# Physics 9702 Doubts | Help Page 184

__Question 900: [Gravitation]__
A binary star consists of two stars
that orbit about a fixed point C, as shown in Fig.1.

The star of mass M

_{1}has a circular orbit of radius R_{1}and the star of mass M_{2}has a circular orbit of radius R_{2}. Both stars have the same angular speed ω, about C.**(a)**State the formula, in terms of G, M

_{1}, M

_{2}, R

_{1}, R

_{2}and ω for

(i) gravitational force between the
two stars,

(ii) centripetal force on the star
of mass M

_{1}.**(b)**The stars orbit each other in a time of 1.26 × 10

^{8}s (4.0 years). Calculate angular speed ω for each star.

**(c)**

(i) Show that the ratio of masses of
the stars is given by the expression

M

_{1}/ M_{2}= R_{2}/ R_{1}
(ii) The ratio M

_{1}/ M_{2}is equal to 3.0 and the separation of the stars is 3.2 × 10^{11}m.**(d)**

(i) By equating the expressions you
have given in (a) and using the data calculated in (b) and (c), determine the
mass of one of the stars.

(ii) State whether answer in (i) is
for the more massive or for the less massive star.

**Reference:**

*Past Exam Paper – June 2004 Paper 4 Q3*

__Solution 900:__**(a)**

(i) Gravitational force = GM

_{1}M_{2}/ (R_{1}+ R_{2})^{2}
(ii) Centripetal force = M

_{1}R_{1}ω^{2}OR = M_{2}R_{2}ω^{2}**(b)**

Angular speed ω = 2π / (1.26 × 10

^{8}) OR 2π / T
Angular speed ω = 4.99 × 10

^{-8}rad s^{-1}
allow 2 s.f.: 1.59π × 10

^{-8}scores 1/2**(c)**

(i)

Reference to EITHER taking moments
(about C) OR they have the same (centripetal) force

M

_{1}R_{1}= M_{2}R_{2}OR M_{1}R_{1}ω^{2}= M_{2}R_{2}ω^{2}
hence M

_{1}/ M_{2}= R_{2}/ R_{1}
(ii)

{M

_{1}/ M_{2}= 3. Since M_{1}/ M_{2}= R_{2}/ R_{1}, the ratio R_{2}/ R_{1}= 3 giving R_{1}= R_{2}/ 3
The separation of the
stars = R

_{1}+ R_{2}= 3.2 × 10^{11}m.
Since R

_{1}= R_{2}/ 3, R_{2}/3 + R_{2}= 3.2 × 10^{11}m
4R

_{2}/ 3 = 3.2 × 10^{11}m}
R

_{2}= 3/4 × (3.2 × 10^{11}m) = 2.4 × 10^{11}m
R

_{1}= (3.2 × 10^{11}) – R_{2}= 8.0 × 10^{10}m (allow vice versa)**(d)**

(i)

{ Gravitational force = GM

_{1}M_{2}/ (R_{1}+ R_{2})^{2}. The gravitational force provides the centripetal force which is equal to M_{1}R_{1}ω^{2}.
So, GM

_{1}M_{2}/ (R_{1}+ R_{2})^{2 }= M_{1}R_{1}ω^{2}}
M

_{2}= {(R_{1}+ R_{2})^{2}× R_{1}× ω^{2}} / G
M

_{2}= (3.2 × 10^{11})^{2}× 8.0×10^{10}× (4.99 × 10^{-8})^{2}/ (6.67 × 10^{-11}) = 3.06 × 10^{29}kg
(ii) The less massive star (only
award this mark if reasonable attempt at (i))

__Question 901: [Vectors]__
A mass of 2kg is in equilibrium on a
smooth plane inclined at an angle of 30° to the horizontal. Find

(i) the minimum force required so
that the mass is in equilibrium,

(ii) the normal reaction force on
the mass by the plane

**Reference:**

*???*

__Solution 901:__
(i)

Weight of the body = mg = 2 × 10 = 20N
The minimum force required so that
the mass is in equilibrium must be equal in magnitude to the component of the
weight along the inclined plane.

Minimum Force = 20 sin30 = 10N

(ii)

The normal reaction force is
perpendicular to the inclined plane and equal to the component of weight in that
direction.

Normal Force = 20 cos30 = 17.3N

__Question 902: [Kinematics > Air resistance]__
A tennis ball is thrown horizontally
in air from top of a tall building.

If the effect of air resistance is
not negligible, what happens to the horizontal and vertical components of the
ball’s velocity?

horizontal
component of velocity vertical
component of velocity

A constant
constant

B constant
increases
at a constant rate

C decreases
to zero increases
at a constant rate

D decreases
to zero increases
to a maximum value

**Reference:**

*Past Exam Paper – June 2014 Paper 11 Q6*

__Solution 902:__**Answer: D.**

The tennis ball is thrown
horizontally in air from the top of a tall building. Since air resistance is not negligible, the
horizontal component of velocity does not remain constant. It decreases to zero
since air resistance is a form of friction which opposes motion. [A and B are incorrect]

Due to the acceleration due to
gravity, the vertical component would increase. However, the air resistance
would also increase in the opposite direction until the air resistance is equal
to the weight of the object. The object’s vertical component of velocity
reaches a maximum value called the

**terminal velocity**. [C is incorrect]

__Question 903: [Simple harmonic motion]__
A piston moves vertically up and
down in a cylinder, as illustrated in Fig.1.

Piston is connected to a wheel by
means of a rod that is pivoted at the piston and at the wheel. As the piston
moves up and down, the wheel is made to rotate.

**(a)**

(i) State number of oscillations
made by the piston during one complete rotation of the wheel.

(ii) The wheel makes 2400 revolutions
per minute. Determine frequency of oscillation of the piston.

**(b)**Amplitude of the oscillations of the piston is 42 mm.

Assuming that these oscillations are
simple harmonic, calculate the maximum values for the piston of

(i) linear speed,

(ii) the acceleration.

**(c)**On Fig.1, mark a position of the pivot P for the piston to have

(i) maximum speed (mark this
position S),

(ii) maximum acceleration (mark this
position A).

**Reference:**

*Past Exam Paper – June 2006 Paper 4 Q4*

__Solution 903:__**(a)**

(i) Number of oscillations = 1.0

(ii)

{Frequency is the number of
revolutions in 1s. 1min = 60s. In 1s, there are 2400 / 60 = 40 revolutions.}

Frequency = 40Hz

**(b)**

(i) Linear speed = 2πfa = 2π × 40 × (42
× 10

^{-3}) = 10.6 m s^{-1}
(ii) Acceleration = 4π

^{2}f^{2 }a = (80π)^{2}× 42 × 10^{-3}= 2650 m s^{-2}
{4π

^{2}f^{2 }= 2^{2}π^{2}f^{2 }= 2^{2}π^{2}(40)^{2 }= (2×40)^{2}π^{2}= (80π)^{2}}**(c)**

(i) S should be marked correctly (on
‘horizontal line through centre of wheel)

(ii) A should be marked correctly
(on ‘vertical line’ through centre of wheel)

This comment has been removed by the author.

ReplyDeleteSee below

DeleteFor 06/O/N/04 Q.8(a)(i), see solution 904 at

ReplyDeletehttp://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-185.html

For 04/M/J/05 Q.5(b), see solution 907 at

http://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-185.html

why the less massive star? any explanation?

ReplyDeleteThe equation M2 = {(R1 + R2)2 × R1 × ω2} / G above

Deleteshows that the greater the value of R1, the bigger is the mass M2. Since the radius R1 is shorter than R2, it implies that M2 is smaller than M1.

ratio of M1/M2 is said to be 3 i.e. 3/1

DeleteSo M2 is smaller

yea

DeleteWhy cant we use keplers 3rd law???

Deletekepler's law is not exactly in the syllabus but teachers like to include it in notes

DeleteCan you please explain how S is horizontal and A is vertical in question 903??

ReplyDeleteThe piston undergoes simple harmonic motion vertically. For shm, speed is zero at maximum displacement (amplitude) but maximum when displacement is zero (at S). Similarly, acceleration is maximum at maximum displacement (vertically) (at A).

DeleteAbout the same question....

DeleteIt's presumable but how do we know that the no. of oscillation made by the piston during one complete rotation of the wheel is

1.

Please try to reply fast, have exam tomorrow

COnsider 1 oscillation. THe piston is initially up. It goes completely down and then movse up again to its original position. This is one oscillation and this corresponds to one complete rotataion of the wheel.

DeleteIf you are not convinced, try to build a small model and perform the experiment