Answer to Question #257308 in Mechanics | Relativity for mimi

Question

A 2500 kg truck travelling at 40 km/h strikes a brick wall and comes to a stop in 0.2 s. a. What is the change in momentum? (Ans. – 2.78 x 10^4 kgm/s) b. What is the impulse? (Ans. – 2.78 x 10^4 kgm/s) c. What is the average force on the wall during the crash? (Ans. 1.39 x 10^5 N)

Answer 1

Mass m =2500kg

Initial velocity u = 40km/hr

$=40×\frac{5}{18}m/s =11.111m/s$

Final velocity v =0

Time t = 0.2 sec

a)Change in momentum

$=m(v-u) =2500(0-11.111)= -27777.77kgm/s$

$= -2.777×10^4kgm/s$

b) The impulse

$=m(v-u) = -2.777×10^4kgm/s$

c) average force on the wall during the crash

$F =\frac{m(v-u)}{t} =\frac{-27777.77}{0.2}=-138888.85N$

Magnitude of the force $F= -1.388×10^5N$