Physics Question-Indian Institute of Science .

Page 1 of 1 Question 1: A car is moving horizontally at a constant speed of ux = 10 ms−1 towards a cliff edge. The vertical distance from the top of the cliff to the ocean below is 20 m. The car drives horizontally off the cliff and lands in the ocean below. Consider the uniform acceleration equation that describes displacement as a function of time t, 1 S(t) = ut + at2. 2 a) Explain, using words, why the above equation becomes the equation stated below when considering only the horizontal displacement of the car, ⇒ Sx(t) = uxt. [The answer to this part of the question must be typed in own words.] [3 marks] b) Calculate the maximum horizontal distance (Sx, max) that the car reaches when it lands in the ocean (also known as the range). [3 marks] Page 2 of 15 Question 2: Consider an object that has the following acceleration as a function of time t, a(t) = 2t + 3. a) Explain, using words, why we can not use the uniform acceleration equations to calculate the velocity of this object. [The answer to this part of the question must be typed in own words.] [3 marks] Given that at the time t = 1, the object has a velocity of v(t = 1) = 6 ms−1. Find an expression for the velocity of this object as a function of time in the following form, v(t) = pt2 + qt + k, where p, q and k are constants. b) State the numerical values of the constants p, q and k. [5 marks] Page 3 of 15 Question 3: A mass m is oscillating on a spring, there is no resistance to the motion of the mass. This type of oscillation is described by simple harmonic motion (SHM) and the differential equation that describes the motion of this particular mass is stated below, ẍ(t) = −25x(t). a) Explain, using words, the meaning of each symbol in the differential equation. In your answer make sure to state the meaning of the two dots appearing on the left hand side and the significance of the negative sign appearing on the right hand side. [The answer to this part of the question must be typed in own words.] [4 marks] b) Calculate the time period for the oscillation of the mass. [3 marks] Page 4 of 15 Question 4: A mass of m = 1.5 Kg is attached to the end of a light inextensible string. The string then begins to rotate such that the mass travels in a circular path of radius r and the string makes an angle θ = 300 with the vertical. This type of arrangement is called a conical pendulum. The force T represents the tension in the string. See diagram below. a) Resolve the forces vertically to calculate the magnitude of the tension T in the string. [3 marks] b) Explain, using words, how you would find the magnitude of the circular force keeping the mass moving on the circular path. In your answer make reference to what is providing the circular force and the direction of the circular force. [The answer to this part of the question must be typed in own words.] [3 marks] c) Therefore, calculate the circular acceleration that the mass experiences whilst moving in the circular path. State which one of Newton’s laws you have used in the calculation of the acceleration. [3 marks] Page 5 of 15 Question 5: A ball is initially at rest and at a height of 1.8 m above a flat horizontal surface. The ball is then released and fulls under the influence of gravity vertically downwards towards the surface. The coefficient of restitution between the ball and surface is e = 0.3. a) Explain, using words, how the coefficient of restitution between the ball and the surface can be found by doing an experiment. (Hint: think of the parameters in the mathematical definition of the coefficient of restitution and how these parameters can be found). [The answer to this part of the question must be typed in own words.] [3 marks] b) By using the uniform acceleration equations, calculate the maximum vertical height that the ball reaches after one bounce off the surface. Assume that the surface remains stationary throughout. [4 marks] Page 6 of 15 Question 6: Consider a horizontal bar that is on a pivot. At one end of the bar is a mass M at a distance of d1 = 0.3 m from the pivot point. At the other end of the bar is a mass m at a distance of d2 from the pivot point. The pivot is in the position such that the bar remains horizontal. See diagram. 𝒅𝟏 𝒅𝟐 Consider the scenario that the mass M is larger than the mass m, such that M > m. a) Explain, using words, how by considering the moment at the pivot point caused by each mass we can deduce that the distance d2 must be larger than the distance d1 such that d2 > d1 for the bar to remain horizontal. [The answer to this part of the question must be typed in own words.] [3 marks] b) If the mass M = 4m, find the corresponding distance d2. [4 marks] Page 7 of 15 Question 7: Consider two lamina rectangles A and B which have the corresponding masses mA = 4 Kg and mB = 2 Kg respectively. Assume each rectangle to have a uniform mass density. Rectangle A has dimensions 2 m by 10 m and rectangle B has dimensions 4 m by 2 m. The two rectangles are attached to each other and the point O represents the end of the horizontal line that passes through the centre of gravity of the entire shape. See diagram below. 𝟏𝒎 𝑩 𝑨 𝟐𝒎 𝟒𝒎 𝒐 𝟏𝟎𝒎 𝟐𝒎 a) Explain, using words, how by considering lines of symmetry of each rectangle we are able to find the center of gravity for each rectangle individually. In your answer make reference to the uniform distribution of mass in each rectangle. [The answer to this part of the question must be typed in own words.] [3 marks] b) Calculate the horizontal distance of the centre of mass of the combined shape from the point O. [3 marks] Page 8 of 15 Question 8: A ball is launched from the ground at an angle of θ = 60o and an initial velocity of u = 50 ms−1. The ball then lands a horizontal distance away from where it started at the same initial vertical height. See the diagram below. 𝜗 Consider the uniform acceleration equation that describes the vertical displacement of the ball as a function of time t, 1 gt2. 2 Where g is the acceleration due to gravity taken to be g = 9.8 ms−2. Sy (t) = u yt − a) Use this equation to calculate the time of flight (T ) of the ball. [3 marks] b) Explain, using words, why there are two time solutions to the equation above. What does the other time solution represent physically? [The answer to this part of the question must be typed in own words.] [3 marks] Page 9 of 15 Question 9: A mass of m = 5 Kg is moving on a straight line. Its displacement from the origin as a function of time t is given by, S(t) = 9t − t3. a) Calculate the kinetic energy of the mass at t = 1 s. [4 marks] b) Explain, using words, how to find the times that correspond to the mass passing through the origin. What would be the physical meaning of a negative time solution? [The answer to this part of the question must be typed in own words.] [4 marks] Page 10 of 15 Question 10: A mass is oscillating with simple harmonic motion (SHM). The time period of the oscillation is T = 1.2 s and the amplitude of the mass is a = 0.36 m. The velocity of the mass as a function of displacement from the equilibrium position is given by, √ v(x) = ω a2 − x2. a) Calculate the velocity of the mass when it is passing through the equilibrium position. [3 marks] b) Explain, using words, why the velocity of the mass is at its minimum when the mass is at maximum displacement from the equilibrium position. State the value of the minimum velocity. (Hint: consider the equation stated above). [The answer to this part of the question must be typed in own words.] [4 marks] Page 11 of 15 Question 11: A cube of mass m = 1.5 Kg is stationary on a horizontal rough disk at a radius of r = 0.15 m from the centre of the disk. The disk has a coefficient of friction of µ = 0.8. See diagram. 𝝎 𝝁 The disk begins to rotate at a constant angular speed of ω = 4 s−1. a) Calculate the magnitude of the circular force that the cube experiences whilst the disk is rotating. [3 marks] b) Therefore, what is the magnitude of the frictional force between the mass and the disk? [1 mark] The angular speed of
the disk is increased. The maximum angular speed for which the mass remains on the rotating disk is found to be ωmax = 7.23 s−1. If the angular speed of the disk is increased above ωmax then the mass slides off the rotating disk. c) Explain, using words, why this maximum angular speed of the disk exists. (Hint: think about the force that is providing the circular force). [The answer to this part of the question must be typed in own words.] [5 marks] Page 12 of 15 Question 12: An object A has a mass mA = 2 Kg and initial velocity vA = 3 ms−1. Another object B has a mass mB = 3 Kg and initial velocity vB = 1 ms−1. The objects are initially a horizontal distance d apart from each other. Both objects are moving horizontally on a smooth surface in a straight line and in the same direction. See diagram. 𝒗𝑨 𝒗𝑩 𝒅 Both of the objects A and B collide and merge into one single object C (the objects coalesce) which continues to move in the same direction at the final velocity vC. a) Explain, using words, how we can use the principle of the conservation of momentum to find the final velocity v C. State, using words, a condition that must be assumed when applying this conservation law. [The answer to this part of the question must be typed in own words.] [4 marks] b) Use momentum conservation to calculate the final velocity v C. [3 marks] Page 13 of 15 Question 13: A force of magnitude | P̄| = 10 N is acting on a rigid horizontal rod at an angle θ = 30o. The opposite end of the rod is fixed and is labelled A, this is the point which the rod can rotate around. The distance between the point A and the point at which the force p̄ acts is d = 3 m. See diagram. 𝑨 𝒅 𝜗 𝑝 a) Calculate the moment at the point A. In your answer make sure to indicate the sense of rotation. [3 marks] Now consider another force F̄ acting on the rod. This additional force is passing through the point A on the rod. b) Explain, using words, what the effect this additional force will have when considering the moment at the point A. In your answer make use of the phrase line of action. [The answer to this part of the question must be typed in own words.] [4 marks] Page 14 of 15 Question 14: Three masses m1 = 4 Kg, m2 = 3 Kg and m3 = 2 Kg are situated on a coordinate grid. The coordinates for each of the masses respectively are (0, 8), (2, 0) and (6, 2). See diagram below. 𝒎𝟏 𝒎𝟑 𝒎𝟐 The center of gravity of these three masses is denoted by the point G which has the coordinates (x̄, ȳ). a) Find the numerical values of the coordinates of the point G. [6 marks] END OF EXAM. Page 15 of 15