Answered: Most of us know intuitively that in a…

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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 8.10 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 73.0 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 0.130 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

Expert Solution

Step 1

Given that the drivers of equal masses(m=73kg) are moving with opposite velocities, one in 800kg and other in 4000kg, so the total initial momentum of the system if car moves in positive x direction and truck in negative x direction is

$P_{1} = (m_{c} + m_{d}) v_{c} + (m_{t} + m_{d}) v_{t} v_{c} = - v_{t}, s o P_{1} = [(m_{c} + m_{d}) - (m_{t} + m_{d})] v_{c} P_{1} = [(m_{c} - m_{t}] v_{c} = [(800 - 4000] 8.1 P_{1} = - 25920 (1)$

After collision both car and truck move with same velocity this gives the final momentum as as,

$P_{2} = [(m_{c} + m_{t} + 2 m_{d}] v'_{c} = [(400 + 4000 + 146] v'_{c} P_{2} = 4546 v'_{c} (2)$

Since momentum has to be conserved, so (1)=(2) which gives the final velocity after collision of both car and truck as,

$v'_{c} = - 5.7 (3)$

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