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Conservation of Momentum and Collision

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Mohammed Asif Name : Roll No. : Topic : Ph : 9391326657, 64606657 DAILY – PRACTISE – PAPER TOPIC: - CONSERVATION OF MOMENTUM AND COLLISIONS 1) A small particle is attached to one of the ends of a light inextensible string of length 2 m, and placed on a smooth horizontal surface. The other end of the string is fixed to a wall. If a velocity v = 10 c m/s be given to the particle, find the final velocity with which the particle moves, after the string becom
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    MohammedAsif Name :Roll No. :Topic :Ph : 9391326657,64606657 DAILY – PRACTISE – PAPER TOPIC: - CONSERVATION OF MOMENTUM AND COLLISIONS 1)A small particle is attached to one of the ends of alight inextensible string of length 2 m, and placedon a smooth horizontal surface. The other end of the string is fixed to a wall. If a velocity v = 10 cm/s be given to the particle, find the final velocitywith which the particle moves, after the stringbecomes taut. l = 1m v = 10 2)Two particles A and B of mass 100 g and 200 g areattached to the two ends of a string of length 5 m,resting on a smooth floor. They are separated by adistance of 3 m. A is given a velocity of v = 20 cm/s,along a direction normal to AB. Determine the velocitiesof A and B just after the string becomes taut?   A B V3 m 3)A particle (a mud pallet, say) of mass M strikes asmooth stationary wedge of mass M with a velocity V 0 , at an angle θ with horizontal. If collision is perfectlyinelastic, find thea) velocity of the wedge just after the collisionb) Change in K.E of the system (M+m) in collision   mM θ 0 V 4)Two particles of masses 12 mandmareconnected by a light and inextensible stringwhich passes over a fixed pulley. Initially, theparticle 1 mmoves with a velocity 0 vwhen thestring is not taut. Neglecting friction in allcontacting surfaces, find the velocities of theparticles 12 mandmjust after the string is taut. 0 V 1 m 2 m x Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065.Ph: 040 – 64606657, 9391326657.www.asifiitphysics.vriti.com1  5)Three identical particles A, B and C lie on a smoothhorizontal table. Light in extensible strings which are just taut connect AB and BC and 0 ABCis135 ∠ . Animpulse J is applied to the particle C in the directionBC. Find the initial speed of eachparticle. The mass of each is m. AB CJ    135    0 6)In the movie Aajab Prem ki Ghazab kahani, Say Ranbir kapoor and Katrinakaif each weighing 40 kg are sitting on a friction less platform somedistance d apart. Ranbir rolls a ball of mass 4 kg on the platform towardsKatrina kaif of course she catches it. Then Katrina rolls the ball towardsranbir and he catches it . The ball keeps on moving back and forth betweenranbir and Katrina kaif. The ball has fixed speed of 5 m/s on the platform.a) Find the speed of Ranbir kapoor after he rolls the ball for the first time.b) Find the speed of Ranbir kapoor after he catches the ball for the firsttimec) Find the speed of Ranbir kapoor and Katrina kaif after the ball has made5 round tripsand is held Ranbir kapoor.d) How many times can Ranbir roll the balle) What is the centre of mass of the system (Ranbir + Katrina + ball) at theend of the n th trip. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065.Ph: 040 – 64606657, 9391326657.www.asifiitphysics.vriti.com2  SOLUTIONS 1)  1 m2 mV θ θ Vcos θ Vsin θ 0 1sin230 θ=θ=  The Component normal to the string 0 1Vsin10sin30105cm/s2 θ = × = × = 2) V θ Vcos θ Vsin θ 3 m ABJ V 1 J 1 1 1 4vcos 20 16cm/ s53vsin 20 12cm/ s5m vcos J m v θ = × =θ = × =∴ θ − = ( ) 1 10016J100v1 × − = → ( ) 211 andJmv200v2 = = → Adding (1) and (2) 1 10016v5.33cm/s300 ×= =  The component of velocityvsin θ of particle A remains unchanged. ( ) ( ) 2 2 final velocity of A5.33 1213.13cm/ s ∴= += And final velocity of B = 5.33 cm/s3) MV Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065.Ph: 040 – 64606657, 9391326657.www.asifiitphysics.vriti.com3  Let the system (M+m) moves as a single mass with a velocity V.Conserving the momentum of the system in horizontal, we have ( ) v00 mvcosMmmvcosVMm θ = +θ=+ ( )( ) 2 202020220 1 1(b) k M m v mv2 2mv cos1 1M m mv2 M m 2mv m1 cos2 M m ∆ = + −θ = + − +   =− − θ  +   ( ) ( ) 220 Mmsinmvk2Mm + θ∆ =+ 4)During the impact, a huge tension develops. The impulse of tension 1  Tisequal to the change in momentum of  1 mwhich is given as ( ) ( ) 110  T.dtmvv1 − = − → ∫  Similarly the impulse of tension 2  Tis equalto the change in momentum of  2 mwhich isgiven as  ( ) 2 21 2  T .dt m vSinceT T − = −= ∫    ( ) 1 0 21 01 2 m v v m Vm vVm m ∴ − ==+   1 m 2 m V 2  T dt ∫   Tdt ∫   Tdt ∫  1  T dt ∫  5)The external impulse applied to C causes both strings to jerk exertinginternal impulses 12  JandJ. AB C uu 1 u 2 v 1 A BC J 1 J 2 45 0      J 1      J 2 J From constant relation obviously ( ) 21 uu1 = → For A ( ) 22  Jmu2 = → For B ( )( ) 0111011  Jcos45Jmu3 Jsin45mv4 − = →= → Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065.Ph: 040 – 64606657, 9391326657.www.asifiitphysics.vriti.com4
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