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Collisions between vehicles take place on the roads in our country daily - NSC Physical Sciences - Question 7 - 2016 - Paper 1

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Collisions between vehicles take place on the roads in our country daily. In one of these collisions, a car of mass 1 650 kg, travelling at a velocity of 25 m·s⁻¹ to... show full transcript

Worked Solution & Example Answer:Collisions between vehicles take place on the roads in our country daily - NSC Physical Sciences - Question 7 - 2016 - Paper 1

Step 1

7.1 Calculate the velocity of the two vehicles after the collision.

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Answer

To find the final velocity (vfv_f) of the two vehicles after the collision, we can use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

The equation is given by: mavai+mbvbi=(ma+mb)vfm_a v_{a_i} + m_b v_{b_i} = (m_a + m_b) v_f Where:

  • ma=1,650extkgm_a = 1,650 ext{ kg} (mass of car)
  • vai=25extms1v_{a_i} = -25 ext{ m·s}^{-1} (initial velocity of car, negative since it's to the left)
  • mb=3,050extkgm_b = 3,050 ext{ kg} (mass of minibus)
  • vbi=15extms1v_{b_i} = 15 ext{ m·s}^{-1} (initial velocity of minibus)

Substituting the values: 1,650imes(25)+3,050imes15=(1,650+3,050)vf1,650 imes (-25) + 3,050 imes 15 = (1,650 + 3,050) v_f Thus, 41,250+45,750=4,700vf-41,250 + 45,750 = 4,700 v_f 4,500=4,700vf4,500 = 4,700 v_f v_f = rac{4,500}{4,700} = 0.96 ext{ m·s}^{-1} This means the two vehicles move together at a velocity of approximately 0.96extms10.96 ext{ m·s}^{-1} to the right.

Step 2

7.2 Prove by means of calculations that the collision was inelastic.

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Answer

A collision is considered inelastic if the total kinetic energy after the collision is less than the total kinetic energy before the collision.

  1. Calculate the total kinetic energy before the collision: KE_{initial} = rac{1}{2} m_a v_{a_i}^2 + rac{1}{2} m_b v_{b_i}^2 Substituting values: KE_{initial} = rac{1}{2} imes 1,650 imes (-25)^2 + rac{1}{2} imes 3,050 imes (15)^2

Calculating each term: =0.5imes1,650imes625+0.5imes3,050imes225= 0.5 imes 1,650 imes 625 + 0.5 imes 3,050 imes 225 =515,625+343,125=858,750extJ= 515,625 + 343,125 = 858,750 ext{ J}

  1. Calculate the total kinetic energy after the collision: KE_{final} = rac{1}{2} (m_a + m_b) v_f^2 Substituting values: = rac{1}{2} imes (1,650 + 3,050) imes (0.96)^2 = rac{1}{2} imes 4,700 imes 0.9216 =2,335.68extJ= 2,335.68 ext{ J}

  2. Since the initial kinetic energy (858,750extJ858,750 ext{ J}) is greater than the final kinetic energy (2,335.68extJ2,335.68 ext{ J}), we can conclude that: KEinitial>KEfinalKE_{initial} > KE_{final} Thus, the collision was inelastic.

Step 3

7.3 Explain how crumple zones can reduce the chances of fatal or serious injury.

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Answer

Crumple zones are designed to absorb energy during a collision, which helps to reduce the force transmitted to the occupants of a vehicle.

  1. Energy Absorption: Crumple zones deform upon impact, extending the time over which the deceleration occurs. This increases the time interval (rianglet riangle t) during which the vehicle stops.

  2. Force Reduction: The relationship between force, mass, and acceleration is described by Newton's second law, F=maF = ma. When the time interval increases, the average force experienced by the passengers decreases, leading to fewer injuries.

  3. Safety of Passengers: By managing the way the vehicle collapses, crumple zones ensure that the impact force does not act directly on the passenger compartment, thereby reducing the risk of serious injuries or fatalities.

In conclusion, crumple zones play a critical role in vehicle safety by managing collision energy and reducing harmful forces impacting the passengers.

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