Respuesta :
In the vertical direction
there is no change of momentum.
Momentum in vertical = 1.5 * 9 = 13.5
Speed in vertical = 9 m/s
Momentum of Raven in x + momentum of falcon = initial momentum in x
=> Mx + 600*5 = 600*20
=> Mx = 600*15
=> Vx = 6000
Vy = 9
=> tanθ = 0.6 kg/9 = 0.066667
=> θ = 48.0128 degrees
The angle of raven changes to [tex]\boxed{48.01^\circ}[/tex] from initial direction after impact.
Further Explanation:
A collision is a phenomenon where, two or more object exerts forces on each other or colloid each other in comparatively short time.
An elastic collision is a collision where, after collision the net kinetic energy of the body remains constant. The kinetic energy and momentum is always conserved in the elastic collision.
An inelastic collision is a collision where, after collision the net kinetic energy of the body does not remain constant. The momentum is always conserved but the net kinetic energy does not conserve in an inelastic collision.
Given:
The mass of the is [tex]600\text{ g}[/tex].
Mass of the raven is [tex]1.5\text{ kg}[/tex].
Velocity of the falcon is [tex]20.0\text{ m/s}[/tex].
Velocity of the raven is [tex]9.00\text{ m/s}[/tex].
Concept:
Here the two birds undergo an elastic collision.
During impact the falcon strikes the raven at right angle and after impact the falcon bounce back as shown in the Figure 1.
Before impact the falcon moving towards positive y-axis direction and the raven is flying towards positive x-axis direction and after impact the falcon flying toward the negative y-axis direction.
Apply conservation of linear momentum.
[tex]\fbox{\begin\\{m_1}{\vec v_1} + {m_2}{\vec v_2}={m_1}{\vec v_1}^\prime + {m_2}{\vec v_2}^\prime\end{minispace}}[/tex]
Here, [tex]{m_1}[/tex] is the mass of the falcon; [tex]{v_1}[/tex] is the velocity of falcon, [tex]{m_2}[/tex] is the mass of the raven, [tex]{v_2}}[/tex] is the velocity of raven, [tex]{v_1}^\prime[/tex] is the velocity of falcon after collision and [tex]{v_2}^\prime[/tex] is the velocity of raven after collision.
Substitute for [tex]600\text{ g}[/tex] for [tex]{m_1}[/tex], [tex]20.0\text{ m/s}[/tex] for [tex]{v_1}[/tex], [tex]1.5\text{ kg}[/tex] for[tex]{m_2}[/tex], [tex]9.00\text{ m/s}[/tex] for [tex]{v_2}[/tex] and [tex]5.00\text{ m/s}[/tex] for [tex]{v_1}^\prime[/tex] in the above equation.
[tex]\left( {600{\text{ g}}\left( {\frac{{1{\text{ kg}}}}{{1000{\text{ g}}}}} \right)} \right)\left( {20.0{\text{ m/s }}\hat j} \right) + \left( {1.5{\text{ kg}}} \right)\left( {9.00{\text{ m/s }}\hat i} \right) = \left( {600{\text{ g}}\left( {\frac{{1{\text{ kg}}}}{{1000{\text{ g}}}}} \right)} \right)\left( {5.00{\text{ m/s}}\left( { - {\text{ }}\hat j} \right)} \right) + \left( {1.5{\text{ kg}}} \right){v_2}^\prime[/tex]
Simplify further.
[tex]{v_2}^\prime=9{\text{ m/s }}\hat i + 10{\text{ m/s }}\hat j[/tex]
The magnitude and the angle of raven after impact is given by:
[tex]\begin{aligned}{v_2}^\prime&=\sqrt {{{10}^2} + {9^2}} {\text{ m/s}}\angle{\tan ^{ - 1}}\left({\frac{{10}}{9}} \right)\\ &=\sqrt {181} {\text{ m/s}}\angle 48.01^\circ\\&=13.45{\text{ m/s}}\angle 48.01^\circ\\ \end{aligned}[/tex]
Therefore, the angle of raven changes to [tex]\boxed{48.01^\circ}[/tex] from initial direction after impact.
Learn more:
1. Change in the momentum https://brainly.com/question/9484203
2. Variables of momentum https://brainly.com/question/11604162
3 Change in the momentum. https://brainly.com/question/11943300
Answer Details:
Grade: High school
Subject: Physics
Chapter: Kinematics
Keywords:
Protecting, his, nest, 600 g, peregrine, falcon, rams, marauding, 1.5 kg, raven, moving, 20.0 m/s, 9.00 m/s, moment, impact, strikes, right angle, direction, rebounds, straight, back, 5.00 m/s, angle, 48.01 deg, 13.45 m/s.
