91Ó°ÊÓ

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

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

(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,

The average speed is,

$$\begin{gathered} \left|{\stackrel{→}{v}}_{avg}\right|=\sqrt{{\left(12\right)}^2+{\left(0\right)}^2+{\left(0\right)}^2} \\ =12\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the average speed is $12\mathrm{m}/\mathrm{s}$.

(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.

$$\begin{gathered} ∆t=\frac{∆x}{\left|{\stackrel{→}{v}}_{avg}\right|} \\ =\frac{0.72}{12} \\ =0.06\mathrm{s} \end{gathered}$$

So, the time interval is $0.06\mathrm{s}$.

(c) Using the momentum principle, the net force on the truck is

${\stackrel{→}{F}}_{net}=\frac{∆\stackrel{→}{p}}{∆t}$

we've find the time interval that of collision

and the magnitude of force is

Hence, the magnitude of force is .

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

$$\begin{gathered} F_g=mg \\ =\left(2500\right)\left(9.8\right) \\ =2.45×{10}^4\mathrm{N} \end{gathered}$$

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

(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.