Question

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?

Answer 1

If the given mass value is correct, the magnitude of the average force is `0.6N`. 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(t)` 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, `Deltavecp`.

Impulse in terms of the average force:

`J_x=vecF_(avg)*Deltat`

Because we know that `J_x=Deltavecp`, this becomes:

`Deltap=vecF_(avg)Deltat`

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

`Deltavecp=mv_f-mv_i`

Which is equivalent to `Deltavecp=m(v_f-v_i)`, factoring out the mass term `m`. We now have:

`=>m(vecv_f-vecv_i)=vecF_(avg)Deltat`

Solving for the average force:

`=>vecF_(avg)=(m(vecv_f-vecv_i))/(Deltat)`

We can now plug in the known values:

`=>vecF_(avg)=(3.0*10^-3kg*(0-20.0m/s))/(0.10s)`

`=>vecF_(avg)=-0.6N`

Or `0.6N` 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*10^3`kg. If that value is correct, the magnitude of the average force is `6.0*10^5N`.

Note that the average force will be less than the maximum force, `F_(max)`.