Numericl Problems With Solutions(STD:-XI) Topic:-Uniform Circulr Motion. An irplne executes horizontl loop of rdius 000m with stedy speed of 900kmh -. Wht is its centripetl ccelertion? Ans:- Centripetl ccelertion, c v /r. v900kmh -.50ms -. 50 50 c 6.5ms 000 A prticle of mss g ttched to string of 70cm length is whirled round in horizontl circle. If the period of revolution is s, clculte the tension. Ans:- Tension mrω. mr(π/t). 3 Tension 0 70 0 ( ) 9.68 0 N 7 3 One end of string of length.m is tied to stone of mss 0.kg nd the other end to smll pivot on smooth verticl bord. Clculte the minimum speed of the stone required t its lowest point so tht the string does not slcken t ny point in its motion long the verticl circle? Ans:- v min 5rg 5 9.8. 8. 8ms Clculte the ngle through which the cyclist bends with the verticl when he negotites circulr pth 3.3m in circumference in s. Ans:- Circumference, πr3.3m 3.3 3.3 7 Rdius, r m π 3.3 Velocity, v ms. Let θ be the ngle which the cyclist mkes with the verticl. v 3.3 3.3 tn θ rg 3.3 7 9.8 o θ tn () 5 5 Determine the period of conicl pendulum of length m if it mkes 30 o with the verticl when executing oscilltions. Also find the period of the simple pendulum with the sme length. Tke the vlue of g9.8ms -. Ans: Period of conicl pendulum,
o l cosθ cos30 T π.869s g 7 9.8 The period of simple pendulum, l T π. 007s g 7 9.8 Topic:-Work, Energy nd Power.. A mss of M kg is suspended by weightless string. Clculte the horizontl force tht is required to displce it until the string mkes n ngle of 5 o with the initil verticl direction. O 5 o l A C Mg Ans:- We cn use the lw of conservtion of mechnicl energy: Work done by the horizontl force gin in P.E of the mss M h AB Mgh Mg AB o AC ( l lcos5 ) Mg Mg Mg( ) o AB lsin5 A force 0.5x+0 cts on prticle. Here is in Newton nd x is in meter. Clculte the work done by the force during the displcement of the prticle from 0 to m. Ans:- Work done by vrible force, dw.dx dx cosθ. Since the force nd the displcements re long the X-xis, Cosθ dw dx. Totl work done,
W 0 dx 0 (0.5x + 0) dx 0 0.5xdx + 0 x 0dx+ 0.5 0 + 0 x 0 J 3 A bullet of mss 0g is found to pss two points 30m prt in time intervl of s. Clculte the kinetic energy of the bullet if it moves with constnt speed. Ans:- Distnce trvelled 30m v 7.5ms Time s mv 0.0 (7.5) K. E 0.563J The kinetic energy of body is incresed by %. Wht is the percentge increse in the liner momentum of the body? Ans:- Let K nd K be the kinetic energies of the object before nd fter. P nd P re the moment before nd fter respectively. rom the dt: mv mv K K mv 00 00 P P P 0 P 0 P P P P 00 00 P 0 P 0 % increse 0 mv 0 5 Clculte the velocity of the bob of simple pendulum t its men position if it is ble to rise to verticl height of 0cm. Given g9.8ms -. Ans: Use the lw of conservtion of mechnicl energy: Loss in K.EGin in P.E. mv mgh v 9.8 0. v.ms 6. One tomic mss unit is equl to.66 x 0-7 kg. Convert this mss into equivlent energy (i) in joule (ii) ev nd (iii) MeV. Ans:- E Δm c. E.66 x0-7 x (3x0 8 )..9 x 0-0 J E (.9 x 0-0 )/(.6 x 0-9 )93 x 0 6 ev 3
E 93 MeV Note:- The ccepted scientific vlue for simple clcultion is.m.u 93 MeV. 7. Wht is the power of n engine which cn lift 0 metric ton of col per hour from mine 0 m deep?(tke g9.8m s -.) Ans:- Work mgh Power time t 0 000 9.8 0 3 Power.09 0 W 3600 Power.09kW 8. An engine of 50kW power is drwing trin of totl mss 5 x0 kg up n incline of in 50. The frictionl resistnce is kg wt per 000kg. Clculte the mximum speed. Given g0ms -.(Note:- in 50 mens sinθ/50) Ans:- The engine hs to pull weight of 5 x 0 sinθ x g nd to overcome the force due to friction The weight 5 0 0 30000N 50 0 riction 5 0 6000N 000 Totl force 36000N...() r r Power. v vcosθ v...() U sin g() nd() v.7ms 9. Two bodies of msses 50g nd 30g moving in the sme direction long the sme stright line with velocities 0.50ms - nd 0.30ms - respectively suffer one dimensionl elstic collision. Clculte their velocities fter collision. Ans:- Let v f nd v f be the finl velocities of the first nd the second body fter collision. Then by usul nottion: m m m 50 30 30 v f v i + vi v f 50 + 30 m + m m + m 50 + 30 50 + 30 v f v v f f 35cms m m m vi + v i m + m m + m 30 50 50 30 + 50 55cms 50 + 30 50 + 30
0. A mss of 0.5kg moving with speed of.5ms - on horizontl smooth surfce, collides with nerly weightless spring of force constnt k50nm -. Determine the mximum compression of the spring. Ans:- We my use work-energy theorem in this cse: Kinetic energy of the mss elstic P.E stored in the spring by compression. mv kx 0.5 (.5) 50* x x x 0.5 (.5) 50 0.05 0.05 0.5m. The potentil energy of kg prticle free to move long the x-xis is given by x V ( x) x J. If the totl mechnicl energy of the prticle is J, clculte the mximum speed of the prticle. Ans:- By the conservtion of mechnicl energy: K.E + P.E Totl M.E or mximum speed the kinetic energy should hve mximum vlue. Then the system should hve minimum potentil energy. or the potentil energy to be minimum, the differentil coefficient of potentil energy function must be equl to zero. Differentiting: 3 x x dv ( x) x x V ( x)...() x dx x ±,Substituting the vlue of x in eqution() The minimum vlue of potentil eenrgy - J. Using the conservtion theorem; K.E + (- ) mv 9 i. e v 3 vmx ms K. E + 9 9 3 x 0 5
Topic:-Newton s lws of motion.. r A force 0 iˆ 6 ˆj + 8kˆ N produces n ccelertion of ms - in body Ans:- of mss m. Clculte the vlue of m. r 0ˆ i 6 ˆj + 8kˆ m 0 + ( 6). + 8.kg.N A bullet of mss 0.0kg moving with speed of 90ms - enters hevy wooden block nd is stopped fter distnce of 60cm. Wht is the verge resistive force exerted by the block on the bullet? Ans:- v u S v u 0 (90) 6750ms S 0.60 Retrdtion 6750ms Retrding orce 0.0 6750 70N 3 A bll of mss kg, moving with velocity of 5ms - strikes wll t n ngle of 5 o nd is reflected t the sme ngle. Clculte the chnge in momentum. Ans:- Method-(By the method of resolution of vectors) P P Cos5 o P Cos5 o P Sin5 o P Sin5 o P Since the collision is elstic mgnitude of P mgnitude of P.P By the resolution of P nd P ; the horizontl components of P nd P will be cncelled nd the chnge in momentum is equl to the sum of the verticl components. Chnge of momentum PSin 5 o. *5*/ 7.07 Ns 6
Method- (By using vector method) P A B - P - P P Chnge in momentum inl momentum Initil momentum P Pˆ Pˆ Δ AB P + P 5 + 5 7. 07Ns A pssenger of mss 7. kg is riding in n elevtor while stnding on pltform scle. Wht does the scle red when the elevtor cb is ) descending with constnt velocity (b) scending with ccelertion 3.0ms -? Ans:- () Wht he feels is the rection of the scle, R mg + m.() Since 0 for constnt velocity, The reding of the sclemg 7. x 9.8 707.56N (b) Using eqution (); The reding of the sclemg+m 7. x 9.8+7. x 3. 938. 6N 5 Two blocks of msses 3kg nd kg re plced in contct with ech other on frictionless tble. ind the force on the common crosssectionl re of contct if force of 5N is pplied on (i) bigger block (ii) smller block. 7
Ans: 5N 3kg R kg R 3kg R kg R 5N (i) (ii) Cse(i) The ccelertion of the system, 5 M + M 3 + ms. R is the contct force; drw the free body digrm for the bigger body. R N 5N M g The force equtions (Newton s equtions) for X nd Y xes: M gn.() ; need not use this eqution since the body does not hve ny upwrd motion. -R M.() R M R 5-3*N Cse(ii) The ccelertion of the system in this cse lso the sme s before, 5 M + M 3 + ms. R is the contct force; drw the free body digrm for the smller body. 8
5N N M g R The force equtions (Newton s equtions) for X nd Y xes: M gn.() ; need not use this eqution since the body does not hve ny upwrd motion. -R M.() R M R 5-*3N 6. Three bodies re connected to ech other with strings s shown in figure. The msses of the bodies re m, 3m nd 5m respectively. These bodies re being pulled with horizontl force on frictionless horizontl surfce. The tension P in the first string is 6N. Clculte (i) ccelertion of the bodies (ii) tension Q in the second string (iii) the force m P 3m Q 5m Ans:- irst find the ccelertion of the system:...() m + 3m + 5m 9m Drw.B.D of mss 3m nd formulte Newton s equtions. 9
P6N N 3m 3mg Q P Q 3m...() Q P 3m 6 3m 9m Q 6 3 3Q 8...() To get the vlue of P drw.b.d of mss m nd formulte the force equtions. m N mg P P m...(3) 6 + m 6 + m 9m 6 +. 9 Q9 +...() ie8 8N rom() 3Q 8 8 30 Q 0N 7. A fire crcker is thrown in horizontl direction with velocity of 50ms -. It explodes into two prts of msses 6kg nd 3kg. The hevier frgment continues to move in the horizontl direction with velocity of 80ms -. Clculte the velocity of the lighter frgment. Ans:- Use the lw of conservtio of liner momentum. The sum of moment before impctthe sum of moment fter impct (M +M )x V M *V + M * V 9*50 6*80 +3* V. V 0ms -. The negtive sign shows tht the lighter frgment moves in the opposite 0
direction with velocity of 0ms -. 8. A body of mss 5kg is hnging by string of negligible weight. A horizontl force is pplied t point of string due to which two prts of the string include n ngle of 50 o. If g0ms -, find the force pplied nd lso the tension in the upper prt of the string. T 50 o 0 o T m T mg Ans:- Since the system is in equilibrium, we cn equte the components. Resolving T in to components: T cos 30 o mg T mg/cos30 o. 5*0/Cos30o. 57.73N T sin30 o 57.73*sin30o.8.87N 9. Aof mss m is resting on nother body of mss M s shown in figure. The coefficient of sttic friction between the two surfces is μ. There is no friction between the mss M nd the horizontl surfce. Wht minmum force should be pplied to M so tht the lod m will just begin to slide long it? m M Ans:- If there is no friction between the blocks, when the lower block is moved forwrd the upper block will fll down verticlly. The friction cn hold the two blocks together under suitble conditions.
When M moves towrds right, the reltive motion of m is towrds left. Hence the direction of friction is towrds right. If force is pplied on the lower block, it is ccelerted. If the frictionl force cn provide n equl force in the sme direction, the upper body will remin t the sme plce on the lower body. The ccelertion of the system, /(M+m)..() orce of friction between m nd M μmg..() The ccelertion of mss m du to the frictionl force μmg/m μg.(3) Equting equtions () & (3): /(M+m) μg (M+m) μg. 0. A block slides down rough inclined plne of inclintion θ with constnt velocity. If this block is projected up the plne with velocity v o, then t wht distnce long the inclined plne, the block will come to rest? Ans:- Since the body slids down with constnt velocity, the frictionl force (in the upwrd direction) is equl to the horizontl component(long the incline) of the weight of the block. N f mgsinθ θ θ mgcosθ mg f mg Sinθ.() Method- When the projected block stops;(apply work-enery theorem for nonconservtive forces) K.E lost P.E gined + Work done ginst friction ½ mv o mgh+s mg sinθ; where S is the distnce long the inclined plne where the body stops. v o gh + gs sinθ () But, h S sinθ Eqution () becomes; V o gs sinθ +gs sinθ gs sinθ S V o /g sinθ
Method- Using Kinmtics nd dynmics. The totl force cting on the body when it is projected up; -mg Sinθ - f -mg Sinθ - mg Sinθ -mg Sinθ; using eqution() Accelertion, /m -g sinθ. We hve v u S i.e 0 v o ( g sinθ) S. S v o /g sinθ 3