A ball of mass 1.2 kg moving with a velocity | vedclass

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(A) Given: $m_1 = m_2 = m = 1.2 \ kg$,$u_1 = 12 \ ms^{-1}$,$u_2 = 0$,$e = \frac{1}{\sqrt{2}}$.By the law of conservation of linear momentum:$m u_1 + m

Vedclass · Accepted Answer

$3: 4$

Vedclass · Answer

(A) Given: $m_1 = m_2 = m = 1.2 \ kg$,$u_1 = 12 \ ms^{-1}$,$u_2 = 0$,$e = \frac{1}{\sqrt{2}}$.By the law of conservation of linear momentum:$m u_1 + m u_2 = m v_1 + m v_2$$12 + 0 = v_1 + v_2 \Rightarrow v_1 + v_2 = 12$  ...$(i)$By the definition of the coefficient of restitution $(e)$:$e = \frac{v_2 - v_1}{u_1 - u_2} \Rightarrow \frac{1}{\sqrt{2}} = \frac{v_2 - v_1}{12}$$v_2 - v_1 = \frac{12}{\sqrt{2}} = 6\sqrt{2}$  ...(ii)Adding equations $(i)$ and (ii):$2v_2 = 12 + 6\sqrt{2} \Rightarrow v_2 = 6 + 3\sqrt{2} \ ms^{-1}$Subtracting equation (ii) from $(i)$:$2v_1 = 12 - 6\sqrt{2} \Rightarrow v_1 = 6 - 3\sqrt{2} \ ms^{-1}$Initial kinetic energy $(KE)_i = \frac{1}{2} m u_1^2 = \frac{1}{2} 	imes 1.2 	imes (12)^2 = 0.6 	imes 144 = 86.4 \ J$.Final kinetic energy $(KE)_f = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 = \frac{1}{2} m (v_1^2 + v_2^2)$.$v_1^2 + v_2^2 = (6 - 3\sqrt{2})^2 + (6 + 3\sqrt{2})^2 = (36 + 18 - 36\sqrt{2}) + (36 + 18 + 36\sqrt{2}) = 54 + 54 = 108$.$(KE)_f = \frac{1}{2} 	imes 1.2 	imes 108 = 0.6 	imes 108 = 64.8 \ J$.Ratio $\frac{(KE)_f}{(KE)_i} = \frac{64.8}{86.4} = \frac{648}{864} = \frac{3}{4} = 3:4$.

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