Two identical blocks each of mass m and speed v, travel towards each other on a frictionless surface

Answer/Explanation

Newton’s third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction. Such forces often cause one object to speed up (gain momentum) and the other object to slow down (lose momentum).

Examiners report

Candidates are often under the misapprehension that there are conditions upon Newton’s third law. It should be stressed that it is always true irrespective of energy or momentum considerations; and it does not depend upon the state of motion of the bodies in contact.

Question 584: [Vectors]

Forces of 3 N, 4 N and 5 N act at one point on an object. Angles at which the forces act can vary.

What is the value of minimum resultant force of these forces?

A 0

B between 0 and 2 N

C 2 N

D between 2 N and 4 N

Reference: Past Exam Paper – June 2010 Paper 11 Q13 & Paper 12 Q12 & Paper 13 Q11

Solution 584:

Answer: A.

For a minimum resultant force, we will need to include subtraction as much as possible so that the resultant has the smallest value possible. This is done by making the forces opposes each other in some way. If all the forces point in the same direction, the resultant force would become even bigger than the individual forces.

Consider the resultant magnitude of 3N + 4N such that the 2 forces are perpendicular to each other. Say the 3N force is vertical and the 4N is horizontal. The 3N does not have any component along the horizontal that could add to the horizontal 4N vector. Similarly, the horizontal 4N force does not have add any vertical component to the 3N force.

Resultant (is the hypotenuse of the triangle formed) = (32 + 42)0.5 = 5N.

Put the 5N force in the opposite direction of this resultant calculated above to obtain zero.

Question 585: [Dynamics > Collisions]

Two identical, perfectly elastic spheres have same mass m. They travel towards each other with same speed v along a horizontal frictionless surface.

Which statement about the sum of kinetic energies of the spheres is correct?

A The sum of their kinetic energies before impact is zero.

B The sum of their kinetic energies before impact is ½ mv2.

C The sum of their kinetic energies after impact is zero.

D The sum of their kinetic energies after impact is mv2.

Reference: Past Exam Paper – November 2012 Paper 12 Q13

Solution 585:

Answer: D.

Kinetic energy = ½ mv2

Since the 2 spheres are moving with speed v before the impact, the sum of their kinetic energies is not zero. [A is incorrect]

Kinetic energy of EACH sphere = ½ mv2

Sum of KE of the 2 spheres = 2 (½ mv2) = mv2 [B is incorrect]

The 2 spheres are perfectly elastic and the surface is frictionless, so the collision is also perfectly elastic. In an elastic collision, energy is conserved. So, the sum of their kinetic energy after the impact is equal to the sum before the impact. [C is incorrect]

Question 586: [Simple harmonic motion]

A mass of 78 g is suspended from a fixed point by means of a spring, as illustrated in Fig.1.

The stationary mass is pulled vertically downwards through a distance of 2.1 cm and then released.

Mass is observed to perform simple harmonic motion with a period of 0.69 s.

(a) Mass is released at time t = 0.

For oscillations of the mass,

(i) calculate angular frequency ω

(ii) determine numerical equations for variation with time t of

1. displacement x in cm

2. speed v in ms-1

(b) Calculate total energy of oscillation of mass

Reference: Past Exam Paper – June 2013 Paper 42 Q3

Solution 586:

(a)

(i) Angular frequency ω = 2π/T = 2π/0.69 = 9.1rads-1

(ii)

1. Displacement x = 2.1cos(9.1t)

{Maximum speed v0 = ωa where a is the amplitude (maximum displacement). This is a known formula.}

Maximum speed vo = ωa = 9.1 x (2.1x10-2) = 0.19ms-1

{The displacement is x. Speed v = dx/dt. By differentiating the equation for displacement, we can obtain the speed. OR some students may learn it in this way: if the displacement contains cosine, then speed will contain sine and vice versa.}

Speed v = vo sin(9.1t)

(b)

Energy = either ½ mvo2 or = ½ mω2xo2

= either ½ (0.078) (0.19)2 or = ½ (0.078) (9.1)2 (2.1x10-2)2

Energy = 1.4x10-3J

Question 587: [Kinematics + Energy]

Two planks of wood AB and BC are inclined at angle of 15° to the horizontal. The two wooden planks are joined at point B, as shown in Fig.1.

Small block of metal M is released from rest at point A. It slides down the slope to B and up the opposite side to C. Points A and C are 0.26 m above B. Assume frictional forces are negligible.

(a)

(i) Describe and explain acceleration of M as it travels from A to B and from B to C.

(ii) Calculate time taken for M to travel from A to B.

(iii) Calculate speed of M at B.

(b) Plank BC is adjusted so that the angle it makes with the horizontal is 30°. M is released from rest at point A and slides down slope to B. It then slides a distance along the plank from B towards C.

Use law of conservation of energy to calculate this distance. Explain your working.

Reference: Past Exam Paper – November 2012 Paper 23 Q2

Solution 587:

(a)

(i) The accelerations (from A to B and from B to C) have the same magnitudes {due to the same component of the weight as the angles are similar} but are opposite in directions {from A to B, the acceleration is the same direction as motion but from B to C, the acceleration opposes motion} OR both accelerations are toward B.

In both cases (from A to B and B to C), the component of the weight down the slope provides the acceleration.

(ii)

Acceleration {= component of weight along slope} = g sin15°

Equation for uniformly accelerated motion: s = ut + ½ at2 = 0 + ½ at2

From the triangle formed: s = 0.26 / sin 15 ° = 1.0

{Acceleration a = g sin15°

10 = ½ (9.8sin15°) t2}

t2 = (1.0 × 2) / (9.8 × sin15°)

Time t = 0.89 s

(iii)

{EITHER v = u + at OR v2 = u2 + 2as}

Speed v = 0 + (g sin15)t or v2 = 0 + 2(g sin15) × 1.0

Speed v = 2.26 m s–1

(b)

{From the conservation of energy, the loss of gravitational potential energy (GPE) at A is equal to the gain in GPE at C (which is now at a greater height since the angle has been increased).

Alternatively, B is the lowest position in the system – so the GPE is B is less. B is at a higher position than B, so the GPE at C is more than at B. From the conservation of energy, the loss of kinetic energy (KE) at B is equal to the gain in GPE at C.}

Loss of GPE at A = gain in GPE at C OR loss of KE at B = gain in GPE at C

{For Loss of GPE at A = gain in GPE ,

As stated in the question, A is higher than B by h1 = 0.26m. So, the loss of GPE at A = mgh1. Or put in another way, the initial energy at A is mgh1. This is the maximum energy of M. From the conservation of energy, this amount of energy cannot increase by itself (unless work is done to increase the energy, which is not the case here). So, the gain in energy at C must be equal to mgh1. That is, the height h2 at C is equal to h1 at A, which is 0.26m.}

EITHER Height h1 = h2 = 0.26 m

OR ½ mv2 = mgh2

{As calculated above, the speed at B is 2.26 ms-1.}

giving Height h2 = 0.5 × (2.26)2 / 9.81 = 0.26 m

From the triangle formed: Slope x = 0.26 / sin 30° = 0.52 m