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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

10 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7.5 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $2 s$ ,
$d = u_{c o m} t + \frac{1}{2} a_{c o m} t^{2}$
$= 10 \times 2 + \frac{1}{2} (- 10) \times 2^{2}$
$= 0$

Suggest Corrections

Similar questions

Q. 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

2 s

. [Take

g = 10 {m/s}^{2}

]

Q. From the top of a tower

100 m

in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with velocity of

25 m s^{- 1}

. Find when and where the two balls will meet. Take

g = 10 m s^{- 2}

A ball is projected vertically up from foot of a tower of height of 40 m/s . At the same instant another ball is dropped from top of tower. When and where do they meet each other . ( take g = 10 m/s^2)

Q. From the top of a tower

100 m

in flight ball is dropped and the same time another ball is projected vertically upward from the ground with a velocity of

25 m / s e c

. Find when and where the two balls will meet?

(g = 9.8 m / s e c^{2})

Q. A ball of mass 100 g is projected vertically upward from the ground with a velocity of 49 m/s. At the same time another ball is dropped from a height of 98 m to fall freely along the same path as the first ball. After some time the two balls collide and stick together and finally fall together.Find the time of flight of masses.

Join BYJU'S Learning Program

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 $g = 10 {m/s}^{2}$ ]