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Recent questions tagged telegram-45625

(chatquestion)...Applying colm 0.500x10-m, x10=0+m,v, As it elastic collision v, -0=10-(-10) 3. v, = 20m/s 500 > 0.500x10 -10m, = 20m, = m, = 3 Applying colm during explosion = v=-20i – 20j–10k And separation = umit(:: a = 0) 4. 20/2x2=40/2m, AD = 40 + 20² +10² × 2 = 10/21 Thus AB = CD = 40 3m 5. Energy released in first case = Energy released in second case =-mv +-mv? rec т mv 1+= * mv, = mv т =-mv -mv|1+ 180(20)(300)* (20)v(300)² Thus = 20 1+ = 90/10m/s (300) v(300) =v, 10 %3D 2x10+0= 2V, +4Vc 6. > V, +2V = 10 Given it is elastic collision Vc-V = 10 10 =V = 20/3 m/s,V =- 3 4x5 = 4m /s 4+1 Just after collision V, = 5m/s final level = 7. 8. During any collision KE is not conserved, but momentum is ce Time taken depends on e , not on masses COLM 0.020 u = 0.200 V 9. 10. v ="%0 Sec: Jr. Chaina Page 2 Sri Chaitanya IIT Academy 21-07-19_Sri Chaitanya- Jr.Chaina_JEE-ADV (20 COME : collision afterwards -0.200V? = 0.200×10x 2.5(1-cos53º) v² v? = 50 1-2 = 20

asked Jan 18, 2020 by balaji.thirumalai

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