Answer:
The final velocity of the first glider is 0.27 m/s in the same direction as the first glider
The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.
0.010935 J
0.0858675 J
Explanation:
[tex]m_1[/tex] = Mass of first glider = 0.3 kg
[tex]m_2[/tex] = Mass of second glider = 0.15 kg
[tex]u_1[/tex] = Initial Velocity of first glider = 0.8 m/s
[tex]u_2[/tex] = Initial Velocity of second glider = 0 m/s
[tex]v_1[/tex] = Final Velocity of first glider
[tex]v_2[/tex] = Final Velocity of second glider
As momentum and Energy is conserved
[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}[/tex]
[tex]{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}[/tex]
From the two equations we get
[tex]v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.3-0.15}{0.3+0.15}\times 0.8+\frac{2\times 0.15}{0.3+0.15}\times 0\\\Rightarrow v_1=0.27\ m/s[/tex]
The final velocity of the first glider is 0.27 m/s in the same direction as the first glider
[tex]v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.3}{0.3+0.15}\times 0.8+\frac{0.3-0.15}{0.3+0.15}\times 0\\\Rightarrow v_2=1.067\ m/s[/tex]
The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.
Kinetic energy is given by
[tex]K=\frac{1}{2}m_1v_1^2\\\Rightarrow K=\frac{1}{2}0.3\times 0.27^2\\\Rightarrow K=0.010935\ J[/tex]
Final kinetic energy of first glider is 0.010935 J
[tex]K=\frac{1}{2}m_2v_2^2\\\Rightarrow K=\frac{1}{2}0.15\times 1.07^2\\\Rightarrow K=0.0858675\ J[/tex]
Final kinetic energy of second glider is 0.0858675 J
The magnitude of final velocity of the first glider is 0.27 m/s in positive direction.
The magnitude of final velocity of the second glider is 1.07 m/s in positive direction.
The final kinetic energy of the first glider is 0.011 J.
The final kinetic energy of the second glider is 0.086 J.
The given parameters;
The final velocity of the first glider is calculated by applying the principle of conservation of linear momentum as follows;
[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\0.3(0.8) + 0.15(0) = 0.3v_1 + 0.15v_2\\\\0.24 = 0.3v_1 + 0.15v_2[/tex]
Apply the principle of one-dimensional velocity;
[tex]u_1 + v_1 = u_2 + v_2\\\\0.8 + v_1 = 0 + v_2\\\\v_2 = 0.8 + v_1[/tex]
The final velocity of the first glider is calculated as follows;
[tex]0.24 = 0.3v_1 + 0.15(0.8 + v_1)\\\\0.24 = 0.3v_1 + 0.12 + 0.15v_1\\\\0.12 = 0.45v_1\\\\v_1 = \frac{0.12}{0.45} \\\\v_1 = 0.27 \ m/s[/tex]
Thus, the final velocity of the first glider occurs in the positive direction.
The final velocity of the second glider is calculated as follows;
[tex]v_2 = 0.8 + v_1\\\\v_2 = 0.8 + 0.27\\\\v_2 = 1.07 \ m/s[/tex]
Thus, the final velocity of the second glider occurs in the positive direction.
The final kinetic energy of the first glider is calculated as follows;
[tex]K.E_f_1 = \frac{1}{2} \times 0.3 \times 0.27^2\\\\K.E_f_1 = 0.011 \ J[/tex]
The final kinetic energy of the second glider is calculated as follows;
[tex]K_f_2 = \frac{1}{2} \times (0.15) \times (1.07)^2\\\\K.E_f_2 = 0.086 \ J[/tex]
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