Answer:
No the given statement is not necessarily true.
Explanation:
We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by
[tex]K.E=\frac{1}{2}mv^{2}[/tex]
Similarly the momentum is given by [tex]m\times v[/tex]
For 2 particles with masses [tex]m_{1},m_{2}[/tex]and moving with velocities [tex]v_{1},v_{2}[/tex] respectively the respective kinetic energies is given by
[tex]K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}[/tex]
[tex]K.E_{2}=\frac{1}{2}m_{2}v_{2}^{2}[/tex]
Similarly For 2 particles with masses [tex]m_{1},m_{2}[/tex]and moving with velocities [tex]v_{1},v_{2}[/tex] respectively the respective momenta are given by
[tex]p_{1}=m_{1}\times v_{1}[/tex]
[tex]p_{2}=m_{2}\times v_{2}[/tex]
Now since it is given that the two kinetic energies are equal thus we have
[tex]\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)[/tex]
Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.
