Question

two bodies of mass m1 and m2 have same kinetic energy . Find the ratio of their momentum

Answer 1

hello friend...!!!

⇒ according to the question we should calculate the ratio of momentum ,

⇒ given the kinetic energies are equal 

⇒ $KE_{1}$  = $\frac{ P_{1}^{2} }{ 2m_{1} }$

⇒$KE_{2}$  = $\frac{ P_{2} ^{2} }{ 2m_{2} }$

since $KE_{1}$  = $KE_{2}$

⇒ $\frac{ P_{1} ^{2} }{ 2m_{1} }$  = $\frac{ P_{2} ^{2} }{ 2m_{2} }$

⇒ $\frac{ P_{1} ^{2} }{ P_{2} ^{2} }$  =  $\frac{ 2m_{1} }{ 2m_{2} }$

⇒ $\frac{ P_{1} }{ P_{2} }$  = $\sqrt[]{ \frac{ m_{1} }{ m_{2} } }$

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hope it helps..!!

Answer 2

$Hey \: \: there \\ \\ Let \: KE \: and \: KE' \: are \: the \: two \: kinetic \: energies \: with \: Mass \: M \: and \: m \\ \\ \\ we \: can \: write \: it \: as \\ \\ KE \: \: = \frac{ {p \: }^{2} }{m} \\ and \: \\ \\ KE' \: = \frac{ {P \: }^{2} \: }{M \: } \\ \\ it \: says \: \\ \\ KE \: = KE' \: \\ \\ = \frac{ {p \: }^{2} }{m} \: = \frac{ {P \: }^{2} \: }{M \: } \: \\ \\ = \frac{{p \: }^{2} \: }{{P \: }^{2} \: } = \frac{m}{M} \: \\ \\ = \frac{p}{P \: } \: \: \: \: = \sqrt{ \frac{m}{M} } \: \: \: \: \: \: \: ( \: since \: \frac{ {a}^{2} }{ {b}^{2} } \: = ({ \frac{a}{b} })^{2} )$



Hope this helps ya ☺