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A bag of sand of mass $9.8 \mathrm{~kg}$ is suspended by a rope. A bullet of $200 \mathrm{~g}$ travelling with speed $10 \mathrm{~ms}^{-1}$ gets embedded in it, then loss of kinetic energy will be :
Option: 1
$4.9\: \mathrm{J}$

Option: 2
$9.8\: \mathrm{J}$

Option: 3
$14.7\: \mathrm{J}$

Option: 4
$19.6\: \mathrm{J}$

Answers (1)

best_answer

$\mathrm{M_{1}=9.8kg}$ $\mathrm{v_{0}=10\: m/s}$

$\mathrm{M_{2}=0.2 kg}$

By linear momentum conservation ,

$\mathrm{m_1 v_0+m_2(0)=\left(m_1+m_2\right) v}$

$\mathrm{v=\frac{m_1 v_0}{\left(m_1+m_2\right)}=0.2 \mathrm{~m} / \mathrm{s}}$

The loss of kinetic energy $\mathrm{=K E_i-K E_f}$

$\mathrm{=\left[\frac{1}{2} m_1 v_0^2-\frac{1}{2}\left(m_1+m_2\right) v^2\right]}$

$\mathrm{=\left[\frac{1}{2} \times 0.2 \times 100\frac{-1}{2} \times 10 \times \frac{4}{100}\right]}$

$\mathrm{=10-0.2}$

$\mathrm{=9.8 \mathrm{~J}}$

Hence (2) is correct option

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