ci-il => -8 =-9+3Vg =) 3Vs= -*... Vg't ft-s vft.4 ln.^$ ger - Lxln SxL+ Ln-6 = Lr-Z+ 3V6 2Vn = -tf

Transcription

1 Ht MftY 2ot3 rneenfrrtonr\l 1. Two particles I and B, of mass 2 kg and 3 kg respectively, are moving towards each other in opposite directions along the same straight line on a smooth horizontal surface. The parlicles collide directly. Immediately before the collision the speed ofl is 5 m s I and the speedofb is 6 m s-r. The magnitude of the impulse exertedonbbyl is 14Ns. Find (a) the speed of,4 immediately after the collision, (3) (b) the speed of B immediately after the collision. ft-s (- -bb (3) avp o_vc lylorn A blare-' to e lmou[se- -- r+ ln.^$ ger - Lxln ci-il SxL+ Ln-6 = Lr-Z+ 3V6 2Vn = -tf vft.4 => -8 =-9+3Vg =) 3Vs= -*... Vg't -

2 2- Figure I A particle of weight 8 N is attached at C to the ends of two light inextensible strings lc and.bc. The other ends, I and -8, are attached to a fixed horizontal ceiling. The particle hangs at rest in equilibrium, with the strings in a vertical plane. The string lc is inclined at 35' to the horizontal and the string BC is inclined at25" to the horizontal, as shown in Figure 1. Find (i) the tension in the string lc, (ii) the tension in the string BC. (8),n3S* (ostn'ls Ts (os2s =) TB= TA-C"sS Cos2,S fr+4--o <-) Tn Srn3S+ GSrn2S = E =) Tinb,njS+ r1q'3s CosLS *8 :) o. RSSSS33.-ta= I =) 'i-a = 8.31., Te " + 'Sao.r

3 3. 6 -{ t^" k +".$:"* ag Figure 2 A fixed rough plane is inclined at 30o to the horizontal. A small smooth pulley P is fixed at the top of the p1ane. Two parlicles A and B, of mass 2 kg and 4 kg respectively, are attached to the ends of a light inextensible string which passes over the pulley P. The part of the string from I to P is parallel to a line of greatest slope of the plane and B hangs freely below P, as shown in Figure 2. The coefficient of friction between I and the plane i, *. tritiutty z is held at rest on the plane. The parlicles are released from rest with!3 the string taut and I moves up the plane. Find the tension in the string immediately after the particles are released. (e) 5rP-'$ '> {nnar =7r,tNA - 11'6 =) {"{o-{ Sre }ur^-.{lo- pl"r- = 30. \,Ld4_ s.xstcxfrh: t'6^ 45-T -- k". o +g-r- f3 (r -b): T= *aru T= -: ^, *b

4 4. At time r : 0, two balls I and B are projected vertically upwards. The ball I is projected vertically upwards with speed 2 m s 1 from a point 50 m above the horizontal ground. The ball -B is projected vertically upwards from the ground with speed 20 m s 1. At time l: 7 seconds, the two balls are at the same vertical height, h metres, above the ground. The balls are modelled as particles moving freely under gravity. Find (a) the value of 7, (b) the value of ft. (s) (2) S. - (so-h) Lt, - 2^ V ( = -1.8 L, r s.i t{,20 V At ---q.8 :{ g = 6t+!atl -4 h- 50 = LT - *-\ro o h -- eor- +'1T" =)!/. SO =x +/-rcf :) -SO= -lgr b) h= 2o(?)- q,r(?)', *.1T1,." T=7

5 5. blanl Figure 3 A particle P of mass 0.6 kg slides with constant acceleration down a line of greatest slope of a rough plane, which is inclined at 25" to the horizontal. The particle passes through two points A and B, where AB: 10 m, as shown in Figure 3. The speed of P at A is 2 m sr. The particle P takes 3.5 s to move from I to B. Find (a) the speed of p at B, (3) (b) the acceleration of P, (c) the coefficient of friction between P and the plane. {r*** ltnr ts\ f 0.53tosZs S -- lo (r 2- NK-- s.3z1o6qtr' ^ "o'b59a2s -' {L^r = s'j21oe981^ e) (5).f\ =r,rc. z) L' S' S=tr"t +!al'--) to '++fo.(s.s)' v b)..-^,y avt t =3.S a) V =tr.+at V=2-+(H,X ). + c) a.+8tr11s - o 6(T)= s'321o81trl,r.'- ),\-- O.+l (zsp) c\.

6 6, lln this question i and j are horizontal unit vectors due east and due north respectively. Position vectors ore given with respect to a fixed origin O.) A ship 5 is moving with constant velocity (3i + 3j) km h r. At time t: 0, the position vector of S is (-4i + 2j) km. (a) Find the position vector of S at time I hours. (2) A ship 7 is moving with constant velocity ()i + ni) km h r. At time I : 0, the position vector of 7 is (6i + j) km. The two ships meet at the point P. (b) Find the value ofr. (s) (c) Find the distance OP. (4) ") v ' (e) s= (-I).. (i) b) r=(t)--(l) fi 2+3t ol+tn -) 2-+L= c) P" (-;)- (t ) = (; ) 2b = -S+3t" = 5b t'l -, I +2n i 2a: $ =)fpls 67 =64" = 8.2Sha,. (3sg) ----

7 1 Figure 4 A truck of mass 1750 kg is towing a car of mass 750 kg along a straight horizontal road. The two vehicles are joined by a light towbar which is inclined at an angle d to the road, as shown in Figure 4. The vehicles are travelling at 20 m s I as they enter a zone where the speed limit is 14 m s 1. The truck's brakes are applied to give a constant braking force on the truck. The distance travelled between the instant when the brakes are applied and the instant when the speed ofeach vehicle is 14 m s I is 100 m. (a) Find the deceleration of the truck and the car. (3) The constant braking force on the truck car also experience constant resistances Given that cos d : 0.9, find has magnitude R newtons. The truck and the to motion of 500 N and 300 N respectively. (b) the lorce in the towbar, (c) the value ofr. (4) (4) I u v a\ t 1 16() V 2, (,tz+,l*a.s t1[ - *OD+ZOOa =20 = tg =)?-@a= -2OY +) A. -l'ol?''- r-oz 4- r. ol Note - 6lls* bre tl'ng tprr,rtorr- rs'th(i)st () urt\dlz-s")s+e$r R+ S@ o.qt -O-tr.2S00y1.02 b) Car 3oO + o.1t =?E>Q-oz) =1 (tt6o 2 2SSo :. [2, i45on.-..._-...- =) T-hcust. 5l+., (ssp) -_

8 t' 8. A C A < > 0.2m 2m tr D A J, so3.xm R Figure 5 A uniform rod lb has length 2 m and mass 50 kg. The rod is in equilibrium in a horizontal position, resting on two smooth supports at C and D, where AC : O.2 metres and DB: x metres, as shown in Figure 5. Given that the magnitude of the reaction on the rod at D is twice the magnitude of the reaction on the rod at C, (a) find the value ofx. (6) The support at D is now moved to the point E on the rod, where EB : 0.4 metres. A particle of mass m kg is placed on the rod at B, and the rod remains in equilibrium in a horizontal position. Given that the magnitude of the reaction on the rod at E is four times the magnitude ofthe reaction on the rod at C, (b) find the value of m. a) t= il.,1, + 3r= sga L.!9q 3u C*z a $5x I's = 5o3r t -) *{* = Lo4 z(= rya. 2q s Q.6 f"\ b) so9 16) Bt 4r(yp.k+(x1.8 " SOSxl '-) 3'$e-- SoS.'. k-- Ea l+j r.) t2so. ol= 32OJ + r.^{ t.-v -) $r( = SO5'i n^5 \aul1p, "' t/.=!@ t+ /3,5w5