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 7.40 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 72.0 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 0.100 s, what force does the seat belt exert on each driver?

Explanation / Answer

Given: M= 4000 kg; m = 800kg; mdr = 72 kg; v = 7.4 m/s; Dt = 0.1s

F1 - average force exerted by the seatbelt onthe truck driver

F2 - average force exerted by the seatbelt oncar driver

Find: F1 and F2

Sol.: First of all find the speed V of joint mass M+mafter a perfectly inelastic head-on collision.

Momentum conservation principle gives:

Mv - mv = (M+m)V

then V = v(M -m) / (M+m)

That is V = 7.4(4000 - 800) / (4000 +800) = 4.93333333333 m/s

Now we can find force exerted on the track, knowing thatmomentum change Dp =F*Dt

fortrack: Dp = M(v -V) = 4000 kg *(7.4- 4.93333333333) m/s = 9866.66666667 kg*m/s

this gives F =Dp/Dt= (9866.66666667 kg*m/s) / 0.1 s = 98666.6666667 kg*m/s2

The same force is exerted on the car, since Newton's thirdlaw tells us that both objects experience forces of the samemagnitude.

Truck's acceleration atr = F/M =(98666.6666667kg*m/s2) / (4000 kg) =24.6666666667m/s2

Car's acceleration acar = F/m =(98666.66 kg*m/s2) / (800 kg) = 123.333325m/s2

Finally:

a) F1 = atr*mdr = 24.67m/s2 *72kg =

N

b) F2 =acar* mdr = 123.33m/s2 * 72kg =8879.9994N

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