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A man of mass $80\; kg$ is riding on a small cart of mass $40\; kg$ moving with velocity $2m/s$ . He is running on the cart, so that his velocity relative to the cart is $3 m/s$ in the direction opposite to the motion of the cart. What is the speed of center of mass of the system.

\[\begin {array} {1 1} (a)\;1.5\;m/s & \quad (b)\;1\;m/s \\ (c)\;3\;m/s & \quad  (d)\;zero \end {array}\]

1 Answer

Velocity of man relative to cart $V_r=3m/s$
Velocity of man relative to ground $3-2=1m/s$
$\therefore$ Velocity of COM $\overrightarrow {V}cm=\large\frac{m_1\overrightarrow{V_1}+m_2 \overrightarrow {V_2}}{(m_1+m_2)}$
$\qquad= \large\frac{80 \times 1 -40 \times 2}{120}$
$\qquad= 0$
answered Nov 22, 2013 by meena.p

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