MUMBAI / AKOLA / DELHI / KOLKATA / LUCKNOW / NASHIK / GOA / BOKARO / PUNE / NAGPUR IIT JEE: OLYMPIAD TEST DATE: 7/8/8 PHYSICS SOLUTION. (C) Averae acceleratin, vf vi tan tan a t. (B) ds Resistance kv Eqatins f mtin are d x dx k... d y k... Interatin () and () and sin the initial cnditins, We et dx kt cs. e... 3 kt And k k sin. e i.e., k sin. e kt... 4 dx k kt k sin. e... kt dx dx k cs. e Directin f prjectin was with the hrizntal, when the directin f mtin aain makes the anle with the hrizntal, it really makes the anle with the hrizntal in the sense f the directin f prjectin. If this happens after the time t, we have frm (), kt k sin. e tan kt k cs. e kt k sin e i.e., tan k cs i.e., k sin k sin. e kt r e kt k sin k r t l sin k CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI #
3. (D) This is a straiht frward prblem in calclatin the ttal distance. The nly trick is t make sre that in the crss-ver lap y se the hyptense. The ttal distance mst be precisely 4 m, s we mst have 4 R. R. /. Slvin fr R ives.44 m. 4. (B) P be the pint where the tanent is parallel t the inclined plane. If PN z be perpendiclar frm P n the inclined plane and PM the vertical altitde f P then evidently fr all pints n the path, P is the pint where z is the reatest and cnseqently PM is the reatest. Nw fr the pint P, velcity perpendiclar t the inclined plane is zer. Nw the velcity and sin and cs and this velcity becmes acceleratin perpendiclar t the plane at O is zer at P. sin cs. z sin z cs Fr maximm rane r 4 4 Hence, z sin cs 4 cs 4 cs = sin 4 cs r PM z sec sin 4 cs 4 sin 4 Maximm rane 4 PM. (B) v v v B BA A (maximm rane) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI #
v cs v cs 6. (A) ˆ v j v 3i ˆ SB v vj ˆ 3i ˆ and S S m s v 3 3 m s S Drift = 3 = m 7. (B) The cmpnent f acceleratin de t ravity aln reatest slpe is sin sin R r 3 6.3 ms.73 ' sin 3 ms 8. (B) Here, v sin sin 37 3 v sin 3 sin 3... i v AB Als, t v cs 37 v r 8 6 3 4 cs v 4 v cs 4 3 3 cs CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI # 3
v cs 3 3 v cs v sin 3 3 3 v 6 ms 9. (B) Time taken t reach rnd t h v sin t t h v sin t t t v sin t h... i Time taken by by t h h t t... ii h Fr the time t t t, by and ball fall with same acceleratin ( dwnwards). a They see each ther perfrmin straiht line mtin. Fr time t t t, by will remain statinary n rnd and ball will cme dwn with acceleratin. It will aain trace a parablic path srt by. Als, we can et b frm Eq. (ii) frm iven t and ca ths als find v sin eqatin. (i) As till t by is still n bildin s rf hell sec ball s trajectry as y v sin t t x v cs t x t v cs x y x tan cs. (C) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI # 4
.4 Frm fire, slpe is tan r v t mm.4 4.4 t v. (ABCD) Chane in velcity = final velcity initial velcity cs ˆi cs ˆi sin ˆj sin ˆj (A) is crrect Averae velcity = (ttal displacement)/(time taken) = ( cs iˆ (B) is crrect. Chane in velcity = final velcity initial velcity cs ˆ i sin ˆ j cs ˆ i sin ˆ j sin ˆj (C) is als crrect. Rate f chane f mmentm = frce Cnstant ravitatinal frce is actin n the prjectile. (D) is als crrect.. (AB) Eqatin f mtin is dv 3 dv v i.e. 3 v Interatin it, we have t A... v Initially, when t, v V A V Hence, eqatin () becmes V t t v V V dx V i.e. v V t This prves the secnd ptin. Interatin it, we have / x V V t D V t D V Initially when t, x D V S that, we have x V t V ˆ Ri / Time f fliht) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI #
3. (ABC) Here, OB OA AB r eˆ eˆ nˆ eˆ nˆ AC CB AB AC eˆ OAcs n ˆ eˆ ˆ cs ˆ ˆ e n OA e eˆ cs n ˆ Bt e ˆ ˆ ˆ ˆ cs cs n e n eˆ eˆ eˆ nˆ n ˆ... i (B) cs ˆ ˆ e n cs ˆ ˆ e n (C) Similarly, cs ˆ ˆ e n 4. (ABC) (A) The dwnward cmpnent f parallel t OY is sin, Hence the dwnward cmpnent aln the rve is a = sin sin. Since (B) sin.8 / x y y a 9.8..8 3.9 m s s v t at, where / and sin. s x y m and v. Ths, s 3.9 t r t.97s v 3.9.97 6.6 m s (C) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI # 6
. (ABCD) bt (A) bke b d y bt Differentiatin nce mre b ke Mltiplyin r expressin fr by b and addin yields bt b ke. Ths () is satisfied. Sbstittin t = and recallin that e, we et y. (B) Since the ball is eased frm rest, y at t. Usin r expressin fr frm (A), We have bk, which yields k. b b (C) If b./s and sin 9.8 m s, We have k 98 m. Then at t s, y 98me 98m ss 36m. b v 98m se 9.8m s. s 6 m s (D) At t 6s, 6 98m se 98m s 6 e. S, 9.8m s 6. () tan 8.6 Let v initial velcity f at s v t x x 4 v cs 8.6 4.4 t v Als sy t... sy v t t v sin 8.6 t t y 4.4 4.4 v sin 8. 3. v v (iven) Slvin this, v 3.76 ft/sec. v x 3.76 cs 8.6 9.3 ft sec v y 3.76sin 8.6.66 ft sec Als v v a s y y y y.66 3. CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI # 7
v y v 9.3 ft s, vx 9.3 ft s; tan 4 v y x 7. (4) 8. (3) p s s t t t n p t t 4 km h 8 9. (8) The psitin vectr is r xˆi yˆj The ball is at the reatest distance frm the riin when its velcity is perpendiclar t psitin vectr. y vy x v x t sin t sin t t cs cs 3 sin t t If this is nt happen drin mtin, the discriminant shld be neative. 3 sin 8 sin n 8 8 3. () In fire, pints A, B and C are lyin n the trajectry f the path f the ball. The eqatin f trajectry is CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI # 8
x y x tan cs x Fr pint A, 3 x tan cs Or 3 x tan x tan... i 3 x 8 tan x 8 tan... ii Fr pint B, Fr pint C, x x 6 3 6 tan tan x 6 tan 3 x 6 tan... iii Sbtractin Eq. (i) frm Eq. (ii), 8 tan tan x 6x 64 x 8tan tan 6x 64... iv Sbtractin Eq. (i) frm Eq. (iii), x x x tan 6 tan 3 6... v Frm Eq. (iv) and Eq. (v), 3 tan 64 tan CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI # 9