183_notes:examples:two_students_colliding

Two students are running to make it to class. They turn a corner and collide; coming to a complete stop. What force did they exert on each other.

Facts

Average mass of a person 68kg.

Students come to complete stop.

Choose student 1 as our system.

Lacking

Collision time Δt

Force of collision on student 1 in x-direction.

Approximations & Assumptions

Approximate speed of person 5m/s^-1 (half speed of olympic sprinter)

On average student 1's body compresses a couple of cm on average - assume compression of 2.5cm.

Representations

Δpsys=FextΔt

psys,f=psys,i+FextΔt

vavg=vf+vi2=ΔrΔt

Solution

We use the momentum principle to relate the momentum of the students to the force applied. We have chosen student 1 as our system.

pfx=pix+Fx,collΔt

We are told that the students come to a complete stop and so pfx = 0.

0=MVix+Fx,collΔt

Rearranging the equation we relate Fx,coll to the remaining variables. We are trying to find force.

Fx,coll=MVixΔt

The negative sign means that the force is in ˆx direction.

Now that we have the above relationship we must find the missing variables in order to solve for Fx,coll. First we need to find collision time Δt. We can relate the average velocity to displacement over time.

vavg=vf+vi2=ΔrΔt

In 1D this looks like: vavg=ΔxΔt=vf+vi2

Relate these 3 equations together to solve for Δt

Δt=ΔxVavg=Δxvf+vi2

Fill in the values for the variables from the assumptions and approximations you made previous.

0.0255m/s+0m/s2=0.01s

Having solved for Δt fill this value and the known value for mass and the approximated value for velocity into the equation that we arranged earlier to find Fx,coll

Fx,coll=MVixΔt=(68kg)(5m/s)0.01s

=34,000N

  • 183_notes/examples/two_students_colliding.txt
  • Last modified: 2014/10/02 15:50
  • by caballero