S14 IAL M1 Model red.

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  • i,tt IrtL Sit1. Two smalt smooth balls I and B have mass 0.6 kg and 0.9 kg respectively. They are

    moving ia a straight line towards each other in opposite directions on a smooth horizontalfloor and collide-directly. Immediatety before the collision the speed of i{ is v m s-l andthe speed of B is 2 m 11. The speed of ,,4 is 2 m s-1 immediately after the collision and Bis brought to rest by the collision.

    Find

    (a) the value of v,

    (b) the magoitude of the impulse exerted oa,4 by B in the collision.*s@;)

    f,^oftr 6tvrom 6

    (3)

    (2){A-)ffiJJ

    }l vzl,-l-8

    . r- - lqpslle,

    i_(:0.1)='O

    ; [-.f,-6tr

  • A ball is thrown verticalty upwards with speed 20 m s_l from a point l, which is ft metresabove the ground. The ball moves freely under gravity until it hits the ground 5 s later.

    {a} Find the value of /r. (3)

    A second ball is thrown vertically downwards with speed 14, m s-l from I and moves freelyunder grayityuntil it hits the ground.

    The first ball hits the ground with speed Zm rl and the second ball hits the ground withtI

    speed I rm s-t.

    (b) Findthe value of w.

    *$;*.".S';at+.|aK,,v -h.tOb:lttltrL" k=s

    (s)

    =) -h = -a}-S

    -: h i z!,zsaa2zlJl;- O va*tar. n; t-o-4.9(l) z-t;l,,

    @ S e22.S'-, O o t^,* v = Zt.+s

    T=qo

  • 3" A particle P of mass 1 .5 kg is placed at a pint ,,{ on a rough planre which is inclined at 30oto the horizontal, The coefficient of friction between P and the plane is 0.6

    (a) Show t}rat P rests in equilibrium at l.

    A horizontal force of magnitudeXnewtons is aow applied to P, as shown in Figure 1.force acts in a vertical plane containing a line of greatest slope of the inclined plane.

    Figure IThe particle is on the point of raoving up the plane.

    (b) Find(i) the magnihrde of the norrnal reaction of &e plane on P,(ii) the value ofX

    (s)

    The

    ilG;I;S3bi3o--@,|tr'sgl^rq)

    rh-*l^r,J,tae*lrrn-erarrzr-tlrrlJla-r-lr-Lf {*^qg-_-_

    S itrv. 3--e--t"}-tr.-t*1"J*rurt'rrt-}--*-O*Sr*5- *-O".aJg

    -{^r*w--: 0---*+1ty9 *;:;-(6n6-p{*-'- *a1lr_1'1u*-

    - -;. (i" ^;- 6l.J?-tArT;JG't-5U-

    -- 0-*SS irs?6i3il-

    d."t 7 ; - ? tos$O*rf; :[$g-+& 3Ol-O*gSCar3O

    ; o.t{-;-o,{EeTO.eStDr3op-x-_g!f

    i l\ra t Qi-O*

    +-ie

  • 4. (2 kg) (3 ke)A

    +-2 flR--{6m

    Figure 2

    AplankLB, of length 6 m and mass 4 kg, rests in equilib,rium horizontally on two supportsat Cand D,wberi AC:2m and DB:1m. Abrick of mass 2 kgrests on theplank atA aflda brick of mass 3 kg rests on the plank at B, as shewn in Figrne 2. T\e plank ismodelled as a rmiform rod and all bricks are modelled as particles.

    (a) Find the magnitude of the reaction exerted on the plank(, by the suppcrt at C(i0 by the support at D.

    The 3 kg brick is nour removsd and replaced with a brick of mass x kg at 8- The plankremains horizontal and in equilibrium but the reactions on the plank at C and at D nowhave equal magnitude"

    {b) Fiild the value ofx.

    (6)

    (4)

    9c

    c1, lsla j (r-z-5_sj16i_- _l';t _d-erf 5aT- ::q*1*::"*-:f{

    st 8*L1$[.r : k5rs_J by5'R-. n 45 ---:-@s4-E

    f={u -e&= &-o).t - 1-6s=&-$g--i-.-d*;3:b

  • [In this question i and i are horizontal unit vectors due east and due north respeetively.Positionvectors are given relafive to aftxed origin O-|A boy B is runniag in a field with constant velocity (3i

    - 2j) m s-1. At time f : 0, B is at

    the point wi& position vector 10j m.

    Fiad

    (a) the speed of8,a)

    (b) the directioa in which B is running, glving your arswer as a bearing.(3)

    At time t: 0, agirl G is at the point with position vector (4i -

    2i) m. The girl is running

    wi& constant velocity flt * Zi] * s-' and meets ,B at the point P.'\3 ")(c) Find

    t0 the value of,rwhen they meet(ii) the position vector of P. (o

    c) B). -=s

  • A car starts from rest at a point r4 and moves along a skaight horizontal road. The carmoves with constant acceleration 1.5 m sa for the first 8 s. The car then moves withconstaat acceleration 0.8 m sa for the next 20 s. It then moves with constant speed forI seconds before slowing down with constant deceleration 2.8 m s-2 until it stops at apoint.B.

    (a) Find the speed of the car 28 s after leaving l.(3)

    (b) Sketch, in the space provided a speed-time graph to illustrate the motion of the caras it travels froml to B. (2'

    (c) find the distance travelled by the car during the first 28 s of its journey from r4. (4)The distance from Ato B is 2 km.

    ($ find the value of 1.o) Y zw-t,a,k VzO+ l-S(f) = \Z

    V=w.t^^ V=lL+O'8tz-o)5) ttl *- -:

    LE uq 3[+1

    ,ttftt5 s tB lo.t--t

    (4)

    f

    t((

    t

    ti{.t1

    c)

    4,, =I itrltra) '

    a& u 3 ifzo)[rtz{)%ofir8-2-

    T:-frr = 2{5cj-tPo8-r*(

  • 7.

    Figure 3

    Two particles P atud Q, of mass 2 kg aad 3 kg respectivelg are con*ected by a lightinextsnsible sting. Initially P is held at rest on a fixed smooth plane inclined at 30" tothe horizontal. The string passes over a srnall smooth fixed pulley at the top of the plane.The particle p hangs free1y below the pulley md 0.6 m above the ground, as shown inFigure 3. The part of the sking from P to the pulley is parallel to a line of greatest slopeof the plane. The system is released from rest with the sfring taut

    For the motion bef.ore O hits the ground,

    (a) (r) show that the acceleration of Q ,,24,5.(ii) find the teas on in the striag.

    On hitting the ground Q is immediately brougktto rest by the impact,

    (b) Find the qpeed of P at the insknt when O hits the grormd-

    In its subsequent motion P does nnt reach the pulley-

    (8)

    Q)

    (c) Find the total distance moved up the plme by P bef,me it comes to instantaneous rest.

    (d) Fiod the length of time between g hirine the gro.und and P first coming toinstantaneorc rest.

    {Z)

    "-:*,f@'trt#--_5 _ 3g

    -

    -f3z J+- .i..)_' -----g"o !.s- , ----;i-S.-o;6-];=*a;u,- I' la

    ---n-'.':i#+- -**_*

    -

    -----

  • c)

    S=-?--rtre ,-\g

    ff' tr=rrAa -)= tc^ -'- ^ = -l\rV '

    = teao Lc*s :- {t , +,rnO=*l-3s '-

    rse-46,.^