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Work, Energy and Power
Relation between Kinetic Energy and Momentum
2 questions from 2 papers, each with its answer.
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Appears in
Concepts tested
- E = p squared over 2m
- Momentum proportional to root of KE
All Questions
2004 AIPMT 1 question
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A particle of mass $m_1$ is moving with a velocity $v_1$ and another particle of mass $m_2$ is moving with a velocity $v_2$. Both of them have the same momentum but their different kinetic energies are $E_1$ and $E_2$ respectively. If $m_1 \gt m_2$ then :Kinetic energy in terms of momentum: $E = \dfrac{1}{2}mv^2 = \dfrac{(mv)^2}{2m} = \dfrac{p^2}{2m}$
Both particles have the same momentum p, so $E \propto \dfrac{1}{m}$
$\dfrac{E_1}{E_2} = \dfrac{m_2}{m_1}$
Since $m_1 \gt m_2$, $\dfrac{E_1}{E_2} \lt 1$
$\Rightarrow E_1 \lt E_2$
2002 AIPMT 1 question
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If kinetic energy of a body is increased by 300% then percentage change in momentum will beMomentum and kinetic energy: $P = \sqrt{2mE}$
Increase of 300 percent means $E_2 = E_1 + 3E_1 = 4E_1$.
$P_2 = \sqrt{2m(4E_1)} = 2\sqrt{2mE_1} = 2P_1$
Percentage change $= \dfrac{P_2 - P_1}{P_1} \times 100 = \dfrac{2P_1 - P_1}{P_1} \times 100 = 100$ percent
So the momentum increases by 100%.