FOUNDATION STUDIES EXAMINATIONS March PHYSICS First Paper. February Program

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1 FOUNDATION STUDIES EXAMINATIONS March 2008 HYSICS First aper February rogram Time allowed hour for writing 0 minutes for reading This paper consists of 2 questions printed on 5 pages. LEASE CHECK BEFORE COMMENCING. Candidates should submit answers to ALL QUESTIONS. Marks on this paper total 20 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 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)

3 3 I = 2 v!2 s 2 m = ke2 2a 0 hc ( 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

4 HYSICS: First aper. February rogram A y 4m B 3m 0N frame D 8N C x 9N Figure : Question ( ( ) = 0 marks): Figure shows a frame, ABCD, with sides of lengths 4 m and 3 m, aligned with the x- and y-axes. Forces of 0 N, 8N, and 9 N, act at D, in the directions illustrated. (i) Express each of the forces in terms of the ijk unit vectors. (ii) Hence find the vector sum, F, of these three forces, in terms of the ijk unit vectors. (iii) Find the magnitude, and direction, of F.

5 HYSICS: First aper. February rogram Question 2 ( (5 + 5) = 0 marks): A force, F = 3i +2j +5k N ewton, drags a block through a displacement of r =6i 4j + k metre. (i) Given, that the work, W, done by a force, F, when it drags a body through a displacement. r, is given by the dot product - W = F r find the work, W, done by the force, F, above. (ii) Use the dot product, to find the angle between F and r. END OF EXAM ANSWERS: Q.(i)! 8 =8i N,! 9 = 9j N,! 0=8i +6j N ; (ii) F = 6i 3j N ; (iii) 6.3 N 0.6 deg below +x-axis Q2. (i) 5 J ; (ii) 70.5 deg

6 FOUNDATION STUDIES EXAMINATIONS June 2008 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 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)

8 3 I = 2 v!2 s 2 m = ke2 2a 0 hc ( 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

9 HYSICS: Second aper. February rogram F 3m T H 20 kg E 2m 80 kg 2m Figure : Question ( (3 + 3) = 6 marks): Beam HE, of mass 20 kg and length 4 m, is hinged to a vertical post at H. This beam is held horizontally by a cable EF, attached to its end E. The other end of the cable is secured to the post at F, which is 3 m above H. A mass of 80 kg hangs from the centre point of the beam. Take the acceleration of gravity g = 0 ms 2. (i) Draw a diagram of the beam, showing all the forces that act upon it. (ii) Using the conditions for the equilibrium of the beam, find the tension, T, in the cable, and the vertical and horizontal components of the reaction of the hinge on the beam, at H. Draw a second diagram of the beam, HE, and on it label these forces.

10 HYSICS: Second aper. February rogram m µ =0 m µ 3m a Figure 2: Question 2 ( ( ) = 7 marks): Figure 2 shows a wheeled trolley, of mass 2m, connected to a block, of mass 3m, by means of a string that passes over a pulley. A second block, of mass m, rests on the top surface of the trolley, but is restrained by a second string, that is attached to a fixed post at. All strings are massless, and all wheels and pulleys are massless and frictionless. The system is released from rest. As block 3m falls, it pulls the trolley along the horizontal surface. Because of the string attached to, block m is held at rest, and so slides along the top surface of the trolley. There is negligible friction in the wheels of the trolley. The coe cient of friction between the bottom surface of block m and the top surface of the trolley is µ. (i) Draw three diagrams - one for the trolley, one for the block of mass, 3m, and one for the block of mass m. Label each of these diagrams with the particular forces that act on each body. Label also the acceleration of each body. (ii) Write equations of motion (Newton s second law) for the trolley, and each of the two blocks, in vertical and horizontal directions, as appropriate (five equations). (iii) Using the above equations, derive an expression for the acceleration, a, of the 3m block, in terms of µ, and the acceleration of gravity g.

11 HYSICS: Second aper. February rogram M rest v m C L C m rest (a) M v (b) Figure 3: Question 3 ( (4 + 3) = 7 marks): Two balls, of masses, m and M (M >m), are secured to the ends of a rod of negligible mass. The distance between the balls is L. Initially, this rod is in the vertical position, as illustrated in Figure 3 (a). When released from rest, the rod rotates clockwise about a pivot at its centre point, C. When the rod is next vertical, both balls are traveling with a speed, v, as shown in Figure 3 (b). There is negligible friction at the pivot. (i) Using energy principles, derive an expression for the speed, v, of the balls in Figure 3 (b), in terms of l, m, M, and the acceleration of gravity, g. (ii) Hence, write down an expression for the angular velocity,!, of the system of rod and balls, in Figure 3 (b). END OF EXAM

12 HYSICS: Second aper. February rogram ANSWERS: Q.(ii) T =8.33 kn, R x = kn, R y = 4.00 kn. Q2. (ii) 3mg T =3ma, r mg = 0, +µr t = 0, R r 2mg = 0, +T µr =2ma; (3 µ) (iii) a = g. 5 q q (M m) Q3. (i) v = 2Lg 8g (M m) (M+m) L (M+m)

13 FOUNDATION STUDIES EXAMINATIONS November 2008 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 hc ( 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 Q B 5 z 7 6 (dimensions in m ) x Figure : Question ( (4 + 6) + (2 + 8) = 20 marks): art (a): Figure shows a rectangular box, with a corner at the origin, 0, and its sides aligned along the x-, y-, and z-axes. Dimensions of the box are labeled in metre. A wire, Q, is stretched between corners and Q of the box. A constant force, F, moves a bead, B, along the wire, from to Q. This force is given by - F =2i +3j +4k N (i) Derive an expression for the displacement vector, Q, in terms of unit vectors ijk. (ii) Calculate the work, W, that is done on the bead by the force, F, between and Q.

17 HYSICS: Final aper. February rogram B A y C Before (a) A =2.0kg B =3.0kg C =4.0kg x y B 5 m/s 6 m/s C x After (b) Figure 2: art (b): Figure 2(a) shows three stationary balls, A, B, and C, with masses as labeled, on a flat horizontal, frictionless surface, near the origin of the x- and y-axes. An explosion occurs between the balls, blowing them apart. Figure 2(b) shows balls B and C after the explosion; ball A is not shown. Ball B moves in the +y direction, while ball C moves in the +x direction, with the velocities as labeled. (i) Redraw Figure 2(b), showing the general direction in which you would expect ball A to move, after the explosion. (ii) Using momentum principles, determine the magnitude and direction of the velocity of ball A, after the explosion.

18 HYSICS: Final aper. February rogram y e! y d L d M (a) m M (b) c m Figure 3: Question 2 ( ( ) + (0) = 20 marks): art (a): A mad inventor devises an instrument for the measurement of the acceleration of gravity, g. This invention is illustrated in Figure 3, and consists of a ball of mass m, and a block of mass M, connected together, over a massless, frictionless pulley, of diameter, d, by a massless string. The length of the string from the pulley, to the ball m, is L. The y-axis in the figure passes symmetrically through block M, and is aligned vertically. The instrument is spun, with a slowly increasing angular velocity about the y-axis. As the angular velocity increases, the angle of the string to the vertical, at ball m, increases, from zero. Figure 3 (a) shows the system before it begins to rotate. Figure 3 (b) shows the system as block M is just about to lift o the shelf on which it stands. At this stage, the angular velocity of ball m is!, and the angle of the string to the vertical is, as labeled.! and are then recorded, and the acceleration of gravity, g, calculated. (i) Draw a diagram of each of masses m and M, as in Figure 3 (b). Label on each diagram all forces that act on each of the two masses. Label also, the acceleration of each mass. (ii) What is the radius of ball m s circular path in Figure 3 (b)? (iii) Use Newton s laws of motion, to derive an expression for the acceleration of gravity, g, in terms of!,, L, and d, as in Figure 3 (b).

19 HYSICS: Final aper. February rogram M d D µ k Figure 4: art (b): Figure 4 shows a block, of mass, M, on a slope. The coe cient of friction between the slope and the block is µ. The block is released from rest, slides a distance, D, down the slope, compresses the spring, of spring constant, k, at the bottom of the slope, and is projected back up the slope. It comes momentarily to rest again at a distance of d, up the slope, beyond the end of the uncompressed spring. D and d are labeled on Figure 4. Using energy principles, derive an expression for the distance,, that the spring was compressed, by the block, in terms of the parameters labeled in Figure 4.

20 HYSICS: Final aper. February rogram G µ H M µ =0.0 kg/m Figure 9: Question 5 ( ( ) + (5 + 5) = 20 marks): art (a): In order to measure the mass, M, of a block, a mad invertor designs the apparatus, shown in Figure 9. A string, with mass per unit length, µ, is stretched over a pulley, H, between a wave generator, G, at one end, and the block, of mass M, at the other. The value of µ is labeled on the figure. Generator G transmits an harmonic wave of fixed frequency and amplitude, from G to H, which is measured using a wave analyser. In a particular measurement, the wave function was found to be - y = 0 3 sin(00x 600t) (SI units) Take the acceleration of gravity, g = 0 ms 2. (i) Find the amplitude, frequency, wavelength, and velocity of this harmonic wave. (ii) Find the power that generator G must output to continuously transmit this harmonic wave along the string. (iii) Find the mass, M, of the block, for this particular measurement. art (b): A photoelectric cell has a silver electrode. The work function for silver is 4.73 ev. (i) Calculate the threshold wavelength for this photocell. (ii) Calculate the stopping potential for this photocell, when illuminated with light of wavelength, = 200 nm?