# Physics 9702 Doubts | Help Page 182

__Question 892: [Gravitation]__**(a)**Define gravitational potential.

**(b)**Explain why values of gravitational potential near to an isolated mass are all negative.

**(c)**Earth may be assumed to be an isolated sphere of radius 6.4 × 10

^{3}km with its mass of 6.0 × 10

^{24}kg concentrated at its centre. An object is projected vertically from the surface of the Earth so that it reaches an altitude of 1.3 × 10

^{4}km.

Calculate, for this object,

(i) change in gravitational
potential,

(ii) speed of projection from the
Earth’s surface, assuming air resistance is negligible.

**(d)**Suggest why the equation

v

^{2}= u^{2}+ 2as
is not appropriate for calculation
in (c)(ii).

**Reference:**

*Past Exam Paper – June 2003 Paper 4 Q1*

__Solution 892:__**(a)**Gravitational potential (at a point) is defined as the work done in bringing/moving unit mass from infinity to the point.

**(b)**The potential at infinity is defined as being zero. The forces are always attractive, so work got out in moving to point (work is done on the mass when moving it to infinity).

**(c)**

(i)

Gravitational potential, φ = - GM / R
= - GM × (1/R)

{The distances should be
converted in metre. Initial position is at the surface of the Earth, which is a
distance of 6.4×10

^{6}m from the centre of the Earth. The final distance is 1.3×10^{7}m (altitude = distance above surface) + 6.4×10^{6}m (distance of surface of the centre) = 1.94×10^{7}m}
Change in potential = (6.67×10

^{-11}) (6.0×10^{24}) × ({6.4×10^{6}}^{-1}– {1.94×10^{7}}^{-1})
Change in potential = 4.19 × 10

^{7}J kg^{-1}(ignore sign)
(ii)

The kinetic energy is
converted to gravitational potential energy as the height of the object from
the surface of the Earth increases.

½ mv

^{2}= mΔφ
v

^{2}= 2 × 4.19×10^{7}= 8.38 × 10^{7}
Speed v = 9150 m s

^{-1}**(d)**The acceleration is not constant.

__Question 893: [Kinematics > Graph]__
Graph shows how the velocity v of a
firework rocket changes with time t.

At which point on the graph does the
rocket have the greatest acceleration?

**Reference:**

*Past Exam Paper – November 2013 Paper 13 Q7*

__Solution 893:__**Answer: B.**

The gradient of
a velocity-time graph gives the acceleration.

Thus, the
acceleration is greatest where the value of the gradient is greatest. Visually,
this is be identified by the steepness of the tangent at a point – the steeper
the tangent, the greater is the gradient.

The tangent is
steepest at point B, compared with the other points.

__Question 894: [__

__Dynamics]__
When a golfer hits a ball his club
is in contract with the ball for about 0.0005 s and the ball leaves the club
with a speed of 70 m/s. The mass of the ball is 46 g. Determine the mean force
on the ball?

**Reference:**

*???*

__Solution 894:__
From Newton’s 2

^{nd}law, Force F is equal / proportional to rate of change of momentum. (F = Δp / t)
Change in momentum is given by p =
mΔv

(Assuming all other velocities are
zero – during the contact, the ball is momentarily stationary)

So, Force = mΔv / t = (0.046 × 70) /
0.0005 = 6440N

__Question 895: [Alternating Current > Rectification]__
Alternating supply of frequency 50
Hz and having an output of 6.0 V r.m.s. is to be rectified so as to provide
direct current for a resistor R. The circuit of Fig.1 is used.

The diode is ideal. The Y-plates of
a cathode-ray oscilloscope (c.r.o.) are connected between points A and B.

**(a)**

(i) Calculate maximum potential
difference across the diode during one cycle.

(ii) State potential difference
across R when the diode has maximum potential difference across it. Give a
reason for your answer.

**(b)**Y-plate sensitivity of the c.r.o. is set at 2.0 V cm

^{–1}and the time-base at 5.0 ms cm

^{–1}.

On Fig.2, draw the waveform that is
seen on the screen of the c.r.o.

**(c)**A capacitor of capacitance 180 μF is connected into the circuit to provide smoothing of the potential difference across the resistor R.

(i) On Fig.1, show the position of the
capacitor in the circuit.

(ii) Calculate energy stored in the
fully-charged capacitor.

(iii) During discharge, potential
difference across the capacitor falls to 0.43V

_{0}, where V_{0}is the maximum potential difference across the capacitor.
Calculate fraction of the total
energy that remains in the capacitor after the discharge.

**Reference:**

*Past Exam Paper – November 2006 Paper 4 Q6*

__Solution 895:__**(a)**

(i) Peak voltage = 6√2 = 8.48 V

(ii) The potential difference is zero
because EITHER there is no current in circuit (and V = IR) OR all p.d. is across
the diode

**(b)**

The waveform is that of a half-wave
rectification.

{Peak voltage = 8.48V. 1cm
represents 2.0V. 8.48V is represented by 8.48 / 2.0 = 4.24cm}

The peak height is at about 4.25 cm

{Frequency f is 50Hz.
Period T = 1/f = 1/50 = 0.02s = 20ms. Half-period = 10ms. 1cm represents 5.0ms.
20ms is represented by 20 / 5 = 4cm. Half-period is represented by 2cm.}

The half-period spacing is 2.0 cm

**(c)**

(i) The capacitor should be shown in
parallel with the resistor.

(ii)

EITHER Energy = ½ CV

^{2}OR = ½ QV__and__Q = CV
Energy = ½ ×
(180×10

^{–6}) × (6√2)^{2}= 6.48 × 10^{–3}J
(iii)

{EITHER Energy is
proportional to V

^{2}. Fraction = remaining energy / total energy.
Remaining energy is proportional
to (0.43V

_{0})^{2}. Total energy is proportional to V_{0}^{2}.
So, fraction = (0.43V

_{0})^{2}/ V_{0}^{2}= 0.43^{2}
OR The actual value of the
remaining energy could be calculated and the fraction could be found using the
previously calculated value of the total energy in (c)(ii). But this is too
time-consuming.}

EITHER fraction = 0.43

^{2}OR final energy = 1.2 mJ
Fraction
= 0.18