Given,
The velocity of the joggers, v=-3.50 m/s
The mass of Jim, M=100 kg
The mass of Tom, m=59 kg
(a) The kinetic energy of the system is given by,
[tex]\begin{gathered} E_a=\frac{1}{2}mv^2+\frac{1}{2}Mv^2 \\ =\frac{1}{2}v^2(m+M) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} E_a=\frac{1}{2}\times3.50^2(59.0+100) \\ =973.88\text{ J} \end{gathered}[/tex]Thus the kinetic energy of the system is 973.88 J
(b)
The total momentum of the system is given by,
[tex]\begin{gathered} p_b=mv+Mv \\ =(m+M)v \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} p_b=(59+100)\times3.50 \\ =556.5\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Thus the total momentum of the system is 556.5 kg m/s
(c)
Given that the velocity of Tom is -v
The total kinetic energy of the system is given by,
[tex]\begin{gathered} E_c=\frac{1}{2}Mv^2+\frac{1}{2}m(-v)^2 \\ =\frac{1}{2}(Mv^2+m(-v)^2) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} E_c=\frac{1}{2}(100\times3.50^2+59\times(-3.50)^2) \\ =973.88\text{ J} \end{gathered}[/tex]Thus the total kinetic energy of the system, in this case, is 973.88 J
(d)
The total momentum of the system is given by,
[tex]\begin{gathered} p_d=Mv+m(-v) \\ =v(M-m) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} p_d=3.50(100-59) \\ =143.5\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Thus the total momentum of the system, in this case, is 143.5 kg m/s
