• 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 m² and is stretched by forces of magnitude 1.9 × 10³ 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 = (9.5×10⁸) / 0.013 = 7.3×10¹⁰ 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 = (1.9×10³) / (9.5×10⁸) = 2.0×10^-6 m² 

{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 m².

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 m²

(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