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Sergio039 [100]
3 years ago
12

To protect their young in the nest, peregrine falcons will fly into birds of prey (such as ravens) at high speed. in one such ep

isode, a 600 g falcon flying at 20.0 m/s ran into a 1.50 kg raven flying at 9.00 m/s . the falcon hit the raven at a right angle to its original path and bounced back with a speed of 5.00 m/s . by what angle did the falcon change the raven's direction of motion?
Physics
1 answer:
lozanna [386]3 years ago
7 0
<span>The momentum of the falcon before collision is 0.6 * 20 = 12000 kgm/s which is actually the momenum of the falcon in the x-component. I had converted 600g to kg. After the collision the x-component of the raven is now mv2cos(thetha) where v2 is the final velocity of the raven and theta is the angle at which the falcon hits the raven. So we have that the falcon's final velocity = 600 * 5 * cos (theta). Likewise, after getting hit the the falcon, the raven's final momentum of is = m2v2cos(theta) = 1.5 * 9 * cos(theta). There's no motion along the y-components. So equating we have, momentum before collision = momentum after collision of the raven + momentum after collision of the falcon. So we have 12000 = 3000cos(theta) + 13.5cos(theta). Cos(theta)(3000 + 13.5) = 12000. Theta = cos^-1( 12000/3013.5 = 3.98 So theta =</span>
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If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of F_{\text{a}} displaces this spring by x_{\text{a}}, while a force of F_{\text{b}} displaces this spring by x_{\text{b}}, then:

\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}.

In this question, it is given that a force of F_{\text{a}} = 7.0 \; {\rm N} would stretch this spring by x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm}). Thus, the force F_{\text{b}} required to stretch this spring by x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm}) would satisfy:

\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}.

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\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}.

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