Question

At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a
mass of 0.25 kg, is thrown straight upward from Earth’s surface with an initial speed of 15 m/s.
They move along nearby lines and pass each other without colliding. At the end of 2.0 s the
height above Earth’s surface of the center of mass of the two-ball system is ?

Answer 1

we need to know the position of two balls after 2 s of its release..

Answer 2

for first ball...
\[s=ut+1/2g t ^{2}\]
here u=0 t=2s
so s=1/2*g*4=2g=2*10 =20 meter [lets take g=10 m/s^2]
so after 2 s first ball is at a height (25-20)=5 meter above earth's surface

for 2nd ball...
\[s=ut-1/2g t ^{2}\]
here u=15 m/s,,,t=2s
so s=15*2-1/2*10*4=30-20=10 meter
so after 2s second ball is at a height 10meter above earth's surface..

Answer 3

now find the center of mass using this...
\[M _{_{total}}y _{cm}=m _{1}y _{1}+m _{2}y _{2}\]

Answer 4

Thanx very much for your help ..

Answer 5

welcome...:)

Answer 6

That is a way to do it. But there is another way that uses the power of the concept of c.m.
Let , m1=0.5 ,
x1=initial position of m1 from ground = 25m
u1=initial velocity of m1 = 0
m2=0.25
x2=initial position of m2 wrt ground =0
u2=initial velocity of m2 =15 m/s
Initial position of c.m wrt ground = ho=

(m1*x1+m2*x2)/(m1+m2) =50/3 m

Initial velocity of cm = uo=

(m1*u1+m2*u2)/(m1+m2) =5m/s

Now c.m. is place where we can assume all mass of the system to be concentrated and all the external force to be applied.The total external force on system = -m1g-m2g
Where - means downwards. Total mass =m1+m2

So acceleration of c.m =-g

Now apply ,

h=ho+uo*t -gt^2/2

It will give you the position of c.m. at any time you want

You will get same answer as the above when you put t=2.

Answer 7

Thanx :)