Answer:
a) [tex]v=7.32m/s[/tex]
b) [tex]\alpha =-35[/tex]Âș
c) [tex]ÎK=-1094.62J[/tex]
Explanation:
From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together
So, If we analyze the collision we got
[tex]p_{1x}=p_{2x}[/tex]
To simplify the problem, lets name D for Daniel and R for Rebecca
a) [tex]p_{D1x}+p_{R1x}=p_{D2x}+p_{R2x}[/tex]
Since Daniel's initial velocity is 0
[tex]m_{R}v_{Rx}=m_{D}v_{D2x}+m_{R}v_{R2x}[/tex]
[tex]v_{D2x}=\frac{m_{R}*v_{R1x}-m_{R}*v_{R2x}}{m_{D}}[/tex]
[tex]v_{D2x}=\frac{(45kg)(14m/s)-(45kg)(8cos(55.1)m/s)}{(70kg)}=6m/s[/tex]
Now, lets analyze the movement in the vertical direction
[tex]p_{1y}=p_{2y}[/tex]
Since [tex]p_{1y}=0[/tex]
[tex]0=m_{D}v_{D2y}+m_{R}v_{R2y}[/tex]
[tex]v_{D2y}=-\frac{m_{R}v_{2Ry}}{m_{D}}=-\frac{(45kg)(8sin(55.1)m/s)}{(70kg)}=-4.21m/s[/tex]
Now, we can find the magnitude of Daniel's velocity after de collision
[tex]v_{D}=\sqrt{(6m/s)^2+(-4.21m/s)^2}=7.32m/s[/tex]
b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle
[tex]\alpha =tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}(\frac{-4.21}{6})=-35[/tex]Âș
c) The change in the total kinetic energy is:
ÎK=[tex]K_{2}-K_{1}[/tex]
ÎK=[tex]\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J[/tex]
That means that the kinetic energy decreases
(a) The magnitude of Daniel's final velocity after the collision is 7.37 m/s.
(b) The direction of Daniel's velocity after the collision is 35â°.
(c) The change in the total kinetic energy of the two skaters as a result of the collision is -1,068.9 J.
The given parameters;
The magnitude of Daniels final velocity after collision is calculated by applying the principle of conservation of linear momentum;
The conservation of the linear momentum in horizontal direction;
[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\70(0) + 45(14) = 70v_1_x+ 45(8\times cos(55.1))\\\\630 = 70v_1 + 205.9\\\\630 - 205.9 = 70v_1_x\\\\v_1_x = \frac{424.1}{70} \\\\v_1 _x= 6.05 m/s[/tex]
The conservation of the linear momentum in vertical direction;
[tex]70(0) +45(0) = 70v_1_y + 45(8\times sin(55.1))\\\\70v_1_y = - 295.2\\\\v_1_y = \frac{-295.2}{70} \\\\v_1_y = -4.217 \ m/s[/tex]
The resultant velocity of Daniel is calculated as follows;
[tex]v_1 = \sqrt{v_1_x^2 + v_1_y^2} \\\\v_1 = \sqrt{6.05^2 + (-4.217)^2} \\\\v_1 = 7.37 \ m/s[/tex]
The direction of Daniel's velocity is calculated as follows;
[tex]\theta = tant^{-1} (\frac{v_1_y}{v_1_x} )\\\\\theta = tant^{-1} (\frac{4.217}{6.05} )\\\\\theta = 35^0[/tex]
The initial kinetic energy of the skaters;
[tex]K.E_i = \frac{1}{2} m_1u_1^2 + \frac{1}{2} m_2u_2^2\\\\K.E_i = (0.5\times 70\times 0^2) + (0.5 \times 45\times 14^2)\\\\K.E_i = 4410 \ J[/tex]
The final kinetic energy of the two skaters;
[tex]K.E_f = \frac{1}{2} m_1v_1^2 + \frac{1}{2} m_2v_2^2\\\\K.E_f = (0.5\times 70\times 7.37^2) + (0.5\times 45 \times 8^2)\\\\K.E_f = 3341.1 \ J[/tex]
The change in kinetic energy of the skaters is calculated as follows;
[tex]\Delta K.E = K.E_f - K.E_i\\\\\Delta K.E = 3341.1- 4410\\\\\Delta K.E = -1,068.9 \ J[/tex]
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