Explanation

(b) [image:6_image81.png]
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
$m_{B}v_{B} + 0 = m_{sys.} \times v_{sys.}$
$\Rightarrow v_{sys.} = \frac{m_{B}v_{B}}{m_{sys.}} = \frac{50 \times 10}{50 + 950} = 0.5\ m/s$
Fractional loss in K.E. $= \frac{\frac{1}{2}\ m_{B}v_{B}^{2} - \frac{1}{2}m_{sys.}v_{sys.}^{2}}{\frac{1}{2}m_{B}v_{B}^{2}}$
By substituting $m_{B} = 50 \times 10^{- 3}kg,\ v_{B} = 10\ m/s$
$m_{sys.} = 1kg,\ v_{s} = 0.5\ m/s\ $ we get
Fractional loss $= \frac{95}{100}\ \therefore$ Percentage loss $= 95\%$