• Physics 9702 Doubts | Help Page 25

Question 138: [Dynamics > Momentum]
Diagram shows two spherical masses approaching each other head-on at an equal speed u. One has mass 2m and other has mass m.

Which diagram, showing situation after the collision, shows result of an elastic collision?

Reference: Past Exam Paper – November 2009 Paper 12 Q8 & November 2009 Paper 11 Q9



Solution 138:

Answer: A.

For an elastic collision,

Velocity of approach (before collision) = Velocity of separation (after collision)

{The above result can be obtained by considering that for elastic collision, both momentum and kinetic energy is conserved. Momentum, p = mv. Kinetic energy = ½mv². By equating the sum of momentum before collision to that after collision and by equating the sum of KE before collision to that after collision, 2 equations are obtained which can be simplified into the above stated result: Velocity of approach (before collision) = Velocity of separation (after collision). The proof will not be shown here}

{Approach means that the 2 sphere are coming towards each other and separation means that they are moving away from each other}

Before collision, velocity of approach = u + u = 2u

Consider A:

Velocity of separation = (u/3) + (5u/3) = 6u/3 = 2u

Consider B:

Velocity of separation = (u/6) + (2u/3) = 5u/6

Consider C:

Both spheres are moving in the same direction. So, speed of separation is the difference in the 2 speed here/

Velocity of separation = (2u/3) – (u/6) = 3u/6 = 0.5u

Consider D:

Since the spheres stick together, they are not separating. They move together.

Velocity of separation = 0

Only answer A gives velocity of separation = 2u.

Question 139: [Dynamics > Inelastic collision]
Which quantities are conserved in inelastic collision?

Reference: Past Exam Paper – June 2008 Paper 1 Q17



Solution 139:

Answer: C.

The laws of conservation of energy and conservation of momentum state that, in a system, the total energy and total momentum is always conserved.

Momentum in a system is always conserved in any collision.  Linear momentum of the momentum of a body in 1 dimension (linear motion).

In an inelastic collision, kinetic energy is NOT conserved. Some of the initial kinetic energy is converted into other forms of energy. So, the total energy is still conserved.

 









Question 140: [Waves > Polarization > Intensity]
When plane-polarised light of amplitude A is passed through polarising filter as shown, amplitude of the light emerging is A cosθ.

Intensity of initial beam is I.
What is the intensity of emerging light when θ is 60.0°?
A 0.250 I                     B 0.500 I                     C 0.750 I                     D 0.866 I

Reference: Past Exam Paper – November 2010 Paper 11 Q25 & Paper 13 Q24 & November 2013 Paper 11 & 12 Q30



Solution 140:
Answer: A.
The intensity, I of the light is proportional to the square of the amplitude, A.

Amplitude of emerging light = A cos 60°
Intensity of emerging light is proportional to A² cos²(60°) = 0.250A²
[Since A² ∝ I,] Thus, the intensity of emerging light is equal to 0.250I.









Question 141: [Current of Electricity > Charge]
Current in a component is reduced uniformly from 100 mA to 20 mA over period of 8.0s.
What is the charge that flows during this time?
A 160 mC                   B 320 mC                    C 480 mC                    D 640 mC

Reference: Past Exam Paper – June 2003 Paper 1 Q30 & November 2013 Paper 13 Q32



Solution 141:

Answer: C.

Charge, Q = It where I is the current and t is the time

Since the current I is changing, the average current should be calculated. {The fact that the average current needs to be calculated would have been deduced if a quick sketch graph was drawn.}

Average current = (100 + 20) /2 = 60mA

Charge, Q = 60 x 8 = 480mC

Question 142: [Current of Electricity > Resistance > Non-uniform wire]
Circular cross-sectional area of metal wire varies along its length. There is current in the wire. Narrow end of wire is at a reference potential of zero.

Which graph best represents variation with distance x along wire of the potential difference V relative to the reference zero?

Reference: Past Exam Paper – June 2013 Paper 13 Q32



Solution 142:

Answer: C.

We need the variation with distance x along the wire (the cross-sectional area of which varies with length) of the potential difference V relative to the reference zero (so, at x = 0, the potential is maximum {at x = 0, we are not at zero potential, but at the start of the wide cross-sectional area}).  All the 4 graphs correctly indicate the potential difference to be maximum at x = 0.

Ohm’s law: V = IR

The current is constant along the wire. Only the resistance changes and this causes a change in the potential difference.

Resistance of wire, R = ρL/A

This requires careful thinking.

First, assume that the wire was uniform, with the diameter not changing. When the length of wire L is furthest away from the zero potential reference, the potential difference would be maximum since length L is highest, causing resistance R to maximum and thus, potential difference V = IR is also the highest. As distance x increases, the length L (from the zero potential reference) decreases, and so, R and finally V also decreases. [A is incorrect]

So, as explained above, as x increases, V should be decreasing.

Now, the diameter of the wire is falling linearly (as x increases) but the area of cross-section, A is not falling linearly [since cross-sectional area = π(d / 2)² where d is the diameter].

Since resistance R is inversely proportional to cross-sectional area A, it is also inversely proportional to d² (when d is small, R is big). Therefore, there is a greater percentage fall (in resistance, and thus, potential difference) per unit length at narrow end than at wide end (that is, the decrease is not linear, but depends on d²). [B, which represents a linear decrease, is incorrect]

This means that potential difference per unit length (this is represented by the gradient of the graphs shown which equal to V/x) is less at the wide end than at the narrow end (the gradient at the highest value of x should be greater than the gradient at x = 0 and the gradient keeps on increasing as x increases). [Answer D is wrong since gradient is highest at x = 0]

This makes the correct answer C.

Question 143: [Kinematics > Projectile motion]

Projectile is launched at 45° to horizontal with initial kinetic energy E.

Assuming air resistance to be negligible, what will be the kinetic energy of projectile when it reaches its highest point?

A 0.50 E                      B 0.71 E                      C 0.87 E                      D E

Reference: Past Exam Paper – November 2009 Paper 12 Q13

Solution 143:

Answer: A.

Let the initial velocity = v

Horizontal component of velocity = v cos(45)

Vertical component of velocity = v sin(45)

At the highest point, the vertical component of velocity of the projectile is zero. The projectile only has a horizontal component of velocity which is v cos(45) [since air resistance is negligible, this component is unchanged].

 

Kinetic energy = ½ mv²

Kinetic energy is proportional to (speed)². Initial kinetic energy = E

Therefore, at the highest point, the kinetic energy is proportional to [vcos(45)]² = 0.5v².

So, kinetic energy is halved at the highest point.

Question 144: [Waves > Graph]

Speed v of waves in deep water is given by equation v² = gλ / 2π where λ is the wavelength of waves and g is the acceleration of free fall.

Student measures wavelength λ and frequency f of a number of these waves.

Which graph should he plot to give a straight line through origin?

A f² against λ

B f against λ²

C f against 1 / λ

D f² against 1/ λ

Reference: Past Exam Paper – June 2014 Paper 11 Q24

Solution 144:

Answer: D.

The equation given is:             v² = gλ / 2π

Speed of wave, v = fλ

Substituting v = fλ in the 1^st equation gives (fλ)² = gλ / 2π which is simplified into

f² = g / 2πλ = (g / 2π) (1 / λ)

Equation of a straight line:      y = mx + c

If the straight line passes through the origin, the equation becomes: y = mx

Therefore, a plot of f² against 1/λ would give a straight line through the origin.

Question 145: [Nuclear Physics > Nucleus]

What is the approximate mass of a nucleus of uranium?

A 10^–15kg                    B 10^–20kg                     C 10^–25kg                     D 10^–30kg

Reference: Past Exam Paper – June 2008 Paper 1 Q39

Solution 145:

Answer: C.

The unified atomic mass constant, u = 1.66x10^-27kg

Uranium contains about 238 nucleons.

Therefore, mass of uranium nucleus = 238 x 1.66x10^-27 ≈ 10^-25kg