robert7248
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At a race track, a car of mass 2000 kg crashes into a concrete wall at a speed of 77 m/s.

a. If the car comes to a stop when it this the wall, what is the magnitude of the impulse applied to the car?

b. To make things safer, the race track installs barrels of sand along the wall. If the car crashes into the barrels, the time it takes to stop will be increased by a factor of 3, as compared with hitting a concrete wall. How does this affect the force applied to the car during a crash? Explain your answer using the appropriate equation.

Respuesta :

PART A)

Impulse is defined as change in momentum

here finally after hitting the wall car comes to rest

So final momentum of the car will be ZERO

Initial momentum of the car is given as

[tex]P = mv[/tex]

here m = 2000 kg

v = 77 m/s

now we have

[tex]P = 77 ( 2000) = 1.54 \times 10^5 kg m/s[/tex]

so here impulse will be

[tex]I = P - 0 = 1.54 \times 10^5 kg m/s[/tex]

PART B)

As we know that force of impact is defined as

[tex]F = \frac{dP}{dt}[/tex]

so here it is ratio of impulse and time

so if we increase the time of impact by 3 times

so the force will be decrease by factor of 1/3

so here force is decreased by 1/3

a) Simply p=mv=2000*77=154000 kg*m/s

b) Force in terms of impulse and time is F=p/t. Thus in case of the barrels the crash force will reduce by a factor of 3: F=p/3t, i.e. 3 times less than in the concrete wall case

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