• Physics 9702 Doubts | Help Page 99

Question 508: [Dynamics]

Stone is projected horizontally in a vacuum and moves along path shown.

X is a point on this path. XV and XH are vertical and horizontal lines respectively through X. XT is the tangent to path at X.

Along which directions do forces act on the stone at X?

A XV only                  B XH only                  C XV and XH                        D XT only

Reference: Past Exam Paper – June 2011 Paper 12 Q14

Solution 508:

Answer: A.

Since the stone is projected in a vacuum, there is no air. Therefore, there is no air resistance.

The force of gravity acts downwards. So, at any point along the path, there will be a force downwards – this will along XV in the diagram given.

As this is in a vacuum, there is no air resistance, nor any other forms of friction (forces that oppose motion).

XT represents the direction of motion of the stone at a point on the path.

XH represents the horizontal component of the motion (speed).

Question 509: [Current of Electricity > Potentiometer]

Diagram shows a potentiometer circuit.

Contact T is placed on the wire and moved along wire until the galvanometer reading is zero. The length XT is then noted.

In order to calculate potential difference per unit length on the wire XY, which value must also be known?

A the e.m.f. of the cell E₁

B the e.m.f. of the cell E₂

C the resistance of resistor R

D the resistance of the wire XY

Reference: Past Exam Paper – November 2008 Paper 1 Q37 & November 2012 Paper 11 Q37

Solution 509:

Answer: B.

The potentiometer circuit works as follows.

When the galvanometer reading is zero, this means that p.d. across XT of the wire is the same as the e.m.f. E₂ connected to the galvanometer.

To calculate the potential difference per unit length on the wire XY, we need to know the length of the wire, and the p.d. across it.

The p.d. across XT is known to be equal to the e.m.f. E₂ when the galvanometer reading is zero. So, we need to know the e.m.f. of the cell E₂ to know the p.d. across the wire.

Question 510: [Simple harmonic motion]

Ball of mass 37 g is held between two fixed points A and B by two stretched helical springs, as shown in Fig.

Ball oscillates along line AB with simple harmonic motion of frequency 3.5 Hz and amplitude 2.8 cm.

(a) Show that total energy of the oscillations is 7.0 mJ.

(b) At two points in oscillation of the ball, its kinetic energy is equal to the potential energy stored in the springs. Calculate magnitude of the displacement at which this occurs

(c) On axes of Fig. and using answers in (a) and (b), sketch graph to show the variation with displacement x of

(i) total energy of the system (line T):

(ii) kinetic energy of ball (line K):

(iii) potential energy stored in the springs (line P):

(d) Arrangement in Fig is now rotated through 90° so that line AB is vertical and the ball oscillates in a vertical plane.

Suggest one form of energy, other than those in (c), that must be taken into consideration when plotting new graphs to show energy changes with displacement.

Reference: Past Exam Paper – June 2012 Paper 42 Q2



Solution 510:

(a)

Energy of oscillations = ½ mω²a²       where ω = 2πf

Energy of oscillations = ½ (37x10^-3)(2π x 3.5)²(2.8x10^-2)²

Energy of oscillations = 7.0x10^-3J

(b)

Kinetic energy = Potential energy

Either

{The total energy is 7.0mJ. So, for KE = PE, they should both be halved the value of the total energy.}

KE or PE = 3.5mJ

{In this method, we are using the formula for either KE or PE which are both 3.5mJ to obtained x.

Total energy = ½ mω²a² ; Kinetic energy = at displacement x = ½ mω²(a² – x²)

And Potential energy at displacement x = ½ mω²x².}

KE = ½ mω²(a² – x²)               or PE = ½ mω²x²

Displacement x = 2.0cm

(KE or PE = 7.0mJ scores 0/3)

OR

{Here, since KE = PE, by equating them we can obtain x}

½ mω²(a² – x²) = ½ mω²x²

x = a / (2^0.5) = 2.8 / (2^0.5) = 2.0cm

OR

Allow:

ω = (k/m)^0.5 So, k = ω²m = (2π x 3.5)²(0.037) = 17.9

E = ½ kx² = 3.5x10^-3 J

x = 2.0cm

(c)

(i)

{amplitude is 2.8cm}

Graph: horizontal line with y-intercept 7.0mJ with end-points of line at +2.8cm and -2.8cm

(ii)

Graph: reasonable curve

with maximum at (0,7.0) and end-points of line at (-2.8,0) and (+2.8,0)

(iii) potential energy stored in the springs (line P):

{As calculated in (b), when KE = PE = 3.5mJ, the displacement is -2.0cm or +2.0cm.}

Graph: inverted version of part (ii)

with intersections at (-2.0,3.5) and (+2.0,3.5)

(d) Gravitational potential energy

Question 511: [Matter > Hooke’s law]

Two springs P and Q both obey Hooke’s law. They have spring constants 2k and k respectively.

Springs are stretched, separately, by force that is gradually increased from zero up to a certain maximum value, the same for each spring. Work done in stretching spring P is W_P, and work done in stretching spring Q is W_Q.

How is W_P related to W_Q?

A W_P = ¼ W_Q                         B W_P = ½ W_Q             C W_P = 2W_Q               D W_P = 4W_Q

Reference: Past Exam Paper – November 2002 Paper 1 Q24

Solution 511:

Answer: B.

Hooke’s law: F = ke    where F: force, k: spring constant and e: extension

Extension e = F / k

The work done in stretching the springs is given by the area under the force-extension graph of the spring. Since both springs obey Hooke’s law, the graphs would be straight lines with positive gradient and passing through the origin.

The area under the graph would be that of a triangle with the base being the extension and the height being the force applied.

Consider spring P. Spring constant = 2k

Let the force applied be F.

Extension, e = F / 2k = 0.5 (F/k)

Consider spring Q. Spring constant = k

Let the force applied be F.

Extension, e = F / k = 1 (F/k)

It can be observed that for any force F, the extension for spring Q is twice that of spring P. This means that the area under the graph (which represents the work done) for spring Q is twice that for spring P.

So, W_Q = 2W_P giving W_P = ½ W_Q

Question 512: [Kinematics]
Car is stationary at traffic lights. When traffic lights go green, the driver presses down sharply on the accelerator. The resultant horizontal force acting on car varies with time as shown.

Which graph shows variation with time of the speed of the car?

Reference: Past Exam Paper – June 2013 Paper 11 Q9



Solution 512:

Answer: A.

From the force-time graph given, the force on the accelerator increases sharply to a constant value (indicated by the horizontal line in the graph) at a point. This causes a constant acceleration since F = ma.

Acceleration is the rate of change of velocity.

Since the acceleration is constant, the increase in velocity in uniform with time. This is represented by a straight line graph, showing that the gradient of the speed-time graph (which gives the acceleration) is constant.