Solved Problem on Impulse, Momentum and Collisions

Solved Problem on Impulse

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:

Figure 1

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)

\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{I}=\Delta \vec{p}={\vec{p}}_{f}-{\vec{p}}_{i}} \tag{I} \end{gather} \]

The magnitude of momentum is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {p=mv} \tag{II} \end{gather} \]

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} \[ \bbox[#99CCFF,10px] {\vec{I}=\vec{F}\Delta t} \end{gather} \]

the magnitude of force

\[ \begin{gather} I=F\Delta t\\[5pt] F=\frac{I}{\Delta t}\\[5pt] F=\frac{10\;\mathrm{N.\cancel s}}{1\times 10^{-3}\;\mathrm{\cancel s}}\\[5pt] F=10\times 10^3\;\mathrm N\\[5pt] F=10000\;\mathrm N \end{gather} \]

Figure 3

Force and impulse have the same direction (Figure 3)

Magnitude: 10 000 N;
Direction: Making an angle of 37º with horizontal to the left.

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