Question

A truck weighing $2000kg$ moving with a velocity of $60km/hr$ retards uniformly to rest in $10sec$. Then the change in momentum that take place in $10sec$:

• A. $3,334N$
• B. $12,000N$
• C. $600N$
• D. $2N$

Answer 1

The correct option is A $3,334N$

Given : $m=2000$ kg $v=0$ m/s $t=10$ s

Initial velocity of the truck $u=60$ km/hr $=60\times \frac{5}{18}=16.67$ $m/s$

Thus change in momentum $\Delta P=m(v-u)=2000(0-16.67)=-33340$ $kg$ $m/s$

$\therefore$ Rate of change of momentum $=\frac{|\Delta P|}{t}=\frac{33340}{10}=3334$ $N$