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YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

## Friday, August 27, 2021

### A uniform beam of mass 1.4 kg is pivoted at P as shown. The beam has a length of 0.60 m and P is 0.20 m from one end.

Question 47

A uniform beam of mass 1.4 kg is pivoted at P as shown. The beam has a length of 0.60 m and P is 0.20 m from one end. Loads of 3.0 kg and 6.0 kg are suspended 0.35 m and 0.15 m from the pivot as shown.

What torque must be applied to the beam in order to maintain it in equilibrium?

A 0.010 N m                B 0.10 N m                  C 0.29 N m                 D 2.8 N m

Reference: Past Exam Paper – June 2013 Paper 12 Q14

Solution:

Answer: D.

Each of the masses will exert a force downwards on the beam (due to gravity) which will be equal to their weight.

Weight = mg

We also need to consider the weight of the beam of mass 1.4 kg which acts at the centre.

For equilibrium,

sum of clockwise moments = sum of anti-clockwise moments

The 6.0 kg load exert a clockwise moment.

Moment = Force × perpendicular distance

Clockwise moment = 0.15 × (6.0×9.81) = 8.829 Nm

Since the uniform beam has a total length of 0.60 m, its centre of mass will act at the 0.30 m mark (centre of the beam), a distance of (0.30 – 0.20 =) 0.10 m from the pivot. This causes an anticlockwise moment.

The 3.0 kg load also exerts an anticlockwise moment.

Anticlockwise moments = [0.10 × (1.4×9.81)] + [0.35 × (3.0×9.81)] = 11.6739 Nm

The torque that must be applied should act in such a way that the 2 moments (clockwise and anticlockwise) calculated would become equal.

Therefore, a (clockwise) torque that must be applied = 11.6739 – 8.829 = 2.8449 2.8 Nm.

## Sunday, July 18, 2021

### A ball is thrown horizontally from the top of a building, as shown in Fig. 2.1.

Question 39

A ball is thrown horizontally from the top of a building, as shown in Fig. 2.1.

Fig. 2.1

The ball is thrown with a horizontal speed of 8.2 m s-1. The side of the building is vertical. At point P on the path of the ball, the ball is distance x from the building and is moving at an angle of 60° to the horizontal. Air resistance is negligible.

(a) For the ball at point P,

(i) show that the vertical component of its velocity is 14.2 m s-1, [2]

(ii) determine the vertical distance through which the ball has fallen, [2]

(iii) determine the horizontal distance x. [2]

(b) The path of the ball in (a), with an initial horizontal speed of 8.2 m s-1, is shown again in Fig. 2.2.

Fig. 2.2

On Fig. 2.2, sketch the new path of the ball for the ball having an initial horizontal speed

(i) greater than 8.2 m s-1 and with negligible air resistance (label this path G), [2]

(ii) equal to 8.2 m s-1 but with air resistance (label this path A). [2]

Reference: Past Exam Paper – November 2010 Paper 21 Q2

Solution:

(a)

(i)

{The horizontal component of the speed is not affected as the air resistance is negligible. The horizontal component is constant throughout the motion. So, at P, the horizontal component is still 8.2 m s-1.}

Horizontal speed = 8.2 m s-1

tan 60 o = vv / 8.2                     [tan 60 = opp / adj]

Vertical component of speed: vv = 8.2 tan60o = 14.2 m s-1

(ii)

{Consider the vertical motion.

v2 = u2 + 2as = 0 + 2as}

14.22 = 0 + (2×9.8×h)

h = 14.22 / (2×9.8)

Vertical distance fallen: h = 10.3 m

(iii)

{Time is the quantity that we will use to relate the horizontal and vertical motion.

We first need to determine the time taken by the ball fall the distance calculated in part (ii) for the vertical motion.

For the vertical motion,

v = u + at

14.2 = 0 + 9.8t}

Time of descent of ball: t (= v / a) = 14.2 / 9.8 = 1.45 s

{We can now calculate the horizontal distance moved by the ball in this amount of time. Since the horizontal speed is constant, we can use the formula

Speed = distance / time

Distance = Speed × Time}

Horizontal distance: x = 8.2 × 1.45 = 11.9 m

(b)

(i)

The sketch shows a smooth path curved and above the given path. The ball hits the ground at a more acute angle

{The ball would travel a greater horizontal distance at any instant of time as the horizontal speed is greater. Note that vertical component of velocity behaves in the same way as previously. The horizontal component is greater than before (and constant at all time), so the resultant velocity will be at a more acute angle than before.}

(ii)

The sketch shows a smooth path curved and below the given path. The ball hits the ground at a steeper angle.

{Air resistance affects all components of the speed, reducing them, since it opposes motion. So, a smaller horizontal distance is travelled by the ball.

The vertical component of speed undergoes acceleration (due to gravity) until terminal speed is reached (the ball may hit the ground before terminal velocity is reached) but compare to the cases before, the speed increases at a smaller rate due to air resistance.

As for the horizontal component, it decreases due to air resistance. So, the resultant speed will be at a steeper angle to the ground since the horizontal component is smaller than in the previous cases.}

## Saturday, June 26, 2021

### An object of mass m travelling with speed v has a head-on collision with another object of mass m travelling with speed v in the opposite direction.

Question 25

An object of mass m travelling with speed v has a head-on collision with another object of mass m travelling with speed v in the opposite direction. The two objects stick together after the collision.

What is the total loss of kinetic energy in the collision?

A 0                   B ½ mv2                         C mv2                                                 D 2mv2

Reference: Past Exam Paper – November 2015 Paper 12 Q13

Solution:

Answer: C.

Momentum is a vector quantity and we need to consider the direction.

Since the objects are moving in opposite direction, one of them will have a negative value.

Sum of momentum before collision = mv – mv = 0

Kinetic energy is a scalar quantity and so, the direction of motion does not matter.

Sum of KE before collision = ½ mv2 + ½ mv2 = mv2

Momentum is always conserved.

Sum of momentum after collision = Sum of momentum before collision = 0

Sum of momentum after collision = 0

2mvf = 0

Giving the speed after collision, vf = 0

That is, the objects do not move after the collision.

KE after collision = 0

Loss in KE = mv2 – 0 = mv2

## Wednesday, June 16, 2021

### Which statement about electromagnetic radiation is correct?

Question 26

Which statement about electromagnetic radiation is correct?

A Waves of wavelength 5 × 10-9 m are high-energy gamma rays.

B Waves of wavelength 3 × 10-8 m are ultra-violet waves.

C Waves of wavelength 5 × 10-7 m are infra-red waves.

D Waves of wavelength 9 × 10-7 m are light waves.

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q22

Solution:

Answer: B.

Estimates for the different components of the electromagnetic spectrum should be known at A-Level.

All EM waves travel at the same speed of 3.0×108 m s-1.

Gamma rays have the highest frequencies, followed by X-rays and UV.

v = fλ

For the given wavelength of UV waves,

f = v / λ = 3.0×108 / 3.0×10-8 = 1×1016 Hz

This is a correct estimate for the frequency of UV waves.

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