kg} + 12,000 ext{ kg} = 28,000 ext{ kg} ). Using conservation of momentum:

egin{align*} \text{Total momentum after} & = (m_A + m_B) v_C
\Rightarrow 82,000 & = 28,000 v_C
\Rightarrow v_C & = \frac{82,000}{28,000} \approx 2.93 ext{ m s}^{-1} \approx 0.3 ext{ m s}^{-1}. \end{align*}

Step 3

6.3 Calculate the impulse that acts on each truck during the collision. Give an appropriate unit for your answer.

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Answer

The impulse can be calculated using the formula:

Impulse=Δp=mΔvImpulse = \Delta p = m \Delta v

For Truck A:

  • Mass, ( m_A = 16,000 \text{ kg} )
  • Initial velocity, ( v_{A,i} = 2.8 \text{ m s}^{-1} )
  • Final velocity, ( v_{A,f} = 0.3 \text{ m s}^{-1} )

Calculating the change in momentum:

\begin{align*} \Delta p_A & = m_A \times (v_{A,f} - v_{A,i}) \ & = 16,000 \times (0.3 - 2.8) \ & = 16,000 \times (-2.5) \ & = -40,000 \text{ kg m s}^{-1}. \end{align*}

For Truck B:

  • Mass, ( m_B = 12,000 \text{ kg} )
  • Initial velocity, ( v_{B,i} = 3.1 \text{ m s}^{-1} )
  • Final velocity, ( v_{B,f} = 0.3 \text{ m s}^{-1} )

Calculating the change in momentum:

\begin{align*} \Delta p_B & = m_B \times (v_{B,f} - v_{B,i}) \ & = 12,000 \times (0.3 - 3.1) \ & = 12,000 \times (-2.8) \ & = -33,600 \text{ kg m s}^{-1}. \end{align*}

Thus, the impulses on Trucks A and B are (-40,000 \text{ Ns}) and (-33,600 \text{ Ns}) respectively.

Step 4

6.4 Explain, without doing a calculation, how the motion of the trucks immediately after the collision would be different for a collision that is perfectly elastic.

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Answer

In a perfectly elastic collision, both momentum and kinetic energy are conserved. After the collision, the trucks would rebound and separate at different speeds, depending on their masses and initial velocities. Each truck would continue to move forward with kinetic energy being retained in the system, while in an inelastic collision, they stick together, losing kinetic energy to other forms such as heat and sound. The final speed of the trucks after an elastic collision would be higher compared to an inelastic collision, where they come to rest after some distance.

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