The friction coefficient between the horizont | vedclass

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(A) Given: $m_1 = 2 \ kg$,$m_2 = 4 \ kg$,$\mu = 0.2$,initial distance $d = 0.16 \ m$,initial velocity $u_1 = 1 \ m/s$,$u_2 = 0 \ m/s$.Retardation $a =

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$5$

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(A) Given: $m_1 = 2 \ kg$,$m_2 = 4 \ kg$,$\mu = 0.2$,initial distance $d = 0.16 \ m$,initial velocity $u_1 = 1 \ m/s$,$u_2 = 0 \ m/s$.Retardation $a = \mu g = 0.2 	imes 10 = 2 \ m/s^2$.Velocity of $2 \ kg$ block just before collision:$v^2 = u^2 - 2ad \implies v^2 = (1)^2 - 2(2)(0.16) = 1 - 0.64 = 0.36 \implies v = 0.6 \ m/s$.For a perfectly elastic collision:Conservation of momentum: $m_1 v = m_1 v_1 + m_2 v_2 \implies 2(0.6) = 2v_1 + 4v_2 \implies 0.6 = v_1 + 2v_2$.Coefficient of restitution $e = 1$: $v_2 - v_1 = e(u_1 - u_2) = 1(0.6 - 0) = 0.6$.Solving the equations: $v_1 + 2v_2 = 0.6$ and $v_2 - v_1 = 0.6$.Adding them: $3v_2 = 1.2 \implies v_2 = 0.4 \ m/s$.Then $v_1 = v_2 - 0.6 = 0.4 - 0.6 = -0.2 \ m/s$.Distance traveled by $2 \ kg$ block after collision: $S_1 = \frac{v_1^2}{2a} = \frac{(-0.2)^2}{2(2)} = \frac{0.04}{4} = 0.01 \ m$.Distance traveled by $4 \ kg$ block after collision: $S_2 = \frac{v_2^2}{2a} = \frac{(0.4)^2}{2(2)} = \frac{0.16}{4} = 0.04 \ m$.Since $v_1$ is negative,the $2 \ kg$ block moves left and the $4 \ kg$ block moves right. The final separation is $S_1 + S_2 = 0.01 + 0.04 = 0.05 \ m = 5 \ cm$.

The friction coefficient between the horizontal surface and each of the blocks shown in the figure is $\mu = 0.2$. The collision between the blocks is perfectly elastic. What is the separation between the blocks when they come to rest? (in $cm$)

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