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?

1 Answer
Jan 2, 2017

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)#.