Chapter 6, Problem 61AP

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

Chapter
Section

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

# 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 8.00 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 4.00 × 103 kg for the truck. If the collision time is 0.120 s, what force does the seat belt exert on each driver?

To determine
The force on each driver.

Explanation

Section 1:

To determine: The force on truck driver.

Answer: The force on truck driver is 1.78Ã—103â€‰N .

Given info: The initial speed of the vehicles is 8.00 m/s. The mass of the drivers (m) is 80.0 kg. Total mass of car and driver is 800 kg. Total mass of the truck and driver is 4.00Ã—103â€‰kg . The collision time ( Î”t ) is 0.120 s.

From law of conservation of momentum,

mTviâˆ’mCvi=(mT+mC)vf

• vf is the final speed of the truck.
• vi is the initial speed of the truck.
• mT is the total mass of truck and driver.
• mC is the total mass of car and driver.

On Re-arranging the above equation,

vf=vi(mTâˆ’mCmT+mC)

Substitute 800 kg for mC , 4.00Ã—103â€‰kg for mT and 8.00 m/s for vi . vf=(8.00â€‰mâ‹…s-1)(4.00Ã—103â€‰kgâˆ’800â€‰kg4.00Ã—103â€‰kg+800â€‰kg)=5.33â€‰mâ‹…s-1

Formula to calculate the force on the truck driver is,

FT=m|vfâˆ’vi|Î”t

Substitute 5.33 m/s for vf , 8

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