advertisement

AIR UNIVERSITY Mid-Term Exam with Solution (Physics-103) Sections: BEMTS(A,B,C,D) AND BEMEC-A,B,C. MAX.MARKS: 20 Instructors: Dr.Pervaiz Akhtar, Ch Ibrahim, Mr.Mozaffar. Q # 1 a)consider three vectors a ,b c with magnitudes 3m 4m and 10m respectively are shown in figure .The angle between a nd b is 300 , where as between b and c is 900 (i) find x- and y-components of all the three vectors individually. Solution: (ii) find the resultant of all the x-components and y-components. Solution: Resultant of all x components = 3+3.46-5 = 1.46 Resultant of all y components = 0+2+8.66= 10.66 (iii) if c = La + Mb , then find the value of L and M. Solution: In terms of components (first x and then y), we must have -5.00 m = L(3.00 m)+M(3.46 m) 8.66 m = L(0)+M (2.00 m). Solving these equations, we find L = –6.67. and M = 4.33 (b) If a = 3.0i – 4.0j and b = -2,0i + 3.0k (i) find the magnitude of cross product of two vectors. Solution: a×b = (3i -4j)×(-2i + 3k) = -12i -9j-8k (ii) also find the angle between the two vectors. Solution: a.b = (3i -4j).(-2i + 3k) = -6 Also a.b = ab cosθ a 32 4 2 = 5, b 4 9 = 13 6 = 109o Therefore (5)( 13 ) cosθ = -6, Therefore θ = cos 1 5 13 Q #2 (a) A particle moves so that its position (in meters) as a function of time (in seconds) is r =1.0i + 4t2j + t k. Write the expression for its velocity and acceleration as a function of time. Solution: (b) A plane, diving with constant speed at an angle of 530 with the vertical, releases a projectile at an altitude of 730m. The projectile hits the ground 5.00s after release (i) What is the speed of the plane? Solution: Which give vo = 202 m/s (ii) How for does the projectile travel horizontally during its flight? Solution: Q # 3 (a) A 2.0 Kg object is subjected to three forces that give it an acceleration a = -(8.00m/s2)i+ (6.00m/s2 )j. If two of the three forces are F1 = (30.0N) i+ (16.0N)j and F2 = -(12.0N)i+(8.00N)j, find the third force. Solution: m = 2 kg Therefore F3 7iˆ 12 ˆj N (b) Consider a block of mass m = 3.0kgslides along a floor while a force F of magnitude 12.0N is applied to it .at an upward angle θ = 300.The co-efficient of kinetic friction between the block and the floor is μk 0.40. Solution: m = 3 kg F = (12 N , 30o) Resolving F in x and y components Fx= 12 cos 30o = 10.4N, Fy = 12 sin30o = 6 N Net force along y-axis is Net force along x-axis is (i) Calculate the magnitude of the normal force acting on the block. FN = mg- Fsinθ = 29.4-6= 23.4N (ii) Calculate the magnitude of the force of kinetic friction. = (0.4)(23.4) = 9.36N (iii) find the value of acceleration a produced in the block. ma = 12cos30 – 9.36 Therefore a = 0.35 m/s2 Q#4(a) the angular position of a point on a rotating wheel is given by θ = 2.0+ 4.0t2 + 2.0t3 , where θ is in radians and t is in seconds. At t=0 what are i) the points angular position ii) the points angular velocity iii) What is its angular velocity at t=4.0s. iv) What is its angular acceleration at t= 2.0s. Solution: Angular position is given by (i) (ii) (iii) At t = 4 s, the angular velocity is (iv) Angular acceleration is given by: (b) (i) Calculate the rotational inertia of a meter stick about an axis perpendicular to the stick and located at 20cm mark. The mass of the meter stick is 0.56Kg. ( I =1/12ML2 (ii) If the angular velocity of the meter stick is 20rev/mint, then find the rotational kinetic energy of the meter stick. Solution: (i) Using parallel axis theorem, rotational inertia at a given point is Therefore (ii) The rotational K.E is given by K.E = ½ Iω2 Where ω is ω = 20 rev/min = (20)(2π) rad/min so ω = 40π/60 = (2/3)π rad /s. Also I = 9.7×10-2kg.m2 as calculated in part (i) so K.E = (½)(9.7×10-2kg.m2) ((2/3)π rad /s)2. Q#5 (a) Which of the following relationships between the force F on a particle and the particles position x implies simple harmonic motion. i) F = -5x ii) F = -400x2 iii) F = 10x 2 iv) F = 3x Solution: (a) A particle execute S.H.M if F = -kx, where F is the net force acting on a particle at particular instant and x gives the position of particle at that instant. This condition is satisfied for case (i) only. (b) Consider a meter stick that swings about a pivot point O at a distance h from the sticks centre of mass. Treat mater stick as thin meter rod and apply the parallel axis theorem to find o h com (i) its rotational inertia about an axis passing through the pivot. Solution: Using parallel axis theorem, rotational inertia at a given point is Where Icom = (1/12)ML2 and h = L/2 so I = (1/12)ML2 + M (L/2)2 I = (1/3)M L2. As L = 1 m so I = (1/3)M (ii) The time period of oscillation T? Solution: The time period of physical pendulum is given by I T 2 Putting the value of I and h, Mgh T = 1.64 s. (c) In an electric shaver, the blade moves back and forth over a distance of 2.0mm in simple harmonic motion. Find i) the amplitude ii) the maximum blade speed iii) the magnitude of the maximum blades acceleration. Solution: (i) As blade covers 2 mm in moving from one extreme position to other extreme position. So amplitude is half of this value. Amplitude = 1 mm (ii) vmax = xm ω and ω = 2πf = 2π(100) vmax = (1mm)2π(100) = 0.2 π m/s. (iii) the maximum value of the acceleration is given by amax = ω2 xm = (2π(100))2(1mm) = 20π m/s2