Indian Olympiad Qualifier in Physics (IOQP) 2021-2022
conducted jointly by
Homi Bhabha Centre for Science Education (HBCSE-TIFR)
and
Indian Association of Physics Teachers (IAPT)
Part II: Indian National Physics Olympiad (INPhO)
Homi Bhabha Centre for Science Education (HBCSE-TIFR)
Date: 13 March 2022
Time: 10:30-12:30 (2 hours) Maximum Marks: 50
Instructions Roll No.:
1. This booklet consists of 12 pages and total of 5 questions. Write roll number at the top wherever
asked.
2. Booklet to write the answers is provided separately. Instructions to write the answers are on the
Answer Booklet.
3. Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and
the Detailed Answer Sheets in the Answer Booklet. Simple short answers and plots may be
directly entered in the Summary Answer Sheet. Marks may be deducted for absence of detailed
work in questions involving longer calculations.
4. Strike out any rough work that you do not want to be considered for evaluation. You may also
use the space on the Question Paper for rough work – this will NOT be evaluated.
5. Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calcula-
tors.
6. Please submit the Answer Booklet at the end of the examination. You may retain the
Question Paper.
, Page 2 Code 65 Questions Last six digits of Roll No.:
1. A block of mass m = 0.1 kg is attached to a spring (one end fixed to the wall) with spring constant
k = 50 N m−1 . The block slides on a rough horizontal table along the x-axis. Assume that both
the coefficients of kinetic (µk ) and static friction (µs ) are same and constant (µk = µs = µ = 0.25).
The block is initially displaced to x0 = 0.1 m from the unstretched position (normal length of the
spring, x = 0) of the spring and released from rest as shown below. Neglect any air resistance.
Take the acceleration g due to gravity to be 10 m/s2 .
unstretched position
x0
(a) [3 marks] How many times (n) will the block cross the unstretched position before coming
to rest permanently?
Solution:
unstretched position
x0
x1
x2
x3
mẍ = −kx ± µmg (1.1)
where the ± sign is used so that the friction is opposite to the direction of the velocity
of the block. Lets define the ratio of the frictional force to the maximum restoring force
to be α i.e. α = µmg/kx0 . During the first half cycle of the motion, the loss of potential
energy is equal to the work done against friction
kx20 kx21
− = µmg(x0 − x1 ) (1.2)
2 2
where x1 is the displacement after one half cycle. This gives
x1 = −x0 + 2αx0 (1.3)
x2 = −x1 − 2αx0 (1.4)
Thus
xj = −xj−1 − (−1)j 2αx0 (1.5)
xj = (−1) (1 − 2αj)x0
j
(1.6)
conducted jointly by
Homi Bhabha Centre for Science Education (HBCSE-TIFR)
and
Indian Association of Physics Teachers (IAPT)
Part II: Indian National Physics Olympiad (INPhO)
Homi Bhabha Centre for Science Education (HBCSE-TIFR)
Date: 13 March 2022
Time: 10:30-12:30 (2 hours) Maximum Marks: 50
Instructions Roll No.:
1. This booklet consists of 12 pages and total of 5 questions. Write roll number at the top wherever
asked.
2. Booklet to write the answers is provided separately. Instructions to write the answers are on the
Answer Booklet.
3. Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and
the Detailed Answer Sheets in the Answer Booklet. Simple short answers and plots may be
directly entered in the Summary Answer Sheet. Marks may be deducted for absence of detailed
work in questions involving longer calculations.
4. Strike out any rough work that you do not want to be considered for evaluation. You may also
use the space on the Question Paper for rough work – this will NOT be evaluated.
5. Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calcula-
tors.
6. Please submit the Answer Booklet at the end of the examination. You may retain the
Question Paper.
, Page 2 Code 65 Questions Last six digits of Roll No.:
1. A block of mass m = 0.1 kg is attached to a spring (one end fixed to the wall) with spring constant
k = 50 N m−1 . The block slides on a rough horizontal table along the x-axis. Assume that both
the coefficients of kinetic (µk ) and static friction (µs ) are same and constant (µk = µs = µ = 0.25).
The block is initially displaced to x0 = 0.1 m from the unstretched position (normal length of the
spring, x = 0) of the spring and released from rest as shown below. Neglect any air resistance.
Take the acceleration g due to gravity to be 10 m/s2 .
unstretched position
x0
(a) [3 marks] How many times (n) will the block cross the unstretched position before coming
to rest permanently?
Solution:
unstretched position
x0
x1
x2
x3
mẍ = −kx ± µmg (1.1)
where the ± sign is used so that the friction is opposite to the direction of the velocity
of the block. Lets define the ratio of the frictional force to the maximum restoring force
to be α i.e. α = µmg/kx0 . During the first half cycle of the motion, the loss of potential
energy is equal to the work done against friction
kx20 kx21
− = µmg(x0 − x1 ) (1.2)
2 2
where x1 is the displacement after one half cycle. This gives
x1 = −x0 + 2αx0 (1.3)
x2 = −x1 − 2αx0 (1.4)
Thus
xj = −xj−1 − (−1)j 2αx0 (1.5)
xj = (−1) (1 − 2αj)x0
j
(1.6)
