• 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 M₁ has a circular orbit of radius R₁ and the star of mass M₂ has a circular orbit of radius R₂. Both stars have the same angular speed ω, about C.

(a) State the formula, in terms of G, M₁, M₂, R₁, R₂ and ω for

(i) gravitational force between the two stars,

(ii) centripetal force on the star of mass M₁.

(b) The stars orbit each other in a time of 1.26 × 10⁸ 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

M₁ / M₂ = R₂ / R₁

(ii) The ratio M₁ / M₂ is equal to 3.0 and the separation of the stars is 3.2 × 10¹¹ 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 = GM₁M₂ / (R₁ + R₂)²

(ii) Centripetal force = M₁R₁ω²                       OR = M₂R₂ω²

(b)

Angular speed ω = 2π / (1.26 × 10⁸)               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

M₁R₁ = M₂R₂                          OR      M₁R₁ω² = M₂R₂ω²

hence M₁ / M₂ = R₂ / R₁

(ii)

{M₁ / M₂ = 3. Since M₁ / M₂ = R₂ / R₁, the ratio R₂ / R₁ = 3 giving R₁ = R₂ / 3

The separation of the stars = R₁ + R₂ = 3.2 × 10¹¹ m.

Since R₁ = R₂ / 3,        R₂/3 + R₂ = 3.2 × 10¹¹ m

4R₂ / 3 = 3.2 × 10¹¹ m}

R₂ = 3/4 × (3.2 × 10¹¹ m) = 2.4 × 10¹¹ m

R₁ = (3.2 × 10¹¹) – R₂ = 8.0 × 10¹⁰ m (allow vice versa)

(d)

(i)

{ Gravitational force = GM₁M₂ / (R₁ + R₂)². The gravitational force provides the centripetal force which is equal to M₁R₁ω².

So, GM₁M₂ / (R₁ + R₂)² = M₁R₁ω² }

M₂ = {(R₁ + R₂)² × R₁ × ω²} / G

M₂ = (3.2 × 10¹¹)² × 8.0×10¹⁰ × (4.99 × 10^-8)² / (6.67 × 10^-11) = 3.06 × 10²⁹ 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 = 4π²f² a = (80π)² × 42 × 10^-3 = 2650 m s^-2

{4π²f² = 2²π²f² = 2²π²(40)² = (2×40)²π² = (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)