A person of mass 80 kg is standing on a trol | vedclass

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(C) Let the mass of the person be $m_p = 80 \ kg$ and the mass of the trolley be $m_t = 320 \ kg$.Since there is no external horizontal force on the s

Vedclass · Accepted Answer

$3.2$

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(C) Let the mass of the person be $m_p = 80 \ kg$ and the mass of the trolley be $m_t = 320 \ kg$.Since there is no external horizontal force on the system,the total momentum of the system remains zero.Let $v_p$ be the velocity of the person relative to the trolley $(1 \ m/s)$ and $v_t$ be the velocity of the trolley relative to the ground.The velocity of the person relative to the ground is $v_g = v_p - v_t$.According to the conservation of linear momentum: $m_p v_g + m_t v_t = 0$.$80(1 - v_t) + 320 v_t = 0$.$80 - 80 v_t + 320 v_t = 0$.$240 v_t = -80 \implies v_t = -1/3 \ m/s$ (The trolley moves backward).The velocity of the person relative to the ground is $v_g = 1 - 1/3 = 2/3 \ m/s$.Displacement relative to the ground after $t = 4 \ s$ is $d = v_g 	imes t = (2/3) 	imes 4 = 8/3 \approx 2.67 \ m$.Wait,re-evaluating the standard interpretation: If the person walks at $1 \ m/s$ relative to the trolley,the velocity of the center of mass remains zero. The displacement of the person relative to the ground is $d_p = v_g 	imes t$. Given the options provided,the calculation $0.8 	imes 4 = 3.2$ suggests the velocity of the trolley is $0.2 \ m/s$ derived from $80 	imes 1 = (80+320)v_t$. Thus,$v_g = 1 - 0.2 = 0.8 \ m/s$. Displacement = $0.8 	imes 4 = 3.2 \ m$.

$A$ person of mass $80 \ kg$ is standing on a trolley of mass $320 \ kg$. The trolley is at rest on a frictionless horizontal rail. If the person walks on the trolley at a speed of $1 \ m/s$,what will be their displacement relative to the ground after $4 \ s$?

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