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Question

Two particles A and B of masses 1 kg and 2 kg respectively are projected with speed uA=200 ms1 and uB=50 ms1 in the directions as shown in figure. Initially they were 90 m apart. They collide in the midair and stick with each other. Find the maximum height attained by the centre of mass of the system.
[Assume acceleration due to gravity to be constant as <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> g=10 ms2]


A
60 m
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B
115.55 m
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C
55.55 m
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D
40 m
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Solution

The correct option is B 115.55 m
Given:
mA=1 kg; mB=2 kg
uA=200 ms1; uB=50 ms1
and initial separation between them is,
d=90 m

Using the relation for centre of mass of the system,

rcom=mArA+mBrBmA+mB


Taking rcom as the origin we get,

0=mA(rA)+mBrBmA+mB

mArA=mBrB1×rA=2×rB

rA=2rB .......(1)

and from given data,

rA+rB=90 m .......(2)

From equations (1) and (2), we get,

rA=60 m and rB=30 m

i.e,. COM is at height 60 m from the ground.

Further, velocity of the COM will be,

uCOM=mAuA+mBuBmA+mB

uCOM=(1)(200)(2)(50)1+2

uCOM=1003 ms1 ()

Let h be the height attained by COM beyond 60 m.

Using, kinematic equations of motion,

v2COM=u2COM+2aCOMh

0=(1003)2(2)(10)h

h=(100)2180=55.55 m

Therefore, the maximum height attained by the centre of mass is,

H=60+55.55=115.55 m

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (B) is the correct answer.
Why this Question?

This question is aimed at developing the student's approach towards solving the problems related to the motion of COM



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