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%You need to login to perform this action.
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