A 1 kg ball moving with a speed of 6 ms^{-1 | vedclass

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(C) The energy lost in an inelastic collision is given by the formula: $\Delta KE = \frac{m_1 m_2}{2(m_1 + m_2)} (1 - e^2) (u_1 + u_2)^2$. Given: $m_1

Vedclass · Accepted Answer

$33.3$

Vedclass · Answer

(C) The energy lost in an inelastic collision is given by the formula: $\Delta KE = \frac{m_1 m_2}{2(m_1 + m_2)} (1 - e^2) (u_1 + u_2)^2$. Given: $m_1 = 1 \ kg$,$m_2 = 0.5 \ kg$,$u_1 = 6 \ ms^{-1}$,$u_2 = 9 \ ms^{-1}$ (since they move in opposite directions,the relative velocity is $u_1 + u_2$),and $e = \frac{1}{3}$. Substituting the values: $\Delta KE = \frac{1 	imes 0.5}{2(1 + 0.5)} \left(1 - (\frac{1}{3})^2ight) (6 + 9)^2$ $\Delta KE = \frac{0.5}{3} 	imes (1 - \frac{1}{9}) 	imes (15)^2$ $\Delta KE = \frac{1}{6} 	imes \frac{8}{9} 	imes 225$ $\Delta KE = \frac{8}{54} 	imes 225 = \frac{4}{27} 	imes 225 = 33.33 \ J$.

$A$ $1 \ kg$ ball moving with a speed of $6 \ ms^{-1}$ collides head-on with a $0.5 \ kg$ ball moving in the opposite direction with a speed of $9 \ ms^{-1}$. If the coefficient of restitution is $\frac{1}{3}$,then the energy lost in the collision is (in $J$)

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