University of Malta. Junior College

Transcription

1 University of Malta Junior College Subject: Advanced Applied Mathematics Date: June Time: End of Year Test Worked Solutions

2 Question (a) W µ o 4W At Equilibrium: esolving vertically esolving horizontally + W sin o 4W + W 4W + W 4W W µ o W cos µ ( W ) W µ o Angle of friction λ tan ( µ ) tan

3 (b) T µ N With respect to the diagram above: esolving c N T µ () 5 N Hence T 5 N Applying Hooke s law i.e. T λx, a where a is the natural length cm. m, λ is modulus of elasticity and x is the extension.

4 we get 5 5x. 5(.) Hence x. m 5 The above applies for the closest and furthest positions. In the former it is a compression, while in the latter it is an extension. Furthest point. +.. m from support. Closest point...8 m from support. Question (i) V m/s 5 m The pump is 6% efficient. Hence the actual mechanical work is 6 6% of.85 kw * kw 495 W (a) The volume of water per minute. m.

5 Hence the volume per second Mass Since Density Volume. 6 m.5 m, then Mass Density * Volume Power Energy/s K.E./s + P.E./s 5 kg/ s Energy/s mv + mgh y (a) above, Energy/s 495 J, m 5 kg/s, h 5 m Substituting, we get 495 ( 5) v + 5()5 On simplifying 495.5v v Hence v 98 i.e. v m/s (ii) If V is the velocity of water through the nozzle and A is the cross sectional area of the nozzle, Then volume per second AV.5 A(9.9) Hence A m 5.cm

6 Question µ A T N T 5 N m (i) Consider the block. Given u, t, s, a? Applying the equation of motion: Applying: v u + at v + ( ) s ut + at a ( ) + ( ) a( ) a v m/s m/s

7 (ii) Applying Newton s second law F ma on block, 5 T.5 5 T.5 T N At equilibrium, esolving c on block A N Applying Newton s second law F ma on block A, T µ a Substituting.75 µ ().75 µ () 5 µ.875 (iii) After hits the floor, there is no tension T in the string. Applying Newton s second law F ma on block A, µ a Substituting.875() a a.875 m/s decelerating Applying the equation of motion: v u + as () + (.875)s.75s Hence s.67 m.75

8 Question 4 (a) Applying the equations of motion s u V a g substituting in s ut + at Vt gt V t, or t g t V gt

9 Considering in horizontal direction: s u U g On substituting in the same equation of motion, we get Ut Hence V U g UV g For maximum height, time time of flight V g Applying the equations of motion for maximum height s H u V a t g V g V g substituting in s ut + at V H V g V g g V g

10 (b) (i) Let U be the initial speed In the horizontal direction s using s ut + at u U a, we get Ut.(i) In the vertical direction s.5 using s ut + at u a, we get.5 + ( ) t.5 5t i.e. t or t Substituting in (i), we get U U m/s

11 (ii) In the horizontal direction s using s ut + at u U a, we get This is the time that the ball is over the net. In the vertical direction t using 5 u t t 5 a at, we get s + ( ) s ut m 5 This implies that the ball descended a distance of.9 m from the starting position, which is 5 m above ground. the ball is m above ground.

12 Question 5 C A 6 µ W (i) (ii) Let and be the reactions at points A and C, as shown. Taking moments at A : Thus W l Wcos6 esolving vertically, we get l l l W b + cos6 W + W Substituting the value of, we get Thus W W 6 5W 6 esolving horizontally, we get W + W µ sin6 Substituting the values of and, we get 5W µ 6 µ W i.e. µ 5

13 Question 6 E D 4 m F C 6 m 6 m O m m A (i) The area of the shaded circle 7 π 7 sq.units π The area of the rectangle OAF *6 7 sq.units The area of the rectangle CDEF 4*6 4 sq.units Hence the area of cross section of the concrete block Let M be the mass per unit area and 89 sq. units let ( x, y) be the centre of mass of the block with respect to the origin O. The centre of mass of the circle and rectangles OAF and CDEF are at (, 6); (6, ) and (, 8) respectively. Applying the principle of moments in the x direction, we get 89 Mgx 7Mg(6) + 4Mg() 7Mg() 48Mg 48 x

14 Applying the principle of moments in the y direction, we get 89 Mg y 7Mg() + 4Mg(8) 7Mg(6) 66Mg 66 y (ii) G ( x, y) θ O θ H At equilibrium, the line through the centre of mass pass through O. With respect to triangle OGH, where angle OHG is a right angle x tan θ y θ tan 5.9

15 Question 7 (i) 5m/s 7m/s U A U A A 5 g g efore Impact After Impact Using the principle of conservation of momentum 5(5) (7) 5U A + U i.e. 5U A + U 5U A + U (i) Using the law of estitution, U U A U U A i.e. U U A 9 (ii) 4 5 ( 7) (i) U + 5U A (ii) U U A 8 7U A 7 U A m/s Thus sphere A changes direction and travels to the left. Substituting this value in (ii), we get U + 9 i.e. U 8m/s Consider the impact of sphere with the wall. Using the law of estitution, U U U 4m/s U Sphere travels in the direction of A with speed of 4m/s, while sphere A travels with speed of m/s. Thus there is surely going to be a second collision between the spheres. 8

16 (ii) m/s 4m/s V A V A A 5 g g efore nd Impact After nd Impact 5() + (4) 5V A + V 5V A + V 5V A + V (i) Using the law of estitution, VA V VA V 9 V A V (ii) 4 4 () 4 (i) 5V A + V (ii) 5 5V 5V A V 4 7 V 4 m/s Substituting in (i): 5V A + 4 V 5 A m/s 5 (iii) Initial K.E. ( 5) + ( 7).5 J Final K.E Loss in K.E J J

17 Question 8 (i) Since Power Force Velocity Then, F v, where v is the velocity of the car. F v When the car is going to descend, then, F v F v N F N θ 8 N F θ 8 N Ascent Descent esolving forces along the line of greatest slope At equilibrium and car ascending + 8sinθ v (i) v 4

18 At equilibrium and car descending + 8 v 4 Substituting (ii) in (i), we get + + v v This reduces to 4 v speed of ascent.5m/s + 8sinθ v + (ii) v v.5m/s 8 Substituting v in (i), we get + 6 N.5 (ii) N Acc. v 8 N Applying Newton s second law i.e. F ma, we get 8a 6 8a v.5 Hence a.5m/s

19 Question 9 (a) µ mg At equilibrium esolving perpendicular to the line of greatest slope mgcos (i) esolving along to the line of greatest slope µ mgsin (ii) (ii) (i) µ mg sin µ tan mg cos (b) v r µ mg

20 At equilibrium esolving vertically: cos µ sin mg ( cos sin ) mg esolving horizontally. µ (iii) Using F mv ma for motion in a circle r mv sin + µ cos r mv + µ (iv) r ( sin cos ) (iv) (iii) It simplifies to (sin (cos r mg v rg mv + µ cos ) µ sin ) (sin + µ cos ) v (cos µ sin ) 6() v v 9.6 i.e. v.4m/s

21 Question (i) D C DC C A C AD W T ED T EC W C C A T AE E T E 4W Since the 7 rods are identical, then all the triangles are identical i.e. they are all equilateral triangles. y symmetry C AD C C ; T DE T CE and T AE T E As the external forces are at equilibrium y symmetry A And resolving vertically, we have A + W + W + 4W 6W or A W (ii) Consider the internal forces (Each joint is in equilibrium) At A b A C AD sin 6 W C or AD 6W C AD 6W C C CAD

22 6W At A T AE C AD cos6 T AE W T E TAE At E b TDE sin 6 4W T DE 4W 4W T EC TDE At D C C + T DC AD cos 6 DE cos 6 6W 4W 5W C DC + Hence rods AE and E we have a tension of magnitude W rods AD and C we have a compression of magnitude 4W rods DE and EC we have a tension of magnitude 5W rod DC we have a compression of magnitude 6W