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NSEP Topper Batch (2023)

WPE and Circular Motion Practice Sheet
1. As shown in figure, a block of mass m is 3. As shown in figure, a particle of mass m = 10–10 kg,
projected from wall A with velocity 2𝑣0 on the moving with velocity 𝑣0 = 105 m/s approaches a
rough surface with constant sliding friction to hit stationary fixed target with impact parameter b
the wall B with velocity 𝑣0. With what velocity from a large distance. If the fixed rigid target has
same mass m should be projected to hit the wall K
a core with repulsive central force F(r) =
r3
B with same velocity 𝑣0 if the surface is now
where K > 0 and the particle scatters elastically.
moving upward with an acceleration of a = 4g?
The closest distance of approach (if numerically
(NSEP-2022)
K = b2) is (NSEP-2021)

(A) 2𝑣0 (B) 3𝑣0

(C) 4𝑣0 (D) 5𝑣0 (A) b (B) b 2
(C) b 3 (D) 2b
2. Consider a particle of mass ‘m’ with a total
energy E moving in a one dimensional potential 4. The kinetic energy of a particle moving along a
field. The potential V (x) is plotted against x in circle of radius R depends upon the distance
the figure beside. The plot of momentum – covered ‘s’ as KE = as 2 where a is a constant. The
position graph of this particle is qualitatively best magnitude of the force acting on the particle as a
represented by (all plots are symmetrical about function of ‘s’ is (NSEP-2020)
x – axis) (NSEP-2021) 2as 2
(A)
R
2as 2
(B)
m
(C) 2as
2
 s 
(D) 2as 1 +  
R

5. The resistive force on an airplane flying in a

horizontal plane is given by F f = kv2, where k is
constant and v is the speed of the airplane. When
the power output from the engine is P 0, the plane
flies at a speed v0. If the power output of the
engine is doubled the airplane will fly at a speed
of (NSEP-2019)
(A) 1.12 v0 (B) 1.26 v0
(A) Figure a (B) Figure b
(C) 1.41 v0 (D) 2.82 v0
(C) Figure c (D) Figure d
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6. If the force acting on a body is inversely 9. The potential energy of a mass m in a conservative
proportional to its speed, the KE of the body force field can be expressed as U = x – y where
varies with time t as (NSEP-2019) (x, y) denote the position coordinates of the body.
(A) t 0
(B) t 1 The acceleration of the body is (NSEP-2018)
 − +
(C) t 2
(D) t−1 (A) (B)
m m
7. Position-time graph of a particle moving in a  2 − 2  2 + 2
(C) (D)
potential field is shown beside. If the mass of the m m
particle is 1 kg its total energy is approximately
(NSEP-2019) 10. The strings AB and AC each of length 40 cm,
connect a ball of mass 200 g to a vertical shaft as
shown. When the shaft rotates at a constant angular
speed , the ball travels in a horizontal circle with
the strings inclined at  = 30° to the shaft. If the
tension in the string AC is 4 N, that in the string AB
and the angular speed  respectively are
(NSEP-2018)

(A) 15.45 × 10−4 J (B) 30.78 × 10−4 J

(C) 7.71 × 10−4 J (D) 3.85 × 10−4 J

8. The potential energy (U) of a particle moving in

a potential field varies with its displacement (x) (A) 6.26 N and 11.32 rad/s
as shown below. The variation of force F(x) (B) 7.92 N and 14.32 rad/s
acting on the particle as a function of x can be (C) 7.92 N and 11.32 rad/s
represented by (NSEP-2019) (D) 6.26 N and 14.32 rad/s

11. Motion of a particle in a plane is described by the

non-orthogonal set of co-ordinates (p, q) with unit
ˆ inclined at an angle  as shown in the
ˆ q)
vectors (p,
diagram. If the mass of the particle is m, its kinetic
 dx 
energy is given by  x =  (NSEP-2017)
 dt 

1
(A) m(p 2 + q 2 + p q cos )
2
1
(B) m(p 2 + q 2 + p q(1 − sin ))
2
1
(C) m(p 2 + q 2 + 2 p q cos )
2
1
(A) Fig (i) (B) Fig (ii) (D) m(p 2 + q 2 + p q cot )
2
(C) Fig (iii) (D) Fig (iv)
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12. A racing car moves along a circular track of radius 15. A body of mass 1.0 kg moves in XY plane under
b. The car starts from rest and its speed increases at the influence of a conservative force. Its potential
a constant rate . Let the angle between the velocity energy is given by U = 2x + 3y where (x, y) denote
and the acceleration be  at time t. Then cos  is the coordinates of the body. The body is at rest at
(NSEP-2017) (2, –4) initially. All the quantities have SI units.
(A) 0 Therefore, the body (NSEP-2017)
(B) at2/b (A) moves along parabolic path
(C) b/(b + at2) (B) moves with a constant acceleration
(D) bt/(b2 + 2t4)1/2 (C) never crosses the X axis
(D) has a speed of 2 13 m/s at t = 2 sec
13. In the figure shown below masses of blocks A and
B are 3 kg and 6 kg respectively. The force 16. A block of mass 2 kg drops vertically from a height
constants of springs S1 and S2 are 160 N/m and of 0.4 m onto a spring whose force constant K is
40 N/m respectively. Length of the light string 1960 N/m. Therefore, the maximum compression
connecting the blocks is 8 m. The system is released of the spring is (NSEP-2016)
from rest with the springs at their natural lengths. (A) 0.4 m
The maximum elongation of spring S1 will be (B) 0.25 m
(NSEP-2017) (C) 0.80 m
(D) 0.1 m

17. Two identical strings with fixed ends separated

by height h have their other ends tied to a body P
A B of mass m as shown in figure. When the body
2g
S1 S2 rotates with uniform angular speed 2 in a
h
horizontal plane about the vertical axis, the ratio
(A) 0.294 m (B) 0.490 m
T1
(C) 0.588 m (D) 0.882 m of tensions in the string is (NSEP-2013)
T2

14. A particle starting from rest at the highest point

slides down the outside of a smooth vertical circular
track of radius 0.3 m. When it leaves the track its
vertical fall is h and the linear velocity is v. The
angle made by the radius at that position of the
particle with the vertical is . Now consider the
following observations: (g = 10 m/s2) (NSEP-2017)
(I) h = 0.1 m and cos  = 2/3
(II) h = 0.2 m and cos  = 1/3 3
(A)
(III) v = 2 m/s 5
(IV) after leaving the circular track, the particle will 5
(B)
describe a parabolic path 3
(A) (I) and (III) both are correct 2
(B) only (II) is incorrect (C)
5
(C) only (III) is correct 5
(D) (IV) is correct (D)
2
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18. A bead of mass 5 g can move without friction on 19. When two ends of a spring are pulled apart
a piece of wire bent in the form of a semicircular increasing its length, it produces force equal to
ring of radius 0.10 m as shown in fig. This ring 1
kx at its ends. At a point of its length from one
can freely rotate about the vertical axis OY. At 4
what height will the bead stay above the ground end the force is (NSEP-2013)
level OX, if the semicircular wire revolves with (A) 0.25 kx
angular velocity 10.63 rad/s? (NSEP-2013) (B) 0.75 kx
(C) 1.0 kx
(D) 0.5 kx

(A) 0.013 m (B) 0.087 m

(C) 0.027 m (D) 0.073 m

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