Impulse-Momentum Theorem

Example: A machine drives a 0.05\ kg ball giving the ball an initial horizontal speed of 45\ \frac{m}{s}. What is the magnitude of the average force exerted on the ball if the force is exerted over 2\ ms?


  • m_{b}=0.05\ kg
  • v_{1}=0\ \frac{m}{s}
  • v_{2}=45\ \frac{m}{s}
  • \Delta t=0.0025\ (2\ ms)


  • F_{avg}=\ ?\ N

Relevant Equation:

  • F_{avg}\Delta t=Impulse
  • momentum\ =\ mass\ \times\ velocity


*Note that if you hold Impulse constant, an increase in F_{avg} means a decrease in \Delta t. The reverse is true. For instance, a car accelerates to a speed of 60 mpg. I can get to this speed in 30 seconds out. If you want it to be faster than that, say reach 60 mph in 5 seconds, you need a lot more force to be exerted each second. Same principle here.

Remember that Impulse can be replaced with a change in moments.

F_{avg}\Delta t=mv_{2}-mv_{1}=m(v_{2}-v_{1})


Solve for F_{avg}F_{avg}=\frac{m(v_{2}-v_{1})}{\Delta t}

Plug in values and solve:


F_{avg}=1125\ N