Most of us know intuitively that in a head-on collision between a large dump tru

Question

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 6.80 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4,000 kg for the truck. If the collision time is 0.110 s, what force does the seat belt exert on each driver? (Enter the magnitude of the force.)

force on truck driver ____ N

force on car driver _____N

Explanation / Answer

Whether the car is crushed or not doesn't affect the result.
The change in velocity is always greater for the car so the acceleration of the occupants is greater.
In turn this means that the forces ON THE OCCUPANTS is far greater in the lighter vehicle.

conservation of momentum

=> 4000 * 6.8 - 800*6.8 = (4000+800)*V

=> V = 4.53 m/s
Your car has gone from + 6.8 m/s to - 4.53 m/s a change of 11.33 m/s
The truck has gone from + 6.8 to + 4.53 m/s a change of 2.27 m/s

Both of these changes apply to the occupants of the respective vehicles.

So the force on the car driver = mass *change in velocity / time taken = 79 *11.33/0.11 = 8137 N approx.

And for the truck driver = 79 * 2.27 /0.11 = 1630.27 N

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