Physics - Speech or Presentation Example

Physics (PART A) Question A particle moves with uniform speed v = 2.0 m s along a circular path of radius 5.0 m. Figure shows two points A and B on the circle. The angle subtended by the arc AB at the centre O is π/3 radians. Also shown are the velocity vectors of the particle, vA and vB, when it passes through A and B, respectively.(a) Determine the length of arc from A to B and the time Δt taken for the particle to travel along this arc. (2 marks) length of arc (S) = r · θ = 5.0 * (π / 3) = 5π/3 m. ≈ 5.24 m. (b) Make a similar sketch and draw in the vector representing the change of velocity Δv = vB – vA.

Use trigonometry to obtain the magnitude |Δv|. (Some useful construction lines are shown dashed in Figure 1b.) Hence determine the magnitude of the average acceleration of the particle during its motion from A to B. (6 marks) |Δv| = |v| · Δθ = (2.0) * (π / 3) = 2π/3 m/s ≈ 2.094 m/s Magnitude of average acceleration = (Δv) / (Δt) = (2π/3)/(5π/6) = 4/5 ≈ 0.8 m/s2 Since Δt = (r · Δθ) / v = (5π/3) / 2 = 5π/6 seconds.(c) Calculate the magnitude of the instantaneous acceleration of the particle at any time. (2 marks) a = dv / dt = (2π/3) / tQuestion-2 (a) Write down Bernoulli’s equation, defining all the terms it contains and state the conditions under which it applies.

(4½ marks) P1 + ½ ρv12 + ρgh1 = P2 + ½ ρv22 + ρgh2 Where h1 and h2 refer to elevation of fluid, P1 and P2 pertain to pressures experienced by the fluid which vary inversely as the speed of the fluid given either by v1 or v2. ρ stands for the density of the fluid and Bernoulli’s equation is basically a statement of conservation of energy (relating the pressure energy, PE, and KE of a perfect fluid) that applies to conditions along a streamline. (b) An aircraft has a mass of 4.

0 × 104 kg and is in steady level flight. The area of each wing is 25 m2 and the speed of the air just below the wing is 280 m s−1. Calculate the speed of the air just above the wing. The density of air is 1.2 kg m−3. (5½ marks) Based on the equation, since there’s no change with respect to elevation, then h1 ≈ h2, so that the ρgh terms may cancel out to give: ΔP = ½ ρv12 - ½ ρv22 so that v1 = {[2(15,696 + (0.5)(1.2)(280)2)] / 1.2}1/2 for ΔP = F / A = (4.0 x 104) (9.81) / (25) = 15,696 N/m2 v1 = 323.36 m/s. (speed of air just above the wing)Question-3 A toboggan (A) of mass 35 kg slides down an icy slope which makes an angle of 20° with the horizontal (Figure 2).

The toboggan starts from rest, travels a distance of 15 m down the slope after which the slope levels out to horizontal and the toboggan immediately collides with a second toboggan (B) of mass 25 kg. The coefficient of sliding friction between a toboggan and the icy surface is 0.24 and air resistance is negligible. a) Show that the acceleration of toboggan A on the slope is 1.1 m s−2. (You will need to draw an appropriate diagram of forces.) (5 marks) Summation of forces along FN axis: FN - (35)(9.81)(sin 70°) = 0 Normal force FN = 322.

64 N then, Summation of forces along the x-axis: (-322.64 N)(0.24)(cos 20°) + (322.64)(cos 70°) = 35(a) Therefore, a = 1.074 m/s2 ≈ 1.1 m/s2b) Show that the speed of toboggan A just before the collision is 5.9 m s−1. (2 marks) x = v0t + ½ at2 and v0 = 0 since the toboggan starts from rest 15 = ½ (1.074) t2 --- t = 5.285 seconds and v = v0 + at = 0 + (1.1) (5.285) ≈ 5.9 m/s c) After the collision both toboggans are moving with velocities that are in the same direction that toboggan A was moving immediately before the collision, and the speed of toboggan A is 1.4 m s−1. Calculate the speed of toboggan B immediately after the collision. (3 marks) Before collision: mA = 35 kg, vA = 5.

9 m/s and mB = 25 kg, vB = 0 After collision: mA = 35 kg, vA’ = 1.4 m/s and mB = 25 kg, vB’ = ? By conservation of momentum: mAvA + mBvB = mAvA’ + mBvB’ (35)(5.9) + 0 = (35)(1.4) + (25)vB’ vB’ = 6.3 m/s Question-4 A trainee pilot is on a flour-bombing exercise. He is aiming to hit a target on the ground which is 15 m in diameter whilst he is flying at a speed of 45 m s−1 at a height of 98 m. Air resistance is negligible and the pilot maintains a steady speed in a straight line that passes directly over the target. a) Calculate how long it takes for the flour bomb to hit the ground from the time of release from the aircraft.

(3½ marks) Free fall: v0 = 0, so y = v0t + ½ gt2 --- 98 = 0 + ½ (9.81) t2 t = [ 2(98) / 9.81 ]1/2 = 4.47 secondsb) At what horizontal distance from the point directly above the target must the pilot release the flour bomb? (1½ marks) x = v*t = (45)*(4.47) = 201.15 m.c) Calculate the margin of error that the pilot has in timing the release of the flour bomb. (2 marks) margin of error = (15/2) / 201.15 * 100 % = 3.73 % d) Describe in words how these quantities would change if the pilot flew at the same speed but at half the height.

Give your reasoning. (3 marks)Obviously, the time it takes for the flour bomb to hit the ground would be lessened since the height is in direct proportion to the square of time. Similarly, the distance from the point above the target would be shorter as well.Question-5 The figure below shows an object that has been placed at a distance of 25 cm from a converging thin lens of focal length 15 cm. a) Copy this diagram into your answer book and carefully draw a ray diagram to show where the image is formed. (3 marks)b) Calculate the distance of the image from the lens. (2 marks) By the thin lens equation: 1 / d0 + 1 / di = 1 / f 1 / 25 + 1 / di = 1 / 15 --- 1 / di = (5 - 3) / 75 di = 75 / 2 = 37.5 m. c) By what factor is the image magnified compared with the object?

Show your working. (2 marks) By the magnification formula, M = | di / d0 | M = | 37.5 / 25 | --- M = 1.5d) Explain, in two or three sentences, how the human eye is able to produce a focused image on the retina, of objects at a range of distances. (3 marks) Because light rays diverge in all directions from their source, the set of rays from each point in space that reach the pupil must be focused. The formation of focused images on the photoreceptors of the retina depends on the refraction (bending) of light by the cornea and the lens.

The cornea is responsible for most of the necessary refraction in producing a clear focused image on the retina.(PART B)Question-1 1 A sample of an ideal gas is at atmospheric pressure and a temperature of 300 K. The sample occupies a volume of 2.5 × 10−2 m3. The sample is heated at constant pressure until the temperature is 350 K. What is the final volume of the gas sample? Choose the option that is closest to your answer.A. 2.1 × 10−2 m3 B. 2.5 × 10−2 m3 C. 2.9 × 10−2 m3 D. 3.4 × 10−2 m3 E. 5.0 × 10−2 m3Solution: By Charles’s law, V1 / T1 = V2 / T2 V2 = T2 * (V1 / T1) = (350 K) * (2.

5 x 10-2 m3) / (300 K) V2 = 2.9 x 10-2 m3Question-2 A large flat disc of moment of inertia I is spinning freely, with angular speed 18 rad s−1, about a frictionless axle which is perpendicular to, and at the centre of, the disc. A small object lands on, and adheres to, the disc, increasing the moment of inertia of the system to 1.5I (see Figure 3). What is the new angular speed of the system? Choose the option that is closest to your answer.A. 8 rad s−1 B. 12 rad s−1 C. 15 rad s−1 D.

18 rad s−1 E. 27 rad s−1Solution: L = I*ω --- I1ω1 = I2ω2 I (18) = (1.5 I) * ω2 --- ω2 = 12 rad s-1 Question-3 An object starts from rest and travels in a straight line with a varying velocity vx as shown in the graph in Figure 4. Estimate the total distance traveled by the object over the time period shown. Choose the option that is closest to your answer.A. 6.0 m B. 25 m C. 45 m D. 105 m E. 150 mQuestion-4 A car starts from rest and accelerates in a straight line at a constant rate of 5 m s−2.

Calculate the distance the car has moved after 4 s. Choose the option that is closest to your answer. A. 10 m B. 20 m C. 40 m D. 60 m E. 80 mSolution: x = v0t + ½ at2, since car starts from rest, then v0 = 0 So, x = 0 + ½ (5)(4)2 = 40 m.Question-5 A gardener pulls a heavy sack of soil along a horizontal path for a distance of 15 m. He does this by applying a force F of 80 N at an angle of 70° to the horizontal as shown in Figure 1. Calculate the amount of work done by the gardener during this process.

Choose the option that is closest to your answer.A. 113 J B. 274 J C. 410 J D. 1200 J E. 11760 JSolution: Work = F·d cos θ = (80 N) (15 m) (cos 70°) Work = 410.42 JQuestion-6 A wheel (rather like a bicycle wheel) is rotating at an angular speed of ω = 22 rad s−1 about an axle through its centre as shown in Figure 1. The mass of the wheel is 1.8 kg and can be considered to be concentrated in the rim of the wheel, which is of radius 0.40 m. What magnitude of torque must be applied to this wheel to bring it to rest in 6.0 s? Choose the option that is closest to your answer.A. 2.6 N m B. 3.7 N m C. 6.6 N m D. 12 N m E. 1.1 N mSolution: Torque = I*(ω / t) = m · r2 (ω / t) = (1.8) (0.4)2 * (22 / 6.0) Torque = 1.

056 Nm ≈ 1.1 NmQuestion-7 A car, on a straight track, is initially traveling at a speed of 30 m s−1. The brakes are applied and the car slows down with a constant acceleration of magnitude 2.5 m s−2 and comes to rest. How far does the car travel from the instant of application of the brakes until coming to rest? Choose the option that is closest to your answer.A. 30 m B. 60 m C. 90 m D. 180 m E. 240 m F. 360 mSolution: v = v0 + at --- 0 = 30 + (-2.5) t --- t = 12.

0 seconds v2 = v02 + 2ad --- 0 = (30)2 - 2(2.5)*d d = (302) / (2)(2.5) --- d = 180 m.Question-8 Two objects are traveling directly towards each other. One has a mass of 2.0 kg and a speed of 6.0 m s−1, the other a mass of 3.0 kg and a speed of 4.0 m s−1. They undergo an inelastic collision and coalesce into a single object. Calculate how much kinetic energy is lost during the collision. Choose the option that is closest to your answer.A. 0 J B. 10 J C. 20 J D. 24 J E. 60 JSolution: KE = ½ mv2 and m = m1 + m2 = 2.0 + 3.0 = 5.

0 kg Due to inelastic collision, ½ (m1 + m2)v2 = ½ m1v12 + ½ m2v22 KE = ½ (2.0)(6.0)2 + ½ (3.0)(4.0)2 KE = 36 + 24 = 60 JQuestion-9 The figure below shows a tank of water, with two holes placed one directly below the other a distance d apart. Both holes are below the water surface. At a particular time, water is escaping from the upper hole at a speed of 1.0 m s−1 and from the lower hole at a speed of 2.0 m s−1. Calculate the distance d between the holes.

Choose the option that is closest to your answer. A. 5.1 × 10−2 m B. 0.10 m C. 0.15 m D. 0.41 m E. 0.46 mSolution: v2 = v02 + 2gy --- y = (v2 - v02) / 2g y = (2.02 - 1.02) / 2(9.81) --- y = 0.153 mQuestion-10 A car is moving along a straight horizontal road. Figure 1 shows a graph of the velocity of the car against time. The motion of the car can be divided into three parts, 1, 2 and 3 as shown on the figure. Two of the statements below are correct statements about the car’s motion.

Choose the correct statements.A. The car accelerates in part 1, decelerates (i.e. has negative acceleration) in part 2 and has zero acceleration in part 3. B. The car accelerates in all parts. C. The car moves with constant velocity in parts 1 and 2 and is stationary in part 3. D. The car accelerates in parts 1 and 2 and is stationary in part 3. E. The distance travelled by the car during part 1 is greater than the distance it travels in part 2.