Triangular F(t) graph

Problem

Graph reading: triangular $F(t)$. During a collision, the graph of force versus time is a triangle with $F_\text{max} = 300\;\text{N}$ and duration $\Delta t = 20\;\text{ms}$. Calculate the impulse (as the area under the graph) and the change in velocity of a $100\;\text{g}$ ball.

Solution

Data: $F_\text{max} = 300\;\text{N}$, $\Delta t = 20\;\text{ms} = 0{,}020\;\text{s}$, $m = 100\;\text{g} = 0{,}100\;\text{kg}$.

The impulse is the area under the curve $F(t)$. The graph is a triangle with base $\Delta t$ and height $F_\text{max}$: $J = \text{area} = \frac{1}{2}\cdot \text{base} \cdot \text{height} = \frac{1}{2}\cdot F_\text{max}\cdot \Delta t$ $J = \frac{1}{2}\cdot 300 \cdot 0{,}020 = 3{,}0\;\mathrm{N\,s}$

By the impulse–momentum theorem $J = \Delta p = m\,\Delta v$, hence: $\Delta v = \frac{J}{m} = \frac{3{,}0}{0{,}100} = 30\;\mathrm{m/s}$ Result: $\ev{J = 3{,}0\;\mathrm{N\,s},\quad \Delta v = 30\;\mathrm{m/s}}$

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Topics: Quantità di moto e urti Concepts: Impulso · Quantità di moto Skills: Lettura dei grafici