Monday, March 2, 2015

Physics 9702 Doubts | Help Page 74

  • Physics 9702 Doubts | Help Page 74


Question 397: [Kinematics]
A ball is to be kicked so that, at the highest point of its path, it just clears a horizontal cross-bar on a pair of goal posts. The ground in level and the cross-bar is 2.5m high. The ball is kicked from the ground level with an initial speed of 8m/s. Calculate the angle of projection of the ball and the distance of the point where the ball was kicked from the goal-line.

Reference: “Cambridge International As and A level Physics,” 2nd edition, by Mike Crundell, Geoff Goodwin and Chris Mee, Chapter 3 (Kinematics) Q3a



Solution 397:
Let the initial speed of the ball = v = 8ms-1

Assuming that the angle at which the ball is kicked from the ground is θ, the 2 components of the speed may be written as
Horizontal component = vh = v cosθ = 8 cosθ
Vertical component = vv = v sinθ = 8 sinθ

Consider the vertical motion.
Take the upward direction as the positive direction. The acceleration due to gravity (which is downwards), g = - 9.81ms-2. The maximum height, s reached is 2.5m. At the maximum height, the vertical speed, v is zero.

Equation for uniformly accelerated motion: v2 = u2 +2as
02 = vv2 + 2 (-9.81) (2.5)
Speed vv = √[2(9.81)(2.5)] = 7.00ms-1

vv = 8 sinθ = 7.00ms-1
Angle θ = sin-1 (7.00 / 8) = 61.0°

Acceleration = (final speed – initial speed) / time
Time, t taken to reach maximum height = (0 – 7.00) / (-9.81) = 0.7136s


Now, consider the horizontal motion.
The acceleration of free fall does not affect the horizontal motion.
Initial speed, vh = 8 cos(61) = 3.88ms-1

When time t = 0.7136s,
Distance s = ut + ½ at2 = 3.88 (0.7136) + 0 = 2.78m

Therefore, the ball was kicked at a distance of 2.78m from the goal post.









Question 398: [Current of Electricity]
Two wires P and Q, each of the same length and the same material, are connected in parallel to a battery. The diameter of P is half that of Q.
What fraction of the total current passes through P?
A 0.20                         B 0.25                         C 0.33                         D 0.50

Reference: Past Exam Paper – N94 / I / 16



Solution 398:
Answer: A.
Resistance R of a wire = ρL / A

Since the wires are of the same length and material, ρ and L are similar for both.

Let the diameter of P = d. So, diameter of Q = 2d.
Cross-section area of P = π(d/2)2 = πd2/4 = A            (let πd2/4 be A)
Cross-section area of Q = π(2d/2)2 = πd2 = 4A

In a parallel connection, the p.d. across the junctions is the same, but the current splits.

Resistance of wire P = ρL / A = R
Resistance of wire Q = ρL / 4A = R/4

Let the p.d. across the junctions of the parallel combination be V.
Ohm’s law: V = IR

For wire P: V = IPR giving current IP = V/R
For wire Q: V = IQ(R/4) giving current IQ = 4V/R

Total current I = V/R + 4V/R = 5V/R

Fraction of total current that passes through P = IP / I = (V/R) / (5V/R) = 1/5 = 0.20









Question 399: [Electric field]
Diagram shows two parallel horizontal metal plates held at potential difference V.

Small charged liquid drop, midway between plates, is held in equilibrium by the combination of its weight and the electric force acting on it.
Acceleration of free fall is g and electric field strength is E.
What is the ratio of charge to mass of the drop, and the polarity of the charge on the drop?
For November 2009:

For November 2014:

Reference: Past Exam Paper – November 2009 Paper 11 Q28 & Paper 12 Q27 & November 2014 Paper 11 & 12 Q28



Solution 399:
Answer: B.
The force of gravity acts downwards (towards the ground) and the direction of the electric force (acting on a positive charge) is taken from positive to negative, so it is also downwards.

For the charge to be in equilibrium, the electric force should cause it to move upwards, towards the positive plate. So, the polarity of the oil drop should be negative for it to move up.

For equilibrium,
Magnitude of electric force = Weight of oil drop
Eq = mg
Ratio of q/m = g/E









Question 400: [Electric field]
Which row describes the circumstances under which forces act on charged particle in uniform electric field?
Charged particle                                  Direction of force
A         moving charges only                           parallel to the field
B         stationary charges only                       perpendicular to the field
C         stationary and moving charges           parallel to the field
D         stationary and moving charges           perpendicular to the field

Reference: Past Exam Paper – June 2010 Paper 11 Q26 & Paper 12 Q29 & Paper 13 Q28



Solution 400:
Answer: C.
The uniform field being considered is an ELECTRIC field.

The electric force in an electric field acts on both stationary and moving charges. [A and B incorrect]

The direction of the electric force acting on the charged particle is, of course, parallel to the field. [D is incorrect]









Question 401: [Units]
Which formula could be correct for speed v of ocean waves in terms of density ρ of seawater, the acceleration of free fall g, the depth h of the ocean and the wavelength λ?
A v = √(gλ)                 B v = √(g/h)                C v = √(ρgh)               D v = √(g/ρ)

Reference: Past Exam Paper – June 2007 Paper 1 Q3



Solution 401:
Answer: A.
Unit of velocity: ms-1.

Consider the units of the choices given. The unit of the correct choice should have the units of velocity.

Unit of √(gλ): [ms-2 m]0.5 = [m2s-2]0.5 = ms-1 
Unit of √(g/h): [ms-2 m-1]0.5 = [s-2]0.5 = s-1
Unit of √(ρgh): [kgm-3 ms-2 m]0.5 = [kg m-1 s-2]0.5 = kg0.5 m-0.5 s-1
Unit of √(g/ρ): [ms-2 kg-1m3]0.5 = [m4 s-2 kg-1]0.5 = m2 s-1 kg-0.5



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