question_answer

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                      [MP PET 1994]

A)             100%  

B)             95%

C)             5%      

D)             50%

Correct Answer: B

Solution :

    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 \[{{m}_{B}}{{v}_{B}}+0={{m}_{\text{sys}\text{.}}}\times {{v}_{\text{sys}\text{.}}}\] Þ \[{{v}_{\text{sys}\text{.}}}=\frac{{{m}_{B}}{{v}_{B}}}{{{m}_{\text{sys}\text{.}}}}=\frac{50\times 10}{50+950}=0.5\ m/s\] Fractional loss in K.E.  = \[\frac{\frac{1}{2}{{m}_{B}}v_{B}^{2}-\frac{1}{2}{{m}_{\text{sys}\text{.}}}v_{\text{sys}\text{.}}^{2}}{\frac{1}{2}{{m}_{B}}v_{B}^{2}}\] By substituting \[{{m}_{B}}=50\times {{10}^{-3}}kg,\ {{v}_{B}}=10\ m/s\]                       \[{{m}_{\text{sys}\text{.}}}=1kg,\ {{v}_{s}}=0.5\ m/s\] we get Fractional loss = \[\frac{95}{100}\] \ Percentage loss = 95%