# Physics 9702 Doubts | Help Page 192

__Question 932: [Dynamics > Collisions]__
Two spheres approach each other
along the same straight line. Their speeds are u

_{1}and u_{2}before collision. After collision, the spheres separate with speeds v_{1}and v_{2}in the directions shown below.
Which equation must be correct if
the collision is perfectly elastic?

A u

_{1}– u_{2}= v_{2}+ v_{1}
B u

_{1}– u_{2}= v_{2}– v_{1}
C u

_{1}+ u_{2}= v_{2}+ v_{1}
D u

_{1}+ u_{2}= v_{2}– v_{1}**Reference:**

*Past Exam Paper – November 2008 Paper 1 Q10 & June 2012 Paper 12 Q12*

__Solution 932:__**Answer: D.**

For a perfectly elastic collision, both
KE and momentum are conserved.

The

**relative speed of**__approach____before__collision should be equal to the**relative speed of**__separation____after__collision in an elastic collision.
Since before collision, the 2 spheres
are moving towards each other (they are APPROACHing each other), the relative
speed of approach is the sum of the 2 speeds.

After the collision, the 1

^{st}sphere is moving towards the 2^{nd}one while the 2^{nd}sphere is moving away from the 1^{st}one. If the speed of the 1^{st}sphere (v_{1}) is greater than v_{2}, it would actually be approaching the 2^{nd}sphere, and if the speed of the 2^{nd}sphere (v_{2}) is greater than v_{1}, it would actually be moving away (separating) from the 1^{st}one.
We want to relative speed of
SEPARATION (that is, v

_{2}is assumed to be greater than v_{1}). The relative speed of separation is v2 – v1 since both are moving in the same direction.

__Question 933: [Electric field]__
Two charged parallel metal plates
produce an electric field.

A charged particle moves from X to
Y.

Which graph shows variation of the
force on the particle with distance from X along the line XY?

**Reference:**

*Past Exam Paper – November 2013 Paper 13 Q30*

__Solution 933:__**Answer: C.**

The electric field strength E (= V / d) between the parallel metal plates is constant.

Force on the particle = Eq where q is the charge of the particle

The force does not depend on the distance of the particle from the plates. So, the electric force is constant.

__Question 934: [Simple harmonic motion]__
A student sets up the apparatus
illustrated in Fig.1 in order to investigate the oscillations of a metal cube
suspended on a spring.

Amplitude of the vibrations produced
by the oscillator is constant.

The variation with frequency of the
amplitude of the oscillations of the metal cube is shown in Fig.2.

**(a)**

(i) State the phenomenon illustrated
in Fig.2.

(ii) For the maximum amplitude of
vibration, state the magnitudes of the amplitude and the frequency.

**(b)**The oscillations of the metal cube of mass 150 g may be assumed to be simple harmonic.

Use answers in (a)(ii) to determine,
for the metal cube,

(i) its maximum acceleration,

(ii) maximum resultant force on the
cube

**(c)**Some very light feathers are attached to the top surface of the cube so that the feathers extend outwards, beyond the vertical sides of the cube.

The investigation is now repeated.

On Fig.2, draw a line to show the
new variation with frequency of the amplitude of vibration for frequencies
between 2 Hz and 10 Hz.

**Reference:**

*Past Exam Paper – November 2010 Paper 41 & 42 Q3*

__Solution 934:__**(a)**

(i) Resonance

(ii) Amplitude = 16mm

__and__Frequency = 4.6Hz**(b)**

(i)

a = (-)Ï‰

^{2}x and Ï‰ = 2Ï€f
So, a = (2Ï€ × 4.6)

^{2}× (16×10^{-3}) = 13.4ms^{-2}
(ii) F = ma = (150×10

^{-3}) × 13.4 = 2.0N**(c)**The line is always ‘below’ the given line in the graph and never zero. Its peak is at 4.6Hz (or slightly lower frequency) and it is flatter (peak is not as sharp as the one given).

__Question 935: [Current of Electricity > Potential divider]__
Circuit is designed to trigger an
alarm system when the input voltage exceeds some preset value. It does this by
comparing V

_{out}with a fixed reference voltage, which is set at 4.8 V.
V

_{out}is equal to 4.8 V.
What is the input voltage V

_{in}?
A 4.8 V B 7.2 V C
9.6 V D 12 V

**Reference:**

*Past Exam Paper – November 2007 Paper 1 Q33*

__Solution 935:__**Answer: D.**

This is a potential divider circuit.
The potential divider equation can be written as follows:

V

_{out}= [10 / (10 + 15)] × V_{in}= 10V_{in}/ 25
V

_{in}= 25V_{out}/ 10
When V

_{out}= 4.8V,
V

_{in}= (25 × 4.8) / 10 = 12V