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A particle is travelling at 2300 m/s in the x-direction. It collides with another particle travelling in the opposite direction at 100 m/s. After the collision, the first particle travels at 2000 m/s, at 45 degrees from the x-direction. What is the horizontal and vertical velocities of the second particle after the collision? Both particles have same mass (you actually don’t need to know the number).

Respuesta :

Answer:

[tex]v'_{x} = 785.786 m/s[/tex]

[tex]v'_{y} = 1414.214 m/s[/tex](decreasing)

Given:

[tex]u_{x}[/tex] = 2300 m/s

[tex]u'_{x}[/tex] = 100 m/s

[tex]v_{x}[/tex] = 2000 m/s

Angle made with the horizontal, [tex]\theta =  45^{\circ}[/tex]

The horizontal component of velocity is on the X-axis whereas the vertical one is on the Y-axis

Now, by the law of  conservation of momentum for horizontal axis:

[tex]mu_{x} + m'u'_{x} = mv_{x} + m'v'_{x}[/tex]

[tex]2300 + (- 100) = 2000cos45^{\circ} + v'_{x}[/tex]

(The mass of the particles is same)

[tex]v'_{x} = 2200 - 1414.214 = 785.786 m/s[/tex]

Now, by the law of  conservation of momentum for vertical axis:

[tex]mu_{y} + m'u'_{y} = mv_{y} + m'v'_{y}[/tex]

(The mass of the particles is same)

[tex]u_{y} + u'_{y} = v_{y} + v'_{y}[/tex]

[tex]0 = v_{y} + v'_{y}[/tex]

(since, initially, there's no vertical component of velocity)

[tex]v'_{y} = - 2000sin45^{\circ} = 1414.214 m/s[/tex](decreasing)

velocity, v = [tex]\sqrt{v^{2}_{x} + v^{2}_{y}}[/tex]

              v = [tex]\sqrt{(785.786)^{2} + (1414.214)^{2}} = 1617.856 m/s[/tex]

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