A body with a mass of 2 kg, and a speed of 4 m/s in the horizontal direction, receives the impulse of a
force. Its speed change to 3 m/s to the vertical direction. Knowing that the force acted in the particle in
a time interval that lasted 1 ms, determine:
a) The impulse received by the particle;
b) The force that acts in the particle.
Problem data:
- Mass of body: m = 2 kg;
- Initial speed of body: v1 = 4 m/s;
- Final speed of body: v2 = 3 m/s;
- Interval of time when the force acts: Δt = 1 ms.
Problem diagram:
Solution:
First, we will convert the time interval given in milliseconds (ms) to seconds used in the
International System of Units (SI)
\[
\begin{gather}
\Delta t=1\;\text{ms}=1\times 10^{-3}\;\text{s}=0.001\;\text{s}
\end{gather}
\]
a) Applying the Impulse-Momentum Theorem, the impulse is equal to the change in momentum
(vectorially calculated)
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{\vec{I}=\Delta \vec{p}={\vec{p}}_{f}-{\vec{p}}_{i}} \tag{I}
\end{gather}
\]
The magnitude of momentum is given by
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{p=mv} \tag{II}
\end{gather}
\]
applying the equation (II) to the initial and final momenta
\[
\begin{gather}
p_{1}=mv_{1} \\[5pt]
p_{1}=(2\;\mathrm{kg})\times\left(4\;\mathrm{\frac{m}{s}}\right) \\[5pt]
p_{1}=8\;\text{kg.m/s} \\[10pt]
p_{2}=mv_{2} \\[5pt]
p_{2}=(2\;\mathrm{kg})\times\left(3\;\mathrm{\frac{m}{s}}\right) \\[5pt]
p_{2}=6\;\text{kg.m/s}
\end{gather}
\]
The equation (I) can be represented as in Figure 2. The magnitude of impulse
(
\( |\vec{I}|=I \))
can be calculated by the
Pythagorean theorem
\[
\begin{gather}
I^{2}=p_{1}^{2}+p_{2}^{2} \\[5pt]
I^{2}=\left(8\;\mathrm{\frac{kg.m}{s}}\right)^2+\left(6\;\mathrm{\frac{kg.m}{s}}\right)^2\\[5pt]
I^{2}=64\;\mathrm{\frac{kg^2.m^2}{s^2}}+36\;\mathrm{\frac{kg^2.m^2}{s^2}} \\[5pt]
I=\sqrt{100\;\mathrm{\frac{kg^2.m^2}{s^2}}\;} \\[5pt]
I=10\;\mathrm{kg.m/s}=10\;\mathrm{N.s}
\end{gather}
\]
Figure 2
The angle θ that the impulse vector does with the horizontal will be
\[
\begin{gather}
\tan\theta =\frac{\text{opposite leg}}{\text{adjacent leg}}=\frac{p_{2}}{p_{1}} \\[5pt]
\tan\theta=\frac{6\;\mathrm{\frac{\cancel{kg}.\cancel m}{\cancel s}}}{8\;\mathrm{\frac{\cancel{kg}.\cancel m}{\cancel s}}} \\[5pt]
\tan\theta =0.75\\[5pt]
\theta=\arctan 0.75\\[5pt]
\theta \simeq 37°
\end{gather}
\]
- Magnitude: 10 N.s;
- Direction: Making an angle of 37º with horizontal to the left.
b) The force that acted on the particle is given by
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{\vec{I}=\vec{F}\Delta t}
\end{gather}
\]
the magnitude of force
\[
\begin{gather}
I=F\Delta t\\[5pt]
F=\frac{I}{\Delta t} \\[5pt]
F=\frac{10\;\mathrm{N.\cancel s}}{1\times 10^{-3}\;\mathrm{\cancel s}} \\[5pt]
F=10\times 10^3\;\mathrm N \\[5pt]
F=10000\;\mathrm N
\end{gather}
\]
Force and impulse have the same direction (Figure 3)
- Magnitude: 10 000 N;
- Direction: Making an angle of 37º with horizontal to the left.
DE
ES
FR
PT-BR
