A man of mass 40, kg is standing on a trolley | vedclass

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(D) The total mass of trolley $A$ with the man is $M_A = 140\, kg + 40\, kg = 180\, kg$.The mass of trolley $B$ is $M_B = 60\, kg$.According to the la

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Both $(B)$ and $(C)$

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(D) The total mass of trolley $A$ with the man is $M_A = 140\, kg + 40\, kg = 180\, kg$.The mass of trolley $B$ is $M_B = 60\, kg$.According to the law of conservation of linear momentum,the initial momentum of the system is zero,so $M_A v_A + M_B v_B = 0$,which implies $M_A v_A = -M_B v_B$.Taking magnitudes,$180 	imes v_A = 60 	imes v_B$,which gives $v_B = 3 v_A$.Thus,the speed of trolley $B$ is $3$ times that of trolley $A$ immediately after the interaction.Assuming the coefficient of friction $\mu$ is the same for both,the deceleration $a = \mu g$ is the same for both.Using the equation $v^2 = 2as$,the distance $s = v^2 / (2a)$.Since $v_B = 3 v_A$,the distance $s_B = (3 v_A)^2 / (2a) = 9 (v_A^2 / 2a) = 9 s_A$.Therefore,both statements $(B)$ and $(C)$ are correct.

$A$ man of mass $40\, kg$ is standing on a trolley $A$ of mass $140\, kg$. He pushes another trolley $B$ of mass $60\, kg$,so that they are set in motion. Then:

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