A 50 g bullet moving with velocity 10 m/s strikes a block of mass 950 g at rest

Question

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

Options:

A . 100%

B . 95%

C . 5%

D . 50%

Answer: Option B
:
B

Initial K.E. of system = K.E. of the bullet =

\frac{1}{2} m_{s} v_{s}^{2}

By the law of conservation of linear momentum

m_{s} v_{s} + 0 = m_{s y s} \times v_{s y s}

\Rightarrow v_{s y s} = \frac{m_{s} v_{s}}{m_{s y s}} = \frac{50 \times 10}{50 + 950} = 0.5 m / s

Fractional loss in K.E. =

\frac{\frac{1}{2} m_{s} v_{s}^{2} - \frac{1}{2} m_{s y s} v_{s y s}^{2}}{\frac{1}{2} m_{s} v_{s}^{2}}

By substituting

m_{s} = 50 \times 10^{- 3} k g, v_{s} = 10 m / s

m_{s y s} = 1 k g, v_{s y s} = 0.5 m / s

we get
Fractional loss =

\frac{95}{100}

∴

Percentage loss = 95 %

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