Respuesta :
As we know that kinetic energy is given as
[tex]KE = \frac{1}{2}mv^2[/tex]
Here we can find the initial speed of puck A by momentum conservation
[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]
[tex]0.252\times v + 0 = 0.252\times (-0.115) + 0.374\times ( 0.650)[/tex]
[tex]v = 0.850 m/s[/tex]
now here we will have initial kinetic energy of the mass is given as
[tex]KE_i = \frac{1}{2}m_1v_1^2[/tex]
[tex]KE_i = \frac{1}{2}(0.252)(0.850^2) = 0.091 J[/tex]
[tex]KE_f = \frac{1}{2}(0.252)(-0.115)^2 + \frac{1}{2}(0.374)(0.650^2)[/tex]
[tex]KE_f = 0.081 J[/tex]
now loss of energy is given as
[tex]KE_i - KE_f = 0.091 - 0.081 = 0.010 J[/tex]
The change in the total kinetic energy of the system that occurs during the collision is 0.010Joules
According to the law of collision;
[tex]m_1u_1+m_2u_2 = m_1v_1+m_2v_2[/tex]
Substituting the given parameters into the equation to get the velocity of puck A, we have;
[tex]0.252u_1+0=0.252(-0.115)+(0.650)(0.374)\\0.252u_1=-0.02898 + 0.2431\\0.252u_1=0.21412\\u_1=\frac{0.21412}{0.252}\\u_1= 0.85m/s[/tex]
Change in kinetic energy = kinetic energy before the collision - kinetic energy after the collision
Change in kinetic energy = [tex]\frac{1}{2}(0.252)(0.115)^2+\frac{1}{2}(0.374)(0.650)^2 - \frac{1}{2}(0.252)(0.86)[/tex]
Change in kinetic energy = [tex]0.091 - 0.081[/tex]
Change in kinetic energy = 0.010 Joules
Hence the change in the total kinetic energy of the system that occurs during the collision is 0.010Joules
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