33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Motion in a Plane
Telegram @unacademyplusdiscounts
23
2 2 2
u sin2θ u sin θ 2 2 2
∴ = or,
2u
sinθcosθ u sin θ
=
g 2g
g 2g
tanq = 4 or q = tan –1 (4)
28. (a) : Here, u = 20 m/s, g = 10 m/s 2
For maximum range, angle of projection is
q = 45°
u
2 sin90° u
2 ⎛
⎞
∴ R = = ⎜ R u 2 sin2θ
max
Q = ⎟
g g ⎝ g ⎠
2
( 20 m/s)
= = 40 m
2
( 10 m/s )
29. (c) : Let f be elevation
angle of the projectile at its
highest point as seen from
the point of projection O and
q be angle of projection with
the horizontal.
From figure,tanφ= H
...(i)
R / 2
In case of projectile motion
sin
Maximumheight, H u 2 2
θ
=
2g
sin
Horizontal range, R u 2
2θ
=
g
Substituting these values of H and R in (i), we get
2 2
u sin θ
2 2
2g
sin θ sin θ 1
tanφ
= = = = tanθ
2
u sin2θ
sin2θ
2sinθcosθ
2
2 g
Here, q = 45°
∴ tanφ = 1 1
tan 45°= ( Q tan 45°=
1)
2 2
− ⎛
φ= tan 1 1
⎝
⎜
⎞ 2 ⎠ ⎟
30. (a) : Let v be velocity of a projectile at maximum
height H.
v = ucosq
According to
given problem,
u
v = 2
u
1
∴ = u cosθ ⇒ cosθ= ⇒ θ = 60°
2
2
31. (a) :
The horizontal momentum does not change. The change
in vertical momentum is
1
mv sin θ−( − mv sin θ)
= 2mv = 2mv
2
32. (b) : Horizontal range, R u 2
sin2θ
=
g
For angle of projection (45° – q), the horizontal range is
2 2
2
u sin[ 245 ( °−θ)] u sin( 90°−2θ ) u cos2θ
∴ R1
=
=
=
g
g
g
For angle of projection (45° + q), the horizontal range is
2 2 2
u sin[ 245 ( °+ θ)] u sin( 90°+
2θ) u cos2θ
R2
=
=
=
g
g
g
2
R1
u cos 2θ
/ g 1
∴ = = .
R 2
2 u cos 2θ
/ g 1
33. (c) : Time required to reach the ground is dependent
on the vertical motion of the particle. Vertical motion of
both the particles A and B are exactly same. Although
particle B has an initial velocity, but that is in horizontal
direction and it has no component in vertical (component
of a vector at a direction of 90° = 0) direction. Hence they
will reach the ground simultaneously.
34. (b) : As q 2 = (90 – q 1 ),
So range of projectile,
v0 2 sin2θ v0 2 2sinθcosθ
R1
= =
g
g
v0 2 2sin( 90−θ1)cos( 90 −θ1)
R2
=
g
v0 2 2cosθ1sinθ1
R2
= = R1
g
35. (b) : For the given velocity of projection u, the
horizontal range is the same for the angle of projection
q and 90° – q.
Horizontal range, R u 2
sin2θ
=
g
2
u sin( 2× 30° ) 2
u sin60°
∴ Forbody A,
RA
=
=
g
g
2
u sin( 2× 60°
)
Forbody BR , B =
g
u
2 u
u
RB = sin120° 2 sin( 180°− 60° ) 2 sin60°
= =
g
g
g
The range is the same whether the angle is q or 90° – q.
\ The ratio of ranges is 1 : 1.
36. (c) : Horizontal range, R u 2
sin2θ
=
g
For maximum horizontal range, q = 45°
2
u
Rm = g