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IIT JEE Advanced 2007 Paper 2
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© examsnet.com
Question : 5
Total: 66
A particle moves in the xy-plane under the influence of a force such that its linear momentum is
→
p
(
t
)
=
A
[
^
i
c
o
s
(
k
t
)
−
^
j
s
i
n
(
k
t
)
]
, where A and k are constants. The angle between the force and the momentum is
0°
30°
45°
90°
Validate
Solution:
We have
|
→
F
|
|
→
P
|
cos θ =
→
F
.
→
P
However,
→
F
=
d
→
P
d
t
= Ak [- i sin (kt) - j cos (kt)]
Therefore,
→
F
.
→
P
= Ak [- i sin (kt) - j cos (kt)] . Ak [i cos (kt) - j sin (kt)] = 0
However,
|
→
F
|
and
|
→
P
|
are not zero ⇒ cos θ = 0 or θ = 90°.
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