Question

A man m = 80kg is standing on a trolley of mass 320kg on a smooth surface. If man starts walking on trolley along rails at speed 1m/s, then after 4 seconds, his displacement from the ground.

Answer 1

Hint: Here to make the center of mass stay the same, the trolley will move in the direction opposite to the man walking. Here the man has walked the distance of 4m relative to the trolley as $d = s \times t$, where d = distance, s = speed, t = time. So, $d = 4(s) \times 1(m/s) = 4m$. Let’s suppose the trolley travelled x distance relative to ground so the net displacement of the man would be 4-x. The formula for center of mass is${m_1}{x_1} = {m_2}{x_2}$. Where ${m_1}$= Mass of object 1; ${m_2}$= Mass of object 2; ${x_1}$= distance covered by object 1; ${x_2}$= Distance covered by object 2. Apply this formula and solve.

Formula Used:
The formula for center of mass is given as
${m_1}{x_1} = {m_2}{x_2}$
Where:
${m_1}$= Mass of object 1 (80kg)
${m_2}$ = Mass of object 2 (320kg)
${x_1}$= Distance covered by object 1 (4-x)
${x_2}$= Distance covered by object 2 (x)

Complete step by step answer:
Apply the formula for the center of mass
${m_1}{x_1} = {m_2}{x_2}$
Put the given values in the equation and simplify it.
$\implies$ $80 \times (4 - x) = 320 \times x$
Separate x on the RHS and the numerical value on the LHS
$\implies$ $(4 - x) = \dfrac{{320}}{{80}} \times x$
Solve, for the unknown x, which is the distance covered by the train relative to ground.
$\implies$ $(4 - x) = \dfrac{{320}}{{80}} \times x$
Simplify the above equation
$\implies$ \[(4 - x) = 4 \times x\]
$\therefore $ \[x = \dfrac{4}{5}\]
x = 0.8m;
The net displacement of the man is 4-0.8= 3.2m.

Final Answer: The net displacement of the man is 3.2m

Note:We can also solve this by applying conservation of linear momentum. In this question the total net external force on the system is zero. According to the conservation of momentum the momentum of man walking forward would be equal to the momentum of the entire system in the backward direction then only the conservation of linear momentum would be valid.
${m_1} \times {v_1} = {m_2} \times {v_2}$;
Here:
$\implies$ ${m_1}$= Mass of the object 1;
$\implies$ ${v_1}$= Velocity of the object 1;
$\implies$ ${m_2}$= Mass of the object 2;
$\implies$ ${v_2}$= Velocity of the object 2;
Put the value in the above equation and solve,
$\implies$ $80 \times 1 = (80 + 320) \times {v_2}$; ….(Here ${m_2}$= Mass of the man + Mass of the trolley )
Keep${v_2}$ in the RHS and take the rest to LHS.
$\implies$ $\dfrac{{80}}{{80 + 320}} = {v_2}$;
$\implies$ ${v_2} = 0.2m/s$;
Here the velocity of man with respect to ground is = ${v_1} - {v_2}$;
$\implies$ $1 - 0.2 = 0.8m/s$;
Now we the formula for speed, distance and time:
$\implies$ $S = \dfrac{D}{T}$ ;
Rearrange the above equation and solve for D
$\implies$ $S \times T = D$;
$\implies$ $D = 0.8 \times 4$;
$\implies$ D = 3.2m;
One has to apply the concept of center of mass and apply its formula. Here to conserve the center of mass the train is moving in the opposite direction and that is why we subtract the distance covered by the trolley to the distance covered by the man on the trolley.