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JEE Main 18 March 2021 Shift 2 Solved Paper
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© examsnet.com
Question : 21
Total: 90
The projectile motion of a particle of mass
5
g
is shown in the figure.
The initial velocity of the particle is
5
√
2
m
s
−
1
and the air resistance is assumed to be negligible. The magnitude of the change in momentum between the points
A
and
B
is
x
×
10
−
2
k
g
−
m
s
−
1
. The value of
x
to the nearest integer, is
[18 Mar 2021 Shift 2]
Your Answer:
Validate
Solution:
The given situation is shown below
Initial velocity of the projectile,
u
=
u
cos
45
∘
∧
i
+
u
sin
45
∘
∧
j
Final velocity of the projectile,
v
=
v
cos
45
∘
∧
i
−
v
sin
45
∘
∧
j
At point
B
,
v
cos
45
∘
=
u
cos
45
∘
and
v
sin
45
∘
=
u
sin
45
∘
v
=
u
cos
45
∘
∧
i
−
u
sin
45
∘
∧
j
The magnitude of change in the momentum between the points
A
and
B
|
∆
p
|
=
|
m
(
v
−
u
)
|
⇒
∆
p
=
2
m
u
sin
45
∘
⇒
∆
p
=
2
×
0.005
×
5
√
2
×
1
√
2
⇒
∆
p
=
5
×
10
−
2
k
g
−
m
/
s
Hence, the value of the
x
is 5 .
© examsnet.com
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