A body of mass m moving with velocity v makes | vedclass

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Question

(B) According to the law of conservation of linear momentum,the total momentum before collision equals the total momentum after collision.$mv + (2m)(0

Vedclass · Accepted Answer

$\frac{200}{3}\,\%$

Vedclass · Answer

(B) According to the law of conservation of linear momentum,the total momentum before collision equals the total momentum after collision.$mv + (2m)(0) = (m + 2m)v'$$mv = 3mv'$$v' = \frac{v}{3}$The initial kinetic energy is $KE_i = \frac{1}{2}mv^2$.The final kinetic energy is $KE_f = \frac{1}{2}(m + 2m)(v')^2 = \frac{1}{2}(3m)(\frac{v}{3})^2 = \frac{1}{2}(3m)(\frac{v^2}{9}) = \frac{1}{6}mv^2$.The loss in kinetic energy is $\Delta KE = KE_i - KE_f = \frac{1}{2}mv^2 - \frac{1}{6}mv^2 = \frac{3-1}{6}mv^2 = \frac{2}{6}mv^2 = \frac{1}{3}mv^2$.The percentage loss in kinetic energy is $\frac{\Delta KE}{KE_i} 	imes 100 = \frac{\frac{1}{3}mv^2}{\frac{1}{2}mv^2} 	imes 100 = \frac{2}{3} 	imes 100 = \frac{200}{3}\,\%$.

$A$ body of mass $m$ moving with velocity $v$ makes a head-on perfectly inelastic collision with another body of mass $2m$ which is initially at rest. What is the percentage loss in the kinetic energy?

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