Two bodies m₁ and m₂ (m₁ > m₂) have the same kinetic energy. Then their momentum p₁ and p₂ satisfy

Concept:

Kinetic energy of a mass m moving with velocity v is equal to,

Kinetic energy, \(K.E.\; = \frac{1}{2}m{v^2}\)

Momentum of a mass m moving with velocity v is equal to,

Momentum = mv

Calculation:

Given:

m1 > m2 and,

Two bodies have the same kinetic energy

\(\frac{1}{2}{m_1}v_1^2 = \frac{1}{2}{m_2}v_2^2\)

We can rewrite it as,

\(\frac{1}{2} × \frac{{{{\left( {{m_1} × {v_1}} \right)}^2}}}{{{m_1}}}=\frac{1}{2} × \frac{{{{\left( {{m_2} × {v_2}} \right)}^2}}}{{{m_2}}}\)

\(\frac{{p_1^2}}{{{m_1}}} = \frac{{p_2^2}}{{{m_2}}}\)

\(\frac{{{p_1}}}{{{p_2}}} = \sqrt {\frac{{{m_1}}}{{{m_2}}}}\) ----------(3)

Since m1 > m2 RHS becomes greater than 1, which implies

\(\frac{{{p_1}}}{{{p_2}}}\) > 1

p1 > p2

Hence, for the given condition \(\frac{{{p_1}}}{{{p_2}}}\) > 1 will satisfy the relation between momentums.