A ball of mass $m$ is projected upward from a tower with velocity $30 m/s$ and at the same time, another ball of mass $2 m$ is dropped from rest from the same tower. Find the displacement of the $C O M$ in $2 s$ . [Take $g = 10 {m/s}^{2}$ ]

0 m

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10 m

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5 m

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7.5 m

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Solution

The correct option is A $0 m$
Given, $u_{1} = 30 m/s$ (upwards)
and $u_{2} = 0$

We know,
$u_{c o m} = \frac{u_{1} m_{1} + u_{2} m_{2}}{m_{1} + m_{2}}$
$= \frac{30 \times m + 0 \times 2 m}{m + 2 m} = 10 m/s$
Acceleration of both the balls is $- g$ , therefore acceleration of $C O M$ will be - $g$
Let the displacement of $C O M$ be $d$ . Then in $2 s$ ,
$d = u_{c o m} t + \frac{1}{2} a_{c o m} t^{2}$
$= 10 \times 2 + \frac{1}{2} (- 10) \times 2^{2}$
$= 0$

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A ball of mass m is projected upward from a tower with velocity 30 m/s and at the same time, another ball of mass 2m is dropped from rest from the same tower. Find the displacement of the COM in 2 s. [Take g=10 m/s2]

The correct option is A 0 mGiven, u1=30 m/s (upwards) and u2=0 We know, ucom=u1m1+u2m2m1+m2 =30×m+0×2mm+2m=10 m/s Acceleration of both the balls is −g, therefore acceleration of COM will be -g Let the displacement of COM be d. Then in 2 s, d=ucomt+12acomt2 =10×2+12(−10)×22 =0

A ball of mass $m$ is projected upward from a tower with velocity $30 m/s$ and at the same time, another ball of mass $2 m$ is dropped from rest from the same tower. Find the displacement of the $C O M$ in $2 s$ . [Take $g = 10 {m/s}^{2}$ ]