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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 $2\ ms$ ?

Known:

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

Find:

$F_{avg}=\ ?\ N$

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:

$F_{avg}=\frac{0.05kg(45\frac{m}{s}-0\frac{m}{s})}{0.002s}$

$F_{avg}=1125\ N$