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Safety barriers are used on UK motorways to prevent vehicles crossing from one carriageway to the other carriageway - AQA - A-Level Physics - Question 5 - 2019 - Paper 1

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Safety barriers are used on UK motorways to prevent vehicles crossing from one carriageway to the other carriageway. The barriers also absorb some of the kinetic ene... show full transcript

Worked Solution & Example Answer:Safety barriers are used on UK motorways to prevent vehicles crossing from one carriageway to the other carriageway - AQA - A-Level Physics - Question 5 - 2019 - Paper 1

Step 1

Show that the initial kinetic energy of the test vehicle is 700 kJ.

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Answer

To find the initial kinetic energy (KE) of the test vehicle, we use the formula:

KE=12mv2KE = \frac{1}{2}mv^2

where:

  • m = mass = 1.5 \times 10^2 \text{ kg}
  • v = speed in m/s = 110 \text{ km/h} = \frac{110 \times 1000}{3600} \approx 30.56 \text{ m/s}

Calculating the kinetic energy:

KE=12×1.5×102×(30.56)2KE = \frac{1}{2} \times 1.5 \times 10^2 \times (30.56)^2

This gives:

KE0.75×100×934.553=700.91 kJKE \approx 0.75 \times 100 \times 934.553 = 700.91 \text{ kJ}

So, we show that the initial kinetic energy is approximately 700 kJ.

Step 2

Calculate the component of the momentum of the test vehicle in a direction along the line of the safety barrier.

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Answer

The momentum (p) of the vehicle can be calculated using the formula:

p=mvp = mv

Using the values: m = 1.5 \times 10^2 \text{ kg}, v = 30.56 \text{ m/s}\n Thus,

p=1.5×102×30.564584 kg m/sp = 1.5 \times 10^2 \times 30.56 \approx 4584 \text{ kg m/s}

The component of momentum along the safety barrier (at a 20° angle) is:

pbarrier=pcos(20°)p_{\text{barrier}} = p \cos(20°)

Thus:

pbarrier4584cos(20°)4308.07 kg m/sp_{\text{barrier}} \approx 4584 \cos(20°) \approx 4308.07 \text{ kg m/s}

So, the component of momentum is approximately 4308 kg m/s.

Step 3

Show that the kinetic energy lost in the collision is about 80 kJ.

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Answer

Initially, the vehicle's kinetic energy is approximately 700 kJ as calculated earlier. After the collision, the vehicle continues along the barrier, retaining a portion of its kinetic energy.

  1. The speed along the barrier can be calculated as: v=vcos(20°)\nv30.56cos(20°)28.84 m/sv' = v \cos(20°)\n v' \approx 30.56 \cos(20°) \approx 28.84 \text{ m/s}

  2. The new kinetic energy is: KE=12m(v)2=12×1.5×102×(28.84)2620 kJKE' = \frac{1}{2} m (v')^2 = \frac{1}{2} \times 1.5 \times 10^2 \times (28.84)^2 \approx 620 \text{ kJ}

  3. The energy lost is: ΔKE=KEKE700 kJ620 kJ=80 kJ\Delta KE = KE - KE' \approx 700 \text{ kJ} - 620 \text{ kJ} = 80 \text{ kJ}

Hence, we have shown that the kinetic energy lost in the collision is about 80 kJ.

Step 4

Deduce whether the safety barrier will pass the test.

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Answer

To determine if the safety barrier passes the test, we analyze the movement of the test vehicle after the collision. If the vehicle moves more than 1.5 m towards the other carriageway, the barrier fails.

  1. As the vehicle maintains its path along the barrier, we consider the deceleration due to the barrier's force.

  2. Assuming the average force from the barrier is 60 kN, we can calculate the distance moved before stopping due to stopping force: F=ma, where F=60,000 N and the mass m=150 kg.F = ma, \text{ where } F = 60,000 \text{ N} \text{ and the mass } m = 150 \text{ kg}. Hence, a=Fm=60000150=400 m/s2a = \frac{F}{m} = \frac{60000}{150} = 400 \text{ m/s}^2

  3. Using kinematics, the distance (s) the vehicle could move is: s=v22a with v=28.84 m/ss = \frac{v^2}{2a} \text{ with } v = 28.84 \text{ m/s} s(28.84)22×400extwhichislessthan1.5m.s \approx \frac{(28.84)^2}{2 \times 400} ext{ which is less than } 1.5 m.

Thus, the barrier will indeed pass the test.

Step 5

Discuss which type of barrier would cause less damage to the dummies in the test.

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Answer

When comparing a steel safety barrier and a solid concrete wall, it is pivotal to assess the forces involved during a collision:

  1. The steel barrier deforms upon impact, which absorbs a significant portion of the collision's energy, thereby reducing the forces transmitted to the dummies. This elastic deformation helps to gradually reduce the speed of the vehicle, minimizing injury risk.

  2. Conversely, a solid concrete wall is rigid and does not deform, which would lead to a more abrupt stop for the vehicle. This implies greater forces impinging on the dummies, likely resulting in more severe injuries.

  3. In conclusion, while structures must meet safety standards, a steel barrier would generally be more effective in minimizing harm to dummies during impact as it dissipates energy through deformation.

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