FOUNDATION STUDIES EXAMINATIONS April PHYSICS First Paper February Program 2007
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1 FOUNDATION STUDIES EXAMINATIONS April 2007 HYSICS First aper February rogram 2007 Time allowed hour for writing 0 minutes for reading This paper consists of 3 questions printed on 5 pages. LEASE CHECK BEFORE COMMENCING. Candidates should submit answers to ALL QUESTIONS. Marks on this paper total 50 marks, and count as 0% of the subject. Start each question at the top of a new page.
2 2 INFORMATION a b = ab cos a b = ab sin ĉ = v dr dt a dv dt v = R a dt r = R v dt v = u + at x = ut + 2 at2 v 2 = u 2 +2ax a = gj v = u gtj r = ut 2 gt2 j i j k a x a y a z b x b y b z s = r v = r! a =! 2 r = v2 r p = mv N : if F = 0 then p = 0 N2 : F = ma N3 : F AB = F BA W = mg F r = µr g = acceleration due to gravity = 0 m s 2 r F Fx = 0 Fy = 0 = 0 0 = N m 2 C 2 E lim q!0 F q V W q E = dv dx = H E da = q 0 C q V C = A d E = k q r 2 ˆr E = q 2 = qv = CV 2 2 C 2 2 C = C + C 2 + C 3 C = C + C 2 + C 3 R = R + R 2 + R 3 R = R + R 2 + R 3 V = IR V = E IR V = k q r = VI = V 2 = R I2 R K : In =0 K2 : (IR 0 s)= (EMF 0 s) F = q v B df = i dl B W R r 2 r F dr W = F s F = i l B = ni A B KE = 2 mv2 E = mgh v = E B r = m q E BB 0 r = mv qb dw dt = F v T = 2 m KE Bq max = R2 B 2 q 2 2m F = kx E = 2 kx2 dv v e = dm m v f v i = v e ln( m i m f ) F = v e dm dt F = k q q 2 r 2 k = Nm 2 C 2 db = µ 0 i dl ˆr 4 r 2 H B ds = µ0 I µ0 =4 0 7 NA 2 = R area B da = B A = N d dt = NAB! sin(!t)
3 3 f = k 2 T! 2 f v = f y = f(x vt) y = A sin k(x vt) =A sin(kx!t) = A sin 2 ( x t ) T = 2 µv!2 A 2 v = s = s m sin(kx!t) p = p m cos(kx!t) I = 2 v!2 s 2 m q F µ n(db 0 s) 0 log I I 2 = 0 log I I 0 where I 0 = 0 2 Wm 2 v±v f r = f r s v v s where v speed of sound = 340 m s y = [2A sin(kx)] cos(!t) h i y (x=0) = 2A cos (!! 2 ) t 2 h y = 2A cos (k + ) 2 f beat = f f 2 = n =(n + 2 ) d sin = n sin (! +! 2 ) 2 t i sin(kx!t + k + 2 ) h h x p x E t dn dt = N N = N 0 e t R dn dt T 2 MATH: If ax 2 + bx + c =0 = ln 2 = then x = b±p b 2 4ac 2a R y dy/dx ydx x n (n ) nx n+ xn+ e kx ke kx k ekx sin(kx) k cos(kx) cos kx k cos(kx) k sin(kx) sin kx k where k = constant Sphere: A =4 r 2 CONSTANTS: V = 4 3 r3 u = kg = MeV ev = J c = ms E = hf c = f h = Js KE max = ev 0 = hf E 2 = p 2 c 2 +(m 0 c 2 ) 2 E = m 0 c 2 E = pc = h (p = m p 0v (nonrelativistic)) e electron charge = C particle mass(u) mass(kg) e p n
4 HYSICS: First aper. February rogram Question ( (9 + 8) = 7 marks): A rectangular box aligned along the x-, y-, and z-axes, with sides of 4 m, 4 m, and 3 m, as labeled, is illustrated in Figure. Three forces act at corner,, of the box. 2 N acts up in the direction of the y-axis; 4 N acts along the diagonal, Q, of the top face; 6 N acts along the diagonal, R, of the right face. (i) Express each of the three forces in terms of the ijk unit vectors. Figure : (ii) Hence find the vector sum of the three forces. Again express your answer in terms of ijk unit vectors. Question 2 ( 6 marks): The power,, generated by a wind turbine depends on the total area, A, swept out by its blades, the velocity, v, of the air that passes through the blades, and the density (mass per unit volume),, of that air. Use dimensions to find an expression for in terms of, A and v. Hint: dimensions of are - [ ]=ML 2 T 3
5 HYSICS: First aper. February rogram Question 3 ( = 7 marks): A rectangular box aligned along the x-, y-, and z-axes, with sides of 2 m, 3 m, and 4 m, as labeled, is illustrated in Figure 2. A triangle, QR, is drawn between corners of the box as! illustrated. Displacement Q a and! displacement R b. (i) Express vectors a and b in terms of the ijk unit vectors. (ii) Use the dot product (a b) to calculate the angle, between a and b. Figure 2: (iii) Find an expression for vector area, A, of triangle QR. Give your answer in terms of ijk unit vectors. You are given that the vector area, A, of QR is given by the cross product A = (a b) 2 END OF EXAM ANSWERS: Q.(i)2 = (2j) N, 4 = ( 6i +2k) N, 6 = ( 24j +8k) N. 5 5 (ii) F = ( 6i 4j +2k) N. 5 Q2. = k Av 3. Q3. (i) a = (+3i 2k) m, b = ( 4i + 3j 2k) m. (ii) = 47.9 deg. (iii) A = (+4j +6k) m 2.
6 FOUNDATION STUDIES EXAMINATIONS June 2007 HYSICS Second aper February rogram Time allowed hour for writing 0 minutes for reading This paper consists of 3 questions printed on 6 pages. LEASE CHECK BEFORE COMMENCING. Candidates should submit answers to ALL QUESTIONS. Marks on this paper total 50 marks, and count as 0% of the subject. Start each question at the top of a new page.
7 2 INFORMATION a b = ab cos a b = ab sin ĉ = v dr dt a dv dt v = R a dt r = R v dt v = u + at x = ut + 2 at2 v 2 = u 2 +2ax a = gj v = u gtj r = ut 2 gt2 j i j k a x a y a z b x b y b z s = r v = r! a =! 2 r = v2 r p = mv N : if F = 0 then p = 0 N2 : F = ma N3 : F AB = F BA W = mg F r = µr g = acceleration due to gravity = 0 m s 2 r F Fx = 0 Fy = 0 = 0 0 = N m 2 C 2 E lim q!0 F q V W q E = dv dx = H E da = q 0 C q V C = A d E = k q r 2 ˆr E = q 2 = qv = CV 2 2 C 2 2 C = C + C 2 + C 3 C = C + C 2 + C 3 R = R + R 2 + R 3 R = R + R 2 + R 3 V = IR V = E IR V = k q r = VI = V 2 = R I2 R K : In =0 K2 : (IR 0 s)= (EMF 0 s) F = q v B df = i dl B W R r 2 r F dr W = F s F = i l B = ni A B KE = 2 mv2 E = mgh v = E B r = m q E BB 0 r = mv qb dw dt = F v T = 2 m KE Bq max = R2 B 2 q 2 2m F = kx E = 2 kx2 dv v e = dm m v f v i = v e ln( m i m f ) F = v e dm dt F = k q q 2 r 2 k = Nm 2 C 2 db = µ 0 i dl ˆr 4 r 2 H B ds = µ0 I µ0 =4 0 7 NA 2 = R area B da = B A = N d dt = NAB! sin(!t)
8 3 f = k 2 T! 2 f v = f y = f(x vt) y = a sin k(x vt) =a sin(kx!t) = a sin 2 ( x t ) T = 2 µv!2 a 2 v = s = s m sin(kx!t) p = p m cos(kx!t) I = 2 v!2 s 2 m q F µ n(db 0 s) 0 log I I 2 = 0 log I I 0 where I 0 = 0 2 Wm 2 v±v f r = f r s v v s where v speed of sound = 340 m s y = y + y 2 KE max = ev 0 = hf E 2 = p 2 c 2 +(m 0 c 2 ) 2 E = m 0 c 2 E = pc = h p (p = m 0v (nonrelativistic)) h h x p x E t dn dt = N N = N 0 e t R dn dt T 2 MATH: = ln 2 = ax 2 + bx + c =0! x = b±p b 2 4ac 2a R y dy/dx ydx x n (n ) nx n+ xn+ e kx ke kx k ekx sin(kx) k cos(kx) cos kx k cos(kx) k sin(kx) sin kx k where k = constant y = [2a sin(kx)] cos(!t) Sphere: A =4 r 2 V = 4 3 r3 (m =0,, 2, 3, 4,...) N : x = m( 2 ) AN : x =(m + 2 )( 2 ) y = [2a cos(!! 2 2 )t] sin(! +! 2 2 )t f B = f f 2 y = [2a cos( k 2 )] sin(kx!t + k 2 ) = d sin Max : = m Min : =(m + 2 ) CONSTANTS: u = kg = MeV ev = J c = ms h = Js e electron charge = C particle mass(u) mass(kg) e p n I = I 0 cos 2 ( k 2 ) E = hf c = f
9 HYSICS: Second aper. February rogram a 2M 4m 5m 3m M µ Figure : Question ( = 7 marks): Figure shows blocks, of masses 2M and M, connected by a string of negligible mass, via two pulleys of negligible mass and friction. The system is released from rest. The coe cient of friction between the block of mass M, and the inclined surface, on which it slides, is µ. Dimensions of the incline are illustrated. (i) Draw a labeled diagram to show all the forces that act on each block. (ii) User Newton s laws of motion to find an expression, in terms of µ, and the acceleration of gravity, g, for the acceleration, a, of the block of mass 2M. (iii) Find a similar expression for the tension, T, in the string.
10 HYSICS: Second aper. February rogram A M horizontal 2M AB =0 m AC = 6 m CB = 8 m C B 3M L Figure 2: Question 2 ( = 6 marks): A beam, AB, of mass M, and length 0 m, is hinged at B, as illustrated in Figure 2. The other end, A, is attached, by means of a light, horizontal rope, over a frictionless pulley, to a block of mass, 3M. In order to lift the block o the ground, a worker, of mass 2M, walks up the beam from B. The dimensions of the system are labeled. (i) Draw a diagram showing all the forces that are acting on the beam AB, just before the block lifts. (ii) Using the conditions for equilibrium, find the distance, L, along the beam AB, that the worker, will need to walk, to just lift the block. (iii) Given that M =30kg, find the reaction of the hinge on the beam, as the block is just lifted. Take acceleration of gravity g = 0 ms 2.
11 HYSICS: Second aper. February rogram rest M µ 2M rest unstretched k L Figure 3: Question 3 ( 7 marks): Two blocks, of masses M and 2M, are connected together, and to an unstretched spring, of spring constant, k, by means of two lengths of massless string, that pass over two massless and frictionless pulleys, as depicted in Figure 3. The block, of mass 2M, is released from rest, and falls a vertical distance of L, before coming momentarily to rest again. The other block, which has a coe cient of friction, µ, with the horizontal surface, is simultaneously pulled along the horizontal surface, as the spring is stretched. Using energy principles, derive an expression for the distance, L, that the 2M block falls, in terms of M, µ, k, and the acceleration of gravity, g. ANSWERS: END OF EXAM Q.(ii) a =( 2 µ 2Mg )g ; (iii) T = (3 + µ) 5 5. Q2. (ii) L =8.75 m ; (iii) R x = 900 N, R y = 900 N. Q3. L = 2Mg (2 µ). k
12 HYSICS: Third aper. February rogram kg Before y-axis 3 kg x-axis 2 kg y-axis v 4 m/s kg 3 kg x-axis 2 kg After 3 m/s Figure : Question ( 7 marks): Three stationary bodies, of masses, 2, and 3 kg, are in contact, on a horizontal frictionless surface, near the origin of x- and y-axes, as shown in Figure (Before). An explosion occurs between the bodies, which blows them apart, on the surface. The kg body moves away in the negative direction of the x-axis, with a velocity of 4 m/s, while the 2 kg body moves away in the negative direction of the y-axis, with a velocity of 3 m/s, as shown in Figure (After). Use momentum principles, to find the magnitude and direction of the velocity, v, of the third (3 kg) body, after the explosion.
13 FOUNDATION STUDIES EXAMINATIONS November 2007 HYSICS Final aper February rogram Time allowed 3 hours for writing 0 minutes for reading This paper consists of 6 questions printed on 3 pages. LEASE CHECK BEFORE COMMENCING. Candidates should submit answers to ALL QUESTIONS. Marks on this paper total 20 marks, and count as 45% of the subject. Start each question at the top of a new page.
14 2 INFORMATION a b = ab cos a b = ab sin ĉ = v dr dt i j k a x a y a z b x b y b z a dv v = R a dt r = R v dt dt v = u + at a = gj x = ut + 2 at2 v = u gtj v 2 = u 2 +2ax r = ut 2 gt2 j s = r v = r! a =! 2 r = v2 r p mv N : if F = 0 then p = 0 N2 : F = ma N3 : F AB = F BA W = mg F r = µr g = acceleration due to gravity = 0 m s 2 = H E da = q 0 C q V C = A d E = q 2 = qv = CV 2 2 C 2 2 C = C + C 2 C = C + C 2 R = R + R 2 R = R + R 2 V = IR V = E IR = VI = V 2 = R I2 R K : In =0 K2 : (IR 0 s)= (EMF 0 s) F = q v B F = i l B df = i dl B = ni A B r F v = E B r = m q E BB 0 r = mv qb Fx = 0 Fy = 0 = 0 T = 2 m KE Bq max = R2 B 2 q 2 2m W R r 2 r F dr W = F s KE = 2 mv2 E = mgh db = µ 0 i dl ˆr 4 r 2 H B ds = µ0 I µ0 =4 0 7 NA 2 dw dt = F v = R area B da = B A F = kx E = 2 kx2 dv v e = dm m v f v i = v e ln( m i m f ) F = v e dm dt F = k q q 2 r 2 k = Nm 2 C 2 0 = N m 2 C 2 E lim q!0 F q E = k q r 2 ˆr = N d dt = NAB! sin(!t) f = k 2 T! 2 f v = f y = f(x vt) y = a sin k(x vt) =a sin(kx!t) = a sin 2 ( x t ) T = 2 µv!2 a 2 v = s = s m sin(kx!t) q F µ V W q E = dv dx V = k q r p = p m cos(kx!t)
15 3 I = 2 v!2 s 2 m = ke2 2a 0 ( n 2 f )=R n 2 H ( i n 2 f ) n 2 i n(db 0 s) 0 log I I 2 = 0 log I I 0 where I 0 = 0 2 Wm 2 v±v f r = f r s v v s where v speed of sound = 340 m s (a 0 = Bohr radius = nm) (R H = m ) (n =, 2, 3...) (k 4 " 0 ) E 2 = p 2 c 2 +(m 0 c 2 ) 2 y = y + y 2 E = m 0 c 2 E = pc y = [2a sin(kx)] cos(!t) N : x = m( 2 ) AN : x =(m + 2 )( 2 ) (m =0,, 2, 3, 4,...) y = [2a cos(!! 2 2 )t] sin(! +! 2 2 )t f B = f f 2 y = [2a cos( k 2 )] sin(kx!t + k 2 ) = d sin Max : = m Min : =(m + 2 ) I = I 0 cos 2 ( k 2 ) E = hf c = f KE max = ev 0 = hf L r p = r mv L = rmv = n( h 2 ) E = hf = E i E f r n = n 2 ( h mke 2 )=n 2 a 0 E n = ke2 2a 0 ( )= 3.6 n 2 n 2 ev = h p (p = m 0v (nonrelativistic)) h h x p x E t dn dt = N N = N 0 e t R dn dt T 2 MATH: = ln 2 = ax 2 + bx + c =0! x = b±p b 2 4ac 2a R y dy/dx ydx x n (n ) nx n+ xn+ e kx ke kx k ekx sin(kx) k cos(kx) cos kx k cos(kx) k sin(kx) sin kx k where k = constant Sphere: A =4 r 2 CONSTANTS: V = 4 3 r3 u = kg = MeV ev = J c = ms h = Js e electron charge = C particle mass(u) mass(kg) e p n
16 HYSICS: Final aper. February rogram y B OA = 4 m OB = 3 m OC = 2 m 2 O 3 4 A x z C Figure : Question ( (2 + 8) + (5 + 5) = 20 marks): art (a): Figure shows a surface, ABC, whose corners are on the x-, y-, and z-axes, as illustrated. (i) Express each of the two vectors, AB! and AC!, in terms of the ijk unit vectors. (ii) Hence express the vector area,! A of the surface, in terms of ijk unit vectors. You are given the following, cross-product expression -! A = (AB 2! AC! )
17 HYSICS: Final aper. February rogram s B u 2m O 6m A C Figure 2: art (b): Figure 2 shows the trajectory of a ball which is projected from a point, O, at ground level, which is 6 m from the base of a 2 m high vertical pole, AB. The ball passes through a loop at the top, B, of the pole, precisely 2 s. later, then continues until it strikes the ground at C. Assume friction e ects of the air to be negligible. (i) Determine the initial velocity, u, with which the ball was projected from point O. You may express your answer in ijk form. (ii) Find the total horizontal distance, OC, between the point of projection, O, and the point C, where the ball hits the ground.
18 HYSICS: Final aper. February rogram m m µ 3m M a Figure 3: Question 2 ( (2 + 8) + ( + 9) = 20 marks): art (a): Figure 3 shows Two blocks of masses, m and M, connected by a massless string, over a massless, frictionless pulley. The dimensions of the system are labeled (tan = 4). 3 There is a coe cient of friction, µ, between the block of mass, m, and the inclined surface on which it slides. The system is released from rest. (i) Draw a labeled diagram, showing all the forces that act on each block. (ii) Use Newton s laws of motion, to derive an expression for the acceleration, a, of the block of mass, M, vertically downward, in terms of M, m, µ, and the acceleration of gravity, g.
19 HYSICS: Final aper. February rogram u m S rest M r (a) bar 2r rest m (b) M v Figure 4: art (b): Figure 4(a) shows a catapult, of length 3r, consisting of a straight bar, which has a block, of mass, M, fixed at one end, and a basket, containing a ball, of mass, m, at the other. The bar is initially horizontal (Figure 4(a)), but can rotate about a fixed pivot,. The bar, and the basket have negligible mass, and there is negligible friction, at the pivot. Relevant dimensions of the figure are labeled. The catapult is released, and because M m, it rotates anticlockwise, until the bar becomes vertical. It then hits stopper S, and the ball is ejected from the basket. Figure 4(b) shows the bar in the vertical position, just before it hits stop, S. At this stage, the block, M, has a horizontal velocity, v, and the ball, m, has a horizontal velocity, u, as labeled. (i) Express the velocity of the block, v, in terms of the velocity of the ball, u, as illustrated in Figure 4(b). (ii) Using energy principles, derive an expression for the velocity, u, of the ball, in Figure 4(b), in terms of M, m, r, and the acceleration of gravity, g.
20 HYSICS: Final aper. February rogram S v M Figure 9: Question 5 ( ( ) + (4 + 6) = 20 marks): art (a): A string, of total length, l =2.0 m, is measured to have a total mass of m = 40 gram. One end of this string is connected to a wave source, S. The string then passes over a frictionless pulley, and is securred, at its other end, to a block, of mass M =0.8 kg, as depicted in Figure 9. The wave source, S, transmits a wave along the string. Take the acceleration of gravity, g = 0 ms 2. (i) Calculate the velocity, v, with which the wave passes along the string. (ii) What output power,, would the wave source, S, need, in order to send a wave, with frequency, f = 00 Hz, and amplitude, a =mm, along the string? (iii) Using your answers above, write down a numerical wave function for the wave transmitted along the string from S, in Figure 9. art (b): Nitrogen isotope 3 7 N has a half-life of 0.0 minutes. A given sample of this isotope is measured to have an activity of 0 MBq. (i) How many atoms of 3 7 N are in this sample? (ii) How long will it take for the activity of this sample to fall to.0 MBq?
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