A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision the second puck travels at a speed of 3.5 m/s at an angle of 30° above the x axis, what is the final velocity of the first puck?

Respuesta :

Answer:

The value of final velocity of the first hockey puck [tex]v_{1} _{f} =[/tex] 2.3 [tex]\frac{m}{s}[/tex]

Explanation:

Mass of hockey puck = m

Initial velocity [tex]v_{1}_{i} = 4.5 \frac{m}{s}[/tex]

From conservation of momentum principal

4.5 m = m [tex]v_{1x} _{f}[/tex] + m × [tex]\cos 30[/tex] × 3.5

[tex]v_{1x} _{f}[/tex] = 1.5 [tex]\frac{m}{s}[/tex]

0 = m [tex]v_{1y} _{f}[/tex] + m × [tex]\sin 30[/tex] × 3.5

[tex]v_{1y} _{f}[/tex] = - 1.75 [tex]\frac{m}{s}[/tex]

Now final velocity of first puck

[tex]v_{1} _{f} = \sqrt{v_{1x} _{f}^{2} + {v_{1y} _{f}^{2} }[/tex]

Put the values in above formula we get

[tex]v_{1} _{f} = \sqrt{(1.5)^{2} + (-1.75)^{2} }[/tex]

[tex]v_{1} _{f} =[/tex] 2.3 [tex]\frac{m}{s}[/tex]

This is the value of final velocity of the first hockey puck.

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