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A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.

Required:
a. Find its x and y components of momentum.
b. Find the magnitude and direction of its momentum.

Respuesta :

Answer:

a) The x and y components of the momentum are [tex]8.731\,\frac{kg\cdot m}{s}[/tex] and [tex]-11.661\,\frac{kg\cdot m}{s}[/tex], respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

[tex]\vec p = m\cdot \vec v[/tex] (1)

Where:

[tex]\vec p[/tex] - Vector momentum, measured in kilogram-meters per second.

[tex]m[/tex] - Mass of the particle, measured in kilograms.

[tex]\vec v[/tex] - Vector velocity, measured in meters per second.

If we know that [tex]m = 2.93\,kg[/tex] and [tex]\vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right][/tex], then the momentum is:

[tex]\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right][/tex]

[tex]\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right][/tex]

The x and y components of the momentum are [tex]8.731\,\frac{kg\cdot m}{s}[/tex] and [tex]-11.661\,\frac{kg\cdot m}{s}[/tex], respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

[tex]\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}[/tex] (2)

[tex]\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)[/tex] (3)

Where:

[tex]\|\vec p\|[/tex] - Magnitude of momentum, measured in kilogram-meters per second.

[tex]\theta[/tex] - Direction of momentum, measured in sexagesimal degrees.

If we know that [tex]p_{x} = 8.731\,\frac{kg\cdot m}{s}[/tex] and [tex]p_{y} = -11.661\,\frac{kg\cdot m}{s}[/tex], then the magnitude and direction of momentum are, respectively:

[tex]\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}[/tex]

[tex]\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}[/tex]

[tex]\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)[/tex]

[tex]\theta \approx 306.823^{\circ}[/tex]

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

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