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Explain how car safety features like airbags and crumple zones reduce passenger injuries during a collision, referencing the impulse-momentum theorem and Newton's Third Law.

Physics

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2 parts
6.2

1.71 m/s WESTWARDS\text{}1.71 \text{ m/s WESTWARDS}\text{}

6.3

5143 N\text{}5143 \text{ N}\text{}

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Car Safety Features

The safety features, such as airbags and crumple zones, contribute to passenger safety by increasing the time of impact during a collision. According to the impulse-momentum theorem (FnetΔt=ΔpF_{net} \Delta t = \Delta p), for a given change in momentum (Δp\Delta p), increasing the time interval (Δt\Delta t) over which the momentum changes results in a decrease in the average net force (FnetF_{net}) exerted on the passengers. A reduced force significantly lowers the risk and severity of injuries. Airbags also spread the impact force over a larger area of the body, further reducing pressure and potential harm.


2008 Prelim Paper 1 Q6

6.1 The weight of truck A is the gravitational force exerted by the Earth on truck A. According to Newton's Third Law, the reaction force to the weight of truck A is the gravitational force exerted by truck A on the Earth.

Step 1: Calculate the magnitude of the weight of truck A.

WA=mAgW_A = m_A g WA=(2000 kg)(9.8 m/s2)W_A = (2000 \text{ kg})(9.8 \text{ m/s}^2) WA=19600 NW_A = 19600 \text{ N}

The magnitude of the reaction force is equal to the magnitude of the weight.
The direction of the reaction force (force of truck A on Earth) is opposite to the direction of the weight (force of Earth on truck A).
The magnitude and direction of the 'reaction force' to the weight of truck A is 19600 N UPWARDS\boxed{\text{}19600 \text{ N UPWARDS}\text{}}

Part 6.2· 1 step

Step 1: Identify the given values and define a positive direction.

Let west be the positive direction.
Mass of truck A mA=2000 kgm_A = 2000 \text{ kg}
Initial velocity of truck A uA=3 m/su_A = 3 \text{ m/s}
Mass of truck B mB=1200 kg+300 kg=1500 kgm_B = 1200 \text{ kg} + 300 \text{ kg} = 1500 \text{ kg}
Initial velocity of truck B uB=0 m/su_B = 0 \text{ m/s}
After the collision, the trucks combine and move with a common final velocity vfv_f.

  1. Apply the principle of conservation of linear momentum.

    mAuA+mBuB=(mA+mB)vfm_A u_A + m_B u_B = (m_A + m_B) v_f (2000 kg)(3 m/s)+(1500 kg)(0 m/s)=(2000 kg+1500 kg)vf(2000 \text{ kg})(3 \text{ m/s}) + (1500 \text{ kg})(0 \text{ m/s}) = (2000 \text{ kg} + 1500 \text{ kg}) v_f 6000 kg⋅m/s+0=(3500 kg)vf6000 \text{ kg}\cdot\text{m/s} + 0 = (3500 \text{ kg}) v_f 6000=3500vf6000 = 3500 v_f vf=60003500 m/sv_f = \frac{6000}{3500} \text{ m/s} vf=127 m/sv_f = \frac{12}{7} \text{ m/s} vf≈1.71 m/sv_f \approx 1.71 \text{ m/s}

    The velocity of truck B after the collision is 1.71 m/s WESTWARDS\boxed{\text{}1.71 \text{ m/s WESTWARDS}\text{}}

Part 6.3

5143 N\text{}5143 \text{ N}\text{}

Step 1: Identify the given values for truck B and the collision time.

Mass of truck B mB=1500 kgm_B = 1500 \text{ kg}
Initial velocity of truck B uB=0 m/su_B = 0 \text{ m/s}
Final velocity of truck B vB=1.714 m/sv_B = 1.714 \text{ m/s} (using the unrounded value 127 m/s\frac{12}{7} \text{ m/s})
Time of collision Δt=0.5 s\Delta t = 0.5 \text{ s}
Force exerted by truck A on truck B F=?F = ?

  1. Apply the impulse-momentum theorem to truck B.

    FΔt=ΔpBF \Delta t = \Delta p_B FΔt=mBvB−mBuBF \Delta t = m_B v_B - m_B u_B F(0.5 s)=(1500 kg)(127 m/s)−(1500 kg)(0 m/s)F (0.5 \text{ s}) = (1500 \text{ kg})\left(\frac{12}{7} \text{ m/s}\right) - (1500 \text{ kg})(0 \text{ m/s}) F(0.5)=180007 kg⋅m/sF (0.5) = \frac{18000}{7} \text{ kg}\cdot\text{m/s} F=180007×0.5 NF = \frac{18000}{7 \times 0.5} \text{ N} F=180003.5 NF = \frac{18000}{3.5} \text{ N} F≈5142.86 NF \approx 5142.86 \text{ N}

    The magnitude of the force that truck A exerts on truck B is 5143 N\boxed{\text{}5143 \text{ N}\text{}}

    6.4 Bubble plastic is preferred because it increases the time of impact (Δt\Delta t) during a collision. According to the impulse-momentum theorem (FnetΔt=ΔpF_{net} \Delta t = \Delta p), for a given change in momentum (Δp\Delta p), a longer impact time (Δt\Delta t) results in a smaller average force (FnetF_{net}) exerted on the electronic equipment. This reduced force minimizes damage to the equipment.

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