Q152 / 160 · TS EAMCET 2014 · Physics

A cannon shell fired breaks into two equal parts at its highest point. One part retraces the path to the cannon with kinetic energy $E_1$ and kinetic energy of the second part is $E_2$, relation between $E_1$ and $E_2$ is

  • A
    $E_2=15 E_1$
  • B
    $E_2=E_1$
  • C
    $E_2=4 E_1$
  • Correct answer
    D
    $E_2=9 E_1$

Solution

At highest point $m u \cos \theta=-\frac{m}{2} u \cos \theta+\frac{m}{2} v$ $m u \cos \theta+\frac{m}{2} u \cos \theta=\frac{m}{2} v$ $\frac{2 m u \cos \theta+m u \cos \theta}{2}=\frac{m}{2} v$ $$ \begin{aligned} m v & =3 m u \cos \theta \\ v & =3 u \cos \theta \end{aligned} $$ The kinetic energy $$ \begin{aligned} & E_1=\frac{1}{2} \times \frac{m}{2} u^2 \cos ^2 \theta \\ & E_1=\frac{1}{4} m u^2 \cos ^2 \theta \end{aligned} $$ Similarly, $$ \begin{aligned} & E_2=\frac{1}{2} \times \frac{m}{2} \times 9 u^2 \cos ^2 \theta \\ & E_2=\frac{9}{4} m u^2 \cos ^2 \theta \end{aligned} $$ The relation between the $E_1$ and $E_2$ $$ \begin{aligned} & E_2=9 \times E_1 \\ & E_2=9 E_1 \end{aligned} $$ (from the Eq. (ii))

Hence, the correct answer is option: 'D'.

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