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EAMCET Engineering 2004 Solved Paper
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© examsnet.com
Question : 6
Total: 160
Two particles of equal mass have velocities
→
v
1
=
4
^
i
and
→
v
2
=
4
^
j
ms
−
1
. First particle has an acceleration
→
a
1
=
(
5
^
i
+
5
^
j
)
ms
−
2
while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of
straight line
parabola
circle
ellipse
Validate
Solution:
v
CM
=
m
1
v
1
+
m
2
v
2
m
1
+
m
2
=
v
1
+
v
2
2
(
as
m
1
=
m
2
)
and
a
CM
=
m
1
a
1
+
m
2
a
2
m
1
+
m
2
=
a
1
+
0
2
=
a
1
2
The centre of mass of two particles will move with the mean velocity of two particles having common acceleration
a
2
. Hence, path of CM will be a straight line.
© examsnet.com
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