# A 3.0*10^-3 kg truck traveling at 20.0 m/s in a test laboratory collides into a wall and comes to rest in 0.10 s. What is the magnitude of the average force acting on the truck during the collision?

Jan 2, 2017

If the given mass value is correct, the magnitude of the average force is $0.6 N$. See below.

#### Explanation:

Impulsive force (${J}_{x}$) is defined as the force exerted during a small, defined time interval. It is given by the area under the ${F}_{x} \left(t\right)$ curve between ${t}_{i}$ and ${t}_{f}$. If we perform this integration, we find that the impulsive force is equal to the change in momentum, $\Delta \vec{p}$.

Impulse in terms of the average force:

${J}_{x} = {\vec{F}}_{a v g} \cdot \Delta t$

Because we know that ${J}_{x} = \Delta \vec{p}$, this becomes:

$\Delta p = {\vec{F}}_{a v g} \Delta t$

Recall that momentum is given by the product of an object's mass and velocity. Therefore, the change in momentum is given by:

$\Delta \vec{p} = m {v}_{f} - m {v}_{i}$

Which is equivalent to $\Delta \vec{p} = m \left({v}_{f} - {v}_{i}\right)$, factoring out the mass term $m$. We now have:

$\implies m \left({\vec{v}}_{f} - {\vec{v}}_{i}\right) = {\vec{F}}_{a v g} \Delta t$

Solving for the average force:

$\implies {\vec{F}}_{a v g} = \frac{m \left({\vec{v}}_{f} - {\vec{v}}_{i}\right)}{\Delta t}$

We can now plug in the known values:

$\implies {\vec{F}}_{a v g} = \frac{3.0 \cdot {10}^{-} 3 k g \cdot \left(0 - 20.0 \frac{m}{s}\right)}{0.10 s}$

$\implies {\vec{F}}_{a v g} = - 0.6 N$

Or $0.6 N$ opposite the direction that the truck was traveling.

This is a small force, but the mass value of the truck was quite small ($3$ grams). This is either an extremely unrealistic value or the mass was meant to be $3.0 \cdot {10}^{3}$kg. If that value is correct, the magnitude of the average force is $6.0 \cdot {10}^{5} N$.

Note that the average force will be less than the maximum force, ${F}_{\max}$.