Anonymous asked Most of us know intuitively that in a head-on collision between

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Anonymous asked



Most of us know intuitively that in a head-on collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that exerted on the truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false; Newton's third law tells us that both objects are acted upon by forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at 9.60 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 80.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4000 kg for the truck. If the collision time is 0.140 s, what force does the seat belt exert on each driver?






force on truck driver

N



force on car driver

N

Explanation / Answer

I did a mistake in the first solution i submitted ..ignore it After collision direction of velocity ofcar will be reversed where as that of truck remains in the same direction but magnitude will be decreased . As car velocity direction is reversed change in momentum of driver is more we know impulsive force = change in momentum/ time as car driver has high change in momentum car driver will experience large force. This can be understood clearly from below equations. Perfectly inelastic means they stick together after collision their common velocity after collision V conservation of momentum equation 4000*9.6 - 800*9.6 = (4000+800)V V = 6.4 m/s momentum change for truck driver = 80(9.6-6.4)=256 Force on truck driver = 256/.140 = 1828.57 N Momentum change for car driver = 80(9.6-(-6.4)) = 1280[initial and final direction of velocities are in opposite directions for car] Force on car driver = 1280/.14 = 9142.86 N

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