Two average forces. A steady stream of 0.250 kgsnowballs is shot perpendicularly into a wall at a speed of 4.00 m/s . Each ball sticks to the wall. Figure 9-49 gives the magnitude F of the force on the wall as a function of time t for two of the snowball impacts. Impacts occur with a repetition time interval $\mathbf{∆}{\mathbf{t}}_{\mathbf{r}}\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{ms}$, last a duration time interval $\mathbf{∆}{\mathbf{t}}_{\mathbf{d}}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{ms}$, and produce isosceles triangles on the graph, with each impact reaching a force maximum ${\mathbf{F}}_{\mathbf{max}}\mathbf{=}\mathbf{20}\mathbf{}\mathbf{N}$.During each impact, what are the magnitudes of (a) the impulse and (b) the average force on the wall? (c) During a time interval of many impacts, what is the magnitude of the average force on the wall?

The mass of each snowball,$\mathrm{m}=0.250\mathrm{kg}$.

The speed of each snowball,$\mathrm{v}=4.00\mathrm{m}/\mathrm{s}$.

The duration of impact,role="math" localid="1661243323595" ${\mathrm{t}}_{\mathrm{d}}=10\mathrm{ms}\left(\frac{1\times {10}^{-3}\mathrm{s}}{1\mathrm{ms}}\right)=1.0\times {10}^{-2}\mathrm{s}$.

The time for repetition of impact is,${\mathrm{t}}_{\mathrm{d}}=50.0\mathrm{ms}\left(\frac{1\times {10}^{-3}\mathrm{s}}{1\mathrm{ms}}\right)=5.00\times {10}^{-2}\mathrm{s}$.

The maximum force of impact is, ${\mathrm{F}}_{\max }=200\mathrm{N}$.

The impulse during each impact can be calculated from the graph using the area under the curve. The average force can be calculated using the equation relating impulse and time. To calculate the average force for many impacts, we assume the number of impacts as ‘N’ and solve the equation of average force. Use the formula given as.

$$\begin{gathered} \mathrm{J}=∆\mathrm{P} \\ ∆\mathrm{P}=\mathrm{m}\times \left({\mathrm{v}}_2-{\mathrm{v}}_1\right) \\ \mathrm{J}={\mathrm{F}}_{\mathrm{avg}}\times \mathrm{t} \end{gathered}$$

The impulse of each impact can be calculated from the area under the F(t) curve.

As it is triangular

$$\begin{gathered} \mathrm{Area}=\frac{1}{2}\mathrm{base}\times \mathrm{height} \\ \mathrm{J}=\frac{1}{2}\times {\mathrm{t}}_{\mathrm{d}}\times {\mathrm{F}}_{\max } \\ =\frac{1}{2}\times 10\times {10}^{-3}\mathrm{s}\times 200\mathrm{N} \\ =1.00\mathrm{N}·\mathrm{s} \end{gathered}$$

The average force is given by

$$\begin{gathered} {\mathrm{F}}_{\mathrm{avg}}=\frac{\mathrm{J}}{{\mathrm{t}}_{\mathrm{d}}} \\ =\frac{1.00\mathrm{N}·\mathrm{s}}{10\times {10}^{-3}\mathrm{s}} \\ =100\mathrm{N} \end{gathered}$$

Let us assume the number of impacts is N . So, the time interval is

$\mathrm{t}=\mathrm{N}\times {\mathrm{t}}_{\mathrm{r}}$

And the total impulse is

${\mathrm{J}}_{\mathrm{total}}=\mathrm{N}\times 1.0\mathrm{N}·\mathrm{s}$

So, the average force for many impacts is

$$\begin{gathered} {\mathrm{F}}_{\mathrm{avg}}=\frac{{\mathrm{J}}_{\mathrm{total}}}{\mathrm{t}} \\ =\frac{\mathrm{N}\times 1.0\mathrm{N}·\mathrm{s}}{\mathrm{N}\times 5\times {10}^{-2}\mathrm{s}} \\ =20.0\mathrm{N} \end{gathered}$$