Question

In a crash test, a truck with mass 2500 kg traveling at 24 m/s smashes head-on into a concrete wall without rebounding. The front end crumples so much that the truck is 0.72 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?
b. About how long does the collision last?
c. What is the magnitude of the average force exerted by the wall on the truck during the collision?

Answer 1

Answer:

Explanation:

Given

mass of truck $m=2500\ kg$

velocity of truck $u=24\ m/s$

Crumbled distance $d=0.72\ m$

using equation of motion

$v^2-u^2=2as$

where u=initial speed

v=final speed

a=acceleration

s=displacement

$0-(24)^2=2\times (a)\times (0.72)$

$a=-400\ m/s^2$

(b)using $v=u+at$

here $v=0$

$-u=-400\times t$

$t=\frac{24}{400}=0.06\ s$

Impulse imparted is    $J=F_{avg}\times \Delta t=\Delta P$

$F_{avg}=\frac{2500\times 24}{0.06}$

$F_{avg}=1000\ kN$

Answer 2

The average speed, time and magnitude of the average force are respectively;

A) v_avg = 12 m/s

B) t = 0.06

C) F_avg = 1000 kN

Momentum

We are given;

• Mass of truck; m = 2500 kg

• Speed of truck; v = 24 m/s

• Crumbled distance; d = 0.72 m

A) Since the concrete is stationary before collision, the average speed of the truck during collision is;

v_avg = (u- v)/2

v_avg = (24 - 0)/2

v_avg = 12 m/s

B) From newtons first equation of motion, we know that;

v = u + at

Thus;

0 = 24 + at

a = -24/t

From newton's second equation of motion;

v² = u² + 2ad

Thus;

0² = 24² + (2 × (-24/t) × 0.72)

0 = 576 - 34.56/t

34.56/t = 576

t = 34.56/576

t = 0.06 s

C) The expression for the magnitude of average force exerted by wall on truck is;

F_avg = mu/t

F_avg = (2500 * 24)/0.06

F_avg = 1000 kN

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