A $50\, g$ bullet moving with velocity $10\, m/s$ strikes a block of mass $950 \,g$ at rest and gets embedded in it. The loss in kinetic energy will be ............... $\%$
A $50\, g$ bullet moving with velocity $10\, m/s$ strikes a block of mass $950 \,g$ at rest and gets embedded in it. The loss in kinetic energy will be ............... $\%$
$100$
$95$
$5$
$50$
A $50\, g$ bullet moving with velocity $10\, m/s$ strikes a block of mass $950 \,g$ at rest and gets embedded in it. The loss in kinetic energy will be ............... $\%$
Initial K.E. of system = K.E. of the bullet =$\frac{1}{2}{m_B}v_B^2$
By the law of conservation of linear momentum
${m_B}{v_B} + 0 = {m_{{\rm{sys}}{\rm{.}}}} \times {v_{{\rm{sys}}{\rm{.}}}}$
==> ${v_{{\rm{sys}}{\rm{.}}}} = \frac{{{m_B}{v_B}}}{{{m_{{\rm{sys}}{\rm{.}}}}}} = \frac{{50 \times 10}}{{50 + 950}} = 0.5\;m/s$
Fractional loss in K.E. = $\frac{{\frac{1}{2}{m_B}v_B^2 - \frac{1}{2}{m_{{\rm{sys}}{\rm{.}}}}v_{{\rm{sys}}{\rm{.}}}^2}}{{\frac{1}{2}{m_B}v_B^2}}$
By substituting ${m_B} = 50 \times {10^{ - 3}}kg,\;{v_B} = 10\;m/s$
${m_{{\rm{sys}}{\rm{.}}}} = 1kg,\;{v_s} = 0.5\;m/s$ we get
Fractional loss = $\frac{{95}}{{100}}$
Percentage loss $= 95\%$
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