# Physics 9702 Doubts | Help Page 214

__Question 1024: [Kinematics > Non-uniform acceleration]__
A trolley of mass 930 g is held on a horizontal surface by means of two
springs, as shown in Fig. 4.1.

The variation with time t of the speed v of the trolley for the first
0.60 s of its motion is shown in Fig. 4.2.

**(a)**Use Fig. 4.2 to determine

(i) the initial acceleration of the trolley,

(ii) the distance moved during the first 0.60 s of its motion.

**(b)**(i) Use your answer to (a)(i) to determine the resultant force acting on the trolley at time t = 0.

(ii) Describe qualitatively the variation with time of the resultant
force acting on the trolley during the first 0.60 s of its motion.

**Reference:**

*Past Exam Paper – November 2005 Paper 2 Q4*

__Solution 1024:__**(a)**

(i)

{The gradient of a speed-time graph gives the
acceleration.

Gradient = Δy / Δx = Δspeed /
Δtime = acceleration

To find the acceleration at a point, the gradient of
the tangent at that point is calculated. (The tangent at a point is a straight
line that touches only that point on the curve.)}

Use of a tangent at time t = 0

(Gradient =) Acceleration = 42 ± 4 cms

^{-2}
(ii)

Use of area of the loop (gives the distance travelled)

Distance = 0.031 ± 0.001 m

**(b)**

(i) F = ma = 0.93 × 0.42 = 0.39N

(ii) The resultant force reduces to zero in the first 0.3s. It then
increases again in the next 0.3s in the opposite direction.

__Question 1025: [Matter > Elastic and Plastic Behaviour]__
A sample of material in the form of
a cylindrical rod has length L and uniform area of cross-section A. The rod
undergoes an increasing tensile stress until it breaks.

Fig. 4.1 shows the variation with
stress of the strain in the rod.

**(a)**State whether the material of the rod is ductile, brittle or polymeric.

**(b)**Determine the Young modulus of the material of the rod.

**(c)**A second cylindrical rod of the same material has a spherical bubble in it, as illustrated in Fig. 4.2.

The rod has an area of cross-section
of 3.2 × 10

^{–6}m^{2}and is stretched by forces of magnitude 1.9 × 10^{3}N.
By reference to Fig. 4.1, calculate
the maximum area of cross-section of the bubble such that the rod does not
break.

**(d)**A straight rod of the same material is bent as shown in Fig. 4.3.

Suggest why a thin rod can bend more
than a thick rod without breaking.

**Reference:**

*Past Exam Paper – November 2007 Paper 2 Q4*

__Solution 1025:__**(a)**Brittle

**(b)**Young modulus = stress / strain = (9.5×10

^{8}) / 0.013 = 7.3×10

^{10 }Pa

**(c)**

Stress = Force / Area

{Since the cross-sectional
area is inversely proportional to the stress, the maximum stress possible (the
value of stress at the breaking point from the graph) would correspond to the
minimum cross-sectional area of the rod.

(minimum) area [of rod] =
force / (stress at breaking point)}

(minimum) area of rod = (1.9×10

^{3}) / (9.5×10^{8}) = 2.0×10^{-6 }m^{2}
{The area calculated above
is the minimum possible cross-sectional area of the rod so that it does not
break.

From the question: The rod
has an area of cross-section of 3.2 × 10

^{–6}m^{2}.
So aside from the minimum
area of rod calculated, the remaining area of the actual rod can be occupied by
the bubble. We thus take the difference between these 2 areas.}

(maximum) area of cross-section of
bubble = (3.2 – 2.0) ×10

^{-6}= 1.2×10^{-6 }m^{2}**(d)**When bent, the ‘top’ and ‘bottom’ edges of the rod have different extensions. With a thick rod, this difference is greater (than with a thin rod). So, the thick rod breaks with less bending.

{Consider the extension of
the ‘outer’ edge and the compression of the ‘inner’ edge as the rod bends. The
thick rod would break more easily because the extension and the compression
would be greater.}

__Question 1026:__

__[Kinematics > Linear motion]__**(a)**Define

*speed*and

*velocity*and use these definitions to explain why one of these quantities is a scalar and the other is a vector.

**(b)**A ball is released from rest and falls vertically. The ball hits the ground and rebounds vertically, as shown in Fig. 2.1.

The variation with time t of the velocity v of the ball is shown in Fig.
2.2.

Air resistance is negligible.

(i) Without calculation, use Fig. 2.2 to describe the variation with
time t of the velocity of the ball from t = 0 to t = 2.1 s.

(ii) Calculate the acceleration of the ball after it rebounds from the
ground. Show your working.

(iii) Calculate, for the ball, from t = 0 to t = 2.1 s,

1. the distance moved,

2. the displacement from the initial position.

(iv) On Fig. 2.3, sketch the variation with t of the speed of the ball.

**Reference:**

*Past Exam Paper – June 2015 Paper 21 Q2*

__Solution 1026:__**(a)**

Speed = distance / time and
Velocity = displacement / time

Speed is a scalar as distance has no direction

**and**velocity is a vector as displacement has direction**(b)**

(i) There is a

__constant__acceleration or linear/uniform increase in velocity until 1.1 s. The ball rebounds or bounces or changes direction and decelerates to zero velocity at the same acceleration as the initial value
(ii)

Acceleration a = (v – u) / t OR
use of gradient implied

Acceleration a = (8.8 + 8.8) / 1.8 OR
appropriate values from line OR =
(8.6 + 8.6) / 1.8

Acceleration a = 9.8 (9.78) m s

^{–2}OR = 9.6 m s^{–2}
(iii)

1.

Distance = first area above graph + second area below graph

Distance = {(1.1 × 10.8) / 2} + {(0.9 × 8.8) / 2} (= 5.94 + 3.96)

Distance = 9.9 m

2.

Displacement = first area above graph – second area below graph

Displacement = {(1.1 × 10.8) / 2} – {(0.9 × 8.8) / 2}

Displacement = 2.0 (1.98) m

(iv)

The sketch should have

correct shape with straight lines and all lines above the time axis or
all below

correct times for zero speeds (0.0, 1.15 s, 2.1 s) and peak speeds (10.8
m s

^{–1}at 1.1 s and 8.8 m s^{–1}at 1.2 s and 3.0 s)

__Question 1027: [Waves > Diffraction]__
Light of wavelength 700 nm is incident on a pair of slits, forming
fringes 3.0 mm apart on a screen. What is the fringe spacing when light of
wavelength 350 nm is used and the slit separation is doubled?

A 0.75 mm B 1.5 mm
C 3.0 mm D 6.0 mm

**Reference:**

*Past Exam Paper – June 2008 Paper 1 Q29*

__Solution 1027:__**Answer: A.**

For double slit: x = λD / a

where x = fringe separation, λ =
wavelength, D = distance between slits and screen and a = slit separation

When λ = 700nm, x = 3.0mm

3.0 = 700 (D / a)

D / a = 3.0 / 700

When λ = 350, split separation = 2a

x = λD / 2a = (350 / 2) × (D
/ a) = (350 / 2) × (3.0 / 700) = 0.75mm

Please consider answering ALL of the following questions:

ReplyDelete4/O/N/02 Q.5(b),Q.6(c)(i)

6/O/N/02 Q.11(a)(b)

6/O/N/03 Q.9

04/M/J/04 Q.8(a),(b)(i),(ii)1.

06/M/J/04 Q.9(b)(iii),Q.11(b)

06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)

04/M/J/05 Q.7(a)

06/O/N/05 Q.8(b),Q.10(a)

04/M/J/06 Q.6(a),(c),Q.7(b)

06/M/J/06 Q.14(b)

04/O/N/06 Q.3(c)

06/O/N/06 Q.3(b)

05/M/J/07 Q.2(d)

04/O/N/07 Q.7(b)(i),(c),Q.10(c)

04/M/J/08 Q.5(b),Q.9(b)

41/O/N/09 Q.10

51/M/J/10 Q.2(d)

For 04/O/N/07 Q.7(b)(i),(c), go to

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Undoubtedly helpful , Thanks !

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