# 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,” 2*

^{nd}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 = v

_{h}= v cosθ = 8 cosθ

Vertical component = v

_{v}= 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: v

^{2}= u

^{2}+2as

0

^{2}= v

_{v}

^{2}+ 2 (-9.81) (2.5)

Speed v

_{v}= √[2(9.81)(2.5)] = 7.00ms

^{-1}

v

_{v}= 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, v

_{h}= 8 cos(61) = 3.88ms

^{-1}

When time t = 0.7136s,

Distance s = ut + ½ at

^{2}= 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}= πd

^{2}/4 = A (let πd

^{2}/4 be A)

Cross-section area of Q = π(2d/2)

^{2}= πd

^{2}= 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 = I

_{P}R giving current I

_{P}= V/R

For wire Q: V = I

_{Q}(R/4) giving current I

_{Q}= 4V/R

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

Fraction of total current that passes through P = I

_{P }/ 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 fieldB 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

Unit of √(g/h):
[ms^{-2}m]^{0.5}= [m^{2}s^{-2}]^{0.5}= ms^{-1}^{-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}= kg

^{0.5}m

^{-0.5}s

^{-1}

Unit of √(g/ρ): [ms

^{-2}kg

^{-1}m

^{3}]

^{0.5}= [m

^{4}s

^{-2}kg

^{-1}]

^{0.5}= m

^{2}s

^{-1}kg

^{-0.5}

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