Question

Two bodies of equal masses move with uniform velocities v and 3v respectively.what is the ratio of their kinetic energies?

Answer 1

given,

$m_{1} = m_{2}$

$v_{1} = v$

$v_{2} = 3v$

$K.E_{1}= \frac{1}{2}m_{1} v_{1} ^{2}$

$K.E_{2}= \frac{1}{2} m_{2} v_{2}^{2}$

$\frac{K.E_{1} }{K.E_{2} } =\frac{\frac{1}{2} m_{1} v_{1}^{2} }{\frac{1}{2} m_{2} v_{2}^{2} }$

$\frac{K.E_{1} }{K.E_{2} } = \frac{v_{1}^{2} }{v_{2}^{2}  }$

$\frac{K.E_{1} }{K.E_{2} } = \frac{v^{2} }{9v^{2} }$

$K.E_{1} :K.E_{2} = 1:9$

Answer 2

Let the mass of each body be m

KE of the first body ,k1=1/2 mv^2

KE of the second body k2=1/2m(3v)^2

Ratio of KE =1/2mv^2÷1/2 m9v^2

1/9

1:9