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IIT JEE Advanced 2012 Paper 1
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© examsnet.com
Question : 8
Total: 60
A small block is connected to one end of a massless spring of un-stretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t = 0. It then executes simple harmonic motion with angular frequency ω = (π /3) rad/s. Simultaneously, at t = 0, a small pebble is projected with speed v from point P at an angle of 45° as shown in the figure. Point O is at a horizontal distance of 10 m from O. If the pebble hits the block at t = 1 s, the value of v is (take g = 10
m
∕
s
2
)
√
50
m/s
√
51
m/s
√
52
m/s
√
53
m/s
Validate
Solution:
Since the block starts executing simple harmonic motion from extreme position, we have
x = acosωt
At 1 s, we have
x =
0.2
c
o
s
(
π
3
)
= 0.1 m
That is, the block is at a distance 5.1 – 0.10 = 5 m from O, which is the range of the bebble as well. Now,
R =
v
2
s
i
n
2
α
g
S =
v
2
s
i
n
2
(
π
4
)
10
Therefore, v =
√
50
m/s.
© examsnet.com
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