# Collisions

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# 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 (in %)

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Solution

## Initial K.E. of system = K.E. of the bullet = $\frac{1}{2}{\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}^{2}$By the law of conservation of linear momentum ${\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}+0={\mathrm{m}}_{\text{sys.}}×{\mathrm{v}}_{\text{sys.}}$$⇒{\mathrm{v}}_{\text{sys.}}=\frac{{\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}}{{\mathrm{m}}_{\text{sys.}}}=\frac{50×10}{50+950}=0.5\text{ }\mathrm{m}\text{​}/\text{​}\mathrm{s}$Fractional loss in K.E.  =  $\frac{\frac{1}{2}{\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}^{2}-\frac{1}{2}{\mathrm{m}}_{\text{sys.}}{\mathrm{v}}_{\text{sys.}}^{2}}{\frac{1}{2}{\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}^{2}}$By substituting  ${\mathrm{m}}_{\mathrm{B}}=50×{10}^{-3}\mathrm{kg},\text{ }{\mathrm{v}}_{\mathrm{B}}=10\text{ }\mathrm{m}\text{​}/\text{​}\mathrm{s}$                      ${\mathrm{m}}_{\text{sys.}}=1\mathrm{kg},\text{ }{\mathrm{v}}_{\mathrm{s}}=0.5\text{ }\mathrm{m}\text{​}/\text{​}\mathrm{s}$ we get Fractional loss =   Percentage loss = 95%

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