Monday, August 31, 2015

Physics 9702 Doubts | Help Page 192

  • Physics 9702 Doubts | Help Page 192

Question 932: [Dynamics > Collisions]
Two spheres approach each other along the same straight line. Their speeds are u1 and u2 before collision. After collision, the spheres separate with speeds v1 and v2 in the directions shown below.

Which equation must be correct if the collision is perfectly elastic?
A u1 – u2 = v2 + v1
B u1 – u2 = v2 – v1
C u1 + u2 = v2 + v1
D u1 + u2 = v2 – v1

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 1st sphere is moving towards the 2nd one while the 2nd sphere is moving away from the 1st one. If the speed of the 1st sphere (v1) is greater than v2, it would actually be approaching the 2nd sphere, and if the speed of the 2nd sphere (v2) is greater than v1, it would actually be moving away (separating) from the 1st one.

We want to relative speed of SEPARATION (that is, v2 is assumed to be greater than v1). 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.

(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:
(i) Resonance

(ii) Amplitude = 16mm           and      Frequency = 4.6Hz

a = (-)ω2x        and      ω = 2πf
So, a = (2π × 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 Vout with a fixed reference voltage, which is set at 4.8 V.

Vout is equal to 4.8 V.
What is the input voltage Vin?
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:
Vout = [10 / (10 + 15)] × Vin = 10Vin / 25
Vin = 25Vout / 10

When Vout = 4.8V,
Vin = (25 × 4.8) / 10 = 12V

Currently Viewing: Physics Reference | August 2015