Chapter 2: Q17P (page 83)

In a crash test, a truck with mass 2500kgtraveling at 24m/ssmashes head-on into a concrete wall without rebounding. The front end crumples so much that the truck is 0.72mshorter 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)?
(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?)
(c) What is the magnitude of the average force exerted by the wall on the truck during the collision?
(d) It is interesting to compare this force to the weight of the tuck. Calculate the ratio of the force of the wall to the gravitational forceon the truck. This large ratio shows why a collision is so damaging.
(e) What approximations did you make in your analysis?

Short Answer

Expert verified

(a) 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) is 12m/s .

(b) The collision lasts about 0.06 s .

(d) The force on the truck by the wall is greater than the gravitational force by about 41 times

(e)The problem is simplified by ignoring many forces that act upon the truck.

Step by step solution

Identification of given data

$m = 2500 kg {\vec{v}}_{i} = <- 24, 0, 0> m / s ∆ x = 0.72 m$

Define net force and give its formula

The change in momentum of a system divided by the time it changes equals the net external force. The difference between the final and starting values of momentum is the change in momentum.

The net force is given by

$F = \frac{D p}{D t}$ where $∆ \vec{p}$ is the change in momentum and $∆ t$ is the time interval.

Evaluate average speed

(a)Let us calculate the average speed of the truck during the collision i,einterval between first contact with the wall and coming to a stop,

${\vec{v}}_{a v g} = \frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{2} = \frac{<0, 0, 0> - <- 24, 0, 0>}{2} = <12, 0, 0> m / s$

The average speed is,

$|{\vec{v}}_{a v g}| = \sqrt{{(12)}^{2} + {(0)}^{2} + {(0)}^{2}} = 12 m / s$

Thus, the average speed is $12 m / s$ .

Evaluate time period

(b) Let us calculate the collision last about.

That is, how long is the interval between first contact with the wall and coming to a stop.

$∆ t = \frac{∆ x}{|{\vec{v}}_{a v g}|} = \frac{0.72}{12} = 0.06 s$

So, the time interval is $0.06 s$ .

Find magnitude of force

${\vec{F}}_{n e t} = \frac{∆ \vec{p}}{∆ t}$

we've find the time interval that of collision

${\vec{F}}_{n e t} = \frac{<- 6 \times 10^{4}, 0, 0>}{0.06} = <- 1 \times 10^{6}, 0, 0> N$

and the magnitude of force is

Hence, the magnitude of force is .

Find magnitude of Weight

(d) The magnitude of the weight of the truck is

$F_{g} = m g = (2500) (9.8) = 2.45 \times 10^{4} N$

Therefore, The force on the truck by the wall is greater than the gravitational force by about 41 times.

Final analysis

(e) Many approximations are made to simplify the solution.For example, ignored the flexibility in the structure of the truck and the wall, ignored the air resistance , the friction and assumed the force to be constant during the collision between the wall and the truck.