FULL MECHANICS SOLUTION

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1 FULL MECHANICS SOLUION. m f For long the tngentil direction m 3g cos 3 sin 3 f N m 3g sin 3 cos3 from soling 3. ( N 4) ( N 8) N gsin 3. = ut + t = ut g sin cos t u t = gsin cos = = s] 3 4 o 4. cos 6 o u i ( u sin 6 gt) j o o u cos 6 j ( u sin 6 gt) k o. ( u sin 6 gt) u cos 6 cos o. ( u cos 6 ) ( u sin 6 gt) 5. In fig. () = 5 In fig.(b) 8 = = 5 =, 3 8 = 8 = 6, 3 6. By constrint motion cos = u = u sec u= m/s h= 9 m or = d = usectn d sin N cos CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA #

2 When contct lees, N= i.e. sin = nd cos = m tn = g or = g cot lso = tn hcot = So nd d h cosec d u sec tn d cos =m nd sin + N = or h cosec u sec tn or tn = or = 6 gh u u sec h cosec or g cot u sec tn or tn= 3 7. (b) Friction force = cos sin 8. (b) l s u = mu l cos 6 o m m o cos 6 l ension will be mimum t the bottom (lower most point) Mg m l m f(s) m 8 g For m N 8g o ensure no slipping, f N 8g s fs CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA #

3 m 4 min 9. (d) u= m R Q M R m gr ill then the bll reches the lowest point, liner momentum will not be consered. Post tht, liner momentum shll lso be consered. m gr M m m gr M m. (c) N =constnt f k F= N fk F N N.5g 5N otl contct force 5 3 N. () ˆ ˆ ˆ ˆ 4 4i 4 3j 8V perfectly inelstic collision 4 34i 3j ˆ 3 i ˆ j 8 9 Loss of mechnicl energy KE KE J 4 CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 3

4 . (c) m/5 4 kg 4 kg m/sec 4t the string will be tut gin fter tencet. he hnging mss will be under free fll. 4t t gt t t t 4 / t 4sec At this instnt the elocities of ech mss. 4m/sec 4m/sec =+g(.4)=6m/sec After the jirk N V V N sec m / sec 3. (c) 3 Gien 3 k where k is constnt of proportionlity d 3 k d Integrting d k d k C [C is constnt of integrtion] When, C k k () Also when, k k CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 4

5 From (), we get When () In ccelertion motion slope ds increses wheres for retrded motion it decreses. 5. (c) Using ector form of kinemtic eqution gh gh g g h u g 6. () Since the mss is sme therefore the length of the plnk should be twice the rnge u sin R 4 3m g 7. () 5 t t t 4 t 5 t So, t sec. i.e., horizontl distnce m () AC sec. So, u sin u sin g m / s g Now y u t t h u sin 3 g 3 y y h 3 9 h 3 45 CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 5

6 h m 9. (b) m mss of the block Applying Newton s Lw on the block in erticl direction N m.. () 3 N (gien reding of spring blnce) N 5 N m s (gien reding of W.M.) N 5 3 m m 4kg g. (b) F.B.D. of mn nd plteform N 5g N 5g N. () he F.B.D. of block A nd B re From constrint, the ccelertion of A & B re nd respectiely. Applying Newton s second lw to blocks A nd B, we we get 4g 4. (i) 5g 5.. (ii) Soling we get ccelertion of A nd B s g 7 Downwrd, g upwrd respectiely 7. (c) CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 6

7 he F.B.D. of pulley is s shown Let, nd FS be the forces Eerted by the horizontl string, erticl String by the support on the mssless pulley Respectiely. hen FS or FS ension in ech string is 3. () 4. (d) m ilblet mn 'sfeet fr fr. N fr fr. 4 4 N ilblet ground 5. (A,C) So net friction on mn is 5 5 N. 6. ime depends only on y-direction ( perpendiculr to rier flow) elocity which is sme for ll the four 7. Suppose only block (A) nd (C) moe = m = m = 9 m = 7 8 g & = 7 9 g CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 7

8 = m. 7g 9 = 4 9 < ] 8. (A,C,D) 9. (C,D) direction of friction does not chnge friction is constnt. 3. (, c, d) As there is no friction between the bll nd the grooe so there will be no centripetl force i.e., No net force on the bll towrds the center hence the bll moes in rdilly outwrd direction w.r.t disc 3. (b, c, d) After Impct e gh cos Before Impct gh cos e gh cos A H gh sin O gh cos N c toa gh cos gc gh sin gh cos c tao e gh cos Due to the impct only horizontl component of elocity gets ffected so totl time of flight remins the sme t t OA AO c c gh sin gh cos e gh cos g c gh sin gh cos e g h sin e c h sin c e c c e h sin c CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 8

9 3. (, b, c) h u g / mu cos m () As N = t ngle of rottion mv So cos () Using eq. () g m cos cos cos cos cos 4 5 cos 6 cos 5/ 6 5 h cos (, c, d) k m F k m k k F d rel d F k k d m m d F k d m m m d F k d d m m m At mimum compression or mimum etension reltie elocity should become zero. F k m m m F/ m or K m m mf K m m CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 9

10 When m m m then m F / K Compression the spring will be zero. 34. (, d) 3 t 33t 8t d 6t 66t 8 d t 66 Direction chnges when t 5, t 6 Minimum speed is when it is zero i.e. t 5 & t 6 Retrded mens elocity is opposite of ccelertion i.e. t 5 & 5.5 t (, c, d) time t t / t / distnce left t / t n n t / s Distnce trelled in time t t t d t d t. t t 36. (, c, d) Slope of displcement-time cure gies elocity. () During OA slope is + e but decresing hence elocity is positie nd ccelertion is negtie. (c) During BC slope is e nd going to zero hence elocity is e but ccelertion is +e. (d) During DE slope is +e nd incresing hence elocity is +e nd incresing +e ccelertion 37. (b, c) Initil elocity of prticle in ector from cn be written s cos ˆi sin ˆ j p.() Velocity of prticle t ny time t will be: ˆ Q cosi sin gtˆj..() Gien tht p Q CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA #

11 . p Q cos sin sin gt Or Or sin gt or t cos ec g Substituting this lue of t in Eq. () we get: ˆ ˆ Q cosi sin j sin Or Q cos sin sin cot 38. (, c) At highest point ngle between nd is ccelertion. n R 9. Hence, totl ccelertion is only norml or rdil g u cos R 39. (b,c) 4. (,c) Sol: u cos Or R g At point of projection component of ccelertion (= g) long elocity ector is g cos 9 or g sin. g tn m g CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA #

12 4. (b,d) Since the pprent weight is incresing, hence ccelertion of the lift should be upwrds. his is possible in cse of (b) nd (d) 4. (,b,c,d) here is no horizontl force on block A, therefore it does not moe in -directing, wheres there is net downwrd fore N is cting on it, mking its ccelertion long negtie y-direction. Block B moes downwrd s well s in negtie -direction. Downwrd ccelertion of A nd B will be equl due to constrin, thus w.r.t. B, A moes in positie -direction. Due to the component of norml eerted by C on B, it moes in negtie -direction 43. (,d) Mimum lue of friction force between 4m nd inclined plne 4 cos45 4 Here pulling force F 4 cos 45 P Block will not moe Accelertion of 4m block, frictionl force on 4m block 44. (b,c) fm.4 8N For equilibrium Minimum lue of m m 8 m kg Mimum lue of m m 8 m 8kg So, (B) nd (c) re correct 45. (CD) 46. (BD) 47. (D) 48. (BC) 49. (AB) 5. (), 5. (b), 5. () Slope of V ersus t grph is g y g t As displcement long y-is is zero k Vy CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA #

13 u y tn u 53. (D) 54. (D) 55. = 5 = 5 = 4g m ' = = r = = 9g m rp = 9g ( g) = g r 5m 56. Let = A Bt d At t, tn A 4 Agin, t = 3 sec, = 3 B3 B So t 4 4 t 3 t d 4 t = 57. [6] f sin + mr µ cos sin + mr r or µ tn + gcos 58. () sin + mrw K = 5 N/m Initilly the spring is reled M If the block moes down by, then the spring is etended by s the string connecting the spring is in etensible. Using energy consertion principle, m K he etension in the spring will be mimum when the block comes to rest momentrily. i.e., = CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 3

14 k k4 k / k Mimum etension in spring cm k 59. (5) If M remins sttionry = Mg Mg cos cos sin m sin m m sin sin m sin Mg m. rd / sec re / sec 6. (5) A V A y B V B C V CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 4 C V Using energy consertion, ma mcg maa mbb mcc g B

15 VA cos V A y VA sin VB sin V B V sin V cos A B VA VB l 6 VA 38 8 VB cos CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 5

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