Question

In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The front end crumples so much that the truck is 0.62 m shorter than before.
(a) What is the average speed of the truck during the collision (that is, during the interval between first contact with the wall and coming to a stop)?

vavg = ______ m/s



(b) About how long does the collision last? (That is, how long is the interval between first contact with the wall and coming to a stop?)

Δt = ______ s



(c) What is the magnitude of the average force exerted by the wall on the truck during the collision?

Fwall, avg = ______ N



(d) It is interesting to compare this force to the weight of the truck. Calculate the ratio of the force of the wall to the gravitational force mg on the truck. This large ratio shows why a collision is so damaging.

Fwall, avg / mg =


(e) What approximations were necessary in making this analysis? (Select all that apply.)

Neglect the horizontal component of the force of the road on the truck tires.

Assume a nearly constant force exerted by the wall, so the speed changes at a nearly constant rate.

The deceleration of the truck is approximately equal to g.

Answer 1

Answer:

a) 11 m/s

b) 0.0564 s

Explanation:

Given:

m = 2100 kg

vi = 22 ..... m/s before collision

vf = 0 ......after collision to stop

Δs = 0.62   distance traveled after collision .. crumpling of truck

Part a

$V_{avg} = \frac{vi- vf}{2}\\\\V_{avg} = \frac{22- 0}{2}\\\\ V_{avg} = 11 m/s$

Part b

$vf = vi + a*t\\vf^2 = vi^2 + 2*a*s\\\\0 = 22 + a*t\\t = -22 /a\\\\0 = 484 +2*a*(0.62)\\a = - 390.3232 m/s^2\\\\t = -22/(-390.3232)\\\\t = 0.0564 s$

Answer 2

Answer:

a) average speed = 11m/s

b) time = 0.056secs

c) Magnitude force = 825000N

d) ratio = 40.05

e) Assumptions ; g = 9.81m/s square

Explanation:

Given the following parameters;

• mass = 2100kg

• Vi (initial velocity) = 22m/s

• distance = 0.62m

• Vf ( final velocity) = 0

a) average speed = final velocity + initial velocity / 2

Vavg = 22 + 0 / 2 = 11m/s

b) how long does the collision last, i.e the time taken,

from average speed = distance / time

time = distance / average speed

t = 0.62m/11m/s

t = 0.056secs

c) Magnitude of average force

from Favg = ma

but a = v/t

Favg = mv/t = 2100 x 22 / 0.056

Average force = 825000N

d) ratio of force to the gravitational force(mg) = (Fwall)avg/mg

= 825000/2100x9.81

=  40.05