Work Energy And Power Ques 2
2. 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 $\mathrm{E}_1$ and $\mathrm{E}_2$ respectively. If $m_1>m_2$ then
[2004]
(a) $E_1=E_2$
(b) $E_1<E_2$
(c) $\frac{E_1}{E_2}=\frac{m_1}{m_2}$
(d) $E_1>E_2$
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Answer:
Correct Answer: 2.(b)
Solution: (b) $E=\frac{p^2}{2 m}$
or, $E_1=\frac{p_1^2}{2 m_1}, E_2=\frac{p_2^2}{2 m_2}$
or, $m_1=\frac{p_1^2}{2 E_1}, m_2=\frac{p_2^2}{2 E_2}$
$m_1>m_2 \Rightarrow \frac{m_1}{m_2}>1$
$\therefore \frac{p_1^2 E_2}{E_1 P_2^2}>1 \Rightarrow \frac{E_2}{E_1}>1 \quad\left[\because p_1=p_2\right]$
or, $E_2>E_1$
Kinetic energy of body of mass $m$ moving with velocity $v$ is
$ \begin{aligned} \text { K.E. } & =\frac{1}{2} m v^2 \\ \Rightarrow \text { K.E. } & =\frac{1}{2} \frac{m v^2 \times m}{m} \end{aligned} $
[Multiplying both numerator and denominator by $m$ ]
$ \begin{aligned} & \Rightarrow \text { K.E. }=\frac{1}{2} \frac{m^2 v^2}{m} \ & \Rightarrow \text { K.E. }=\frac{1}{2} \frac{p^2}{m} \quad[\because \text { Momentum, } P=m v] \\ & \Rightarrow P^2=2 m \text { K.E. } \\ & \Rightarrow P=\sqrt{2 m K . E} \end{aligned} $
BETA
