Tiêu đề 04. MOtion in a plane Hard Ans.pdf ✅

1. (c) Body moves horizontally with constant initial velocity 3 m/s upto 4 seconds  x = ut = 3  4 = 12 m and in perpendicular direction it moves under the effect of constant force with zero initial velocity upto 4 seconds.  2 ( ) 2 1 y = ut + a t 2 2 1 0 t m F       = + 2 4 2 4 2 1       = = 16 m So its distance from O is given by 2 2 2 2 d = x + y = (12) + (16)  d = 20 m 2. (c) Displacement along X- axis : 2 2 1 x u t a t = x + x 6 (4) 48 m 2 1 2 =   = Displacement along Y- axis : y uy t ay t 8 (4) 64 m 2 1 2 1 2 2 = + =   = Total distance from the origin x y (48) (64) 80 m 2 2 2 2 = + = + = 3. (b) At half of the time of flight, the position of the projectile will be at the highest point of the parabola and at that position particle possess horizontal component of velocity only. Given uvertical = u sin  = 80  u m s o 160 / sin 30 80 = =  u u cos 160 cos 30 80 3 m / s. o horizontal =  = = 4. (b) Horizontal velocity at point 'O' = u cos Horizontal velocity at point ' P' = v sin In projectile motion horizontal component of velocity remains constant throughout the motion  v sin = u cos  v = u cot 5. (b) Both particle collide at the highest point it means the vertical distance travelled by both the particle will be equal, i.e. the vertical component of velocity of both particle will be equal 1 30 2 u sin  = u  2 1 2 u u =  u1 = 2u2 6. (b) Let in 2 sec body reaches upto point A and after one more sec upto point B. Total time of ascent for a body is given 3 sec i.e. 3 sin = = g u t   u sin = 10  3 = 30 .....(i) Horizontal component of velocity remains always constant u cos = v cos 30 .....(ii) For vertical upward motion between point O and A v sin 30 = u sin − g  2 o  Using v = u − g t sin 30 = 30 − 20 o v As u sin  = 30   v = 20 m / s. Substituting this value in equation (ii) o u cos = 20 cos 30 = 10 3 .....(iii) From equation (i) and (iii) u = 20 3 and  = 60 7. (d) At point N angle of projection of the body will be 450 . Let velocity of projection at this point is v. If the body just manages to cross the well then Range = Diameter of well u O  v B A 30o u cos 90o u v  O P v sin 90o –  u sin  u cos x O y E F u d
40 sin 2 2 = g v  As  = 45 400 2 v =  v = 20 m / s But we have to calculate the velocity (u) of the body at point M. For motion along the inclined plane (from M to N) Final velocity (v) = 20 m/s, acceleration (a) = – g sin = – g sin 45o , distance of inclined plane (s) = 20 2 m .20 2 2 (20) 2 2 2 g = u − [Using v 2 = u 2 + 2as] 20 400 2 2 u = +  u = 20 2 m /s. 8. (d) Horizontal momentum remains always constant So change in vertical momentum ( p  ) = Final vertical momentum – Initial vertical momentum = 0 − mu sin  o | P | = 0.1  20  sin 30 = 1 kg m / sec . 9. (a) Both masses will collide at the highest point of their trajectory with equal and opposite momentum. So net momentum of the system will be zero. 10. (b) g m u L 2 cos sin 3 2   = = (4 2 ) 3 g mv [As  = 45o ] 11. (b) Let 1 x and 2 x are the horizontal distances travelled by particle A and B respectively in time t. t u x = . cos 30  3 1 .....(i) and x u t o 2 = cos 60  ......(ii) t u t ut u x x o o + = . cos 30  + cos 60  = 3 1 2  x = ut  t = x / u 12. (b) When body projected with initial velocity u by making angle  with the horizontal. Then after time t, (at point P) it’s direction is perpendicular to u . Magnitude of velocity at point P is given by v = u cot . (from sample problem no. 9) For vertical motion : Initial velocity (at point O) = u sin  Final velocity (at point P) = −v cos = −u cot cos Time of flight (from point O to P) = t Applying first equation of motion v = u − g t −u cot cos =u sin − g t  g u u t sin + cot cos =     2 2 sin cos sin = + g u g u cosec = 13. (c) Formula for calculation of time to reach the body on the ground from the tower of height ‘h’ (If it is thrown vertically up with velocity u) is given by         = + + 2 2 1 1 u gh g u t So we can resolve the given velocity in vertical direction and can apply the above formula. Initial vertical component of velocity u sin = 50 sin 30 = 25 m /s.            = + + 2 (25) 2 9.8 70 1 1 9.8 25 t = 6.33 sec. 14. (c) 30o u 70 m u sin30o 90o u P  O v (90 – ) v cos u sin  u cos 40 m 45o M R N u v
g u R sin 2 2 = and g u T 2 sin = 2  R  u and T  u (If  and g are constant). In the given condition to make range double, velocity must be increased upto 2 times that of previous value. So automatically time of flight will becomes 2 times. 15. (c) For first particles angle of projection from the horizontal is . So g u T 2 sin 1 = For second particle angle of projection from the vertical is . it mean from the horizontal is (90 −). g u T 2 sin (90 ) 2 −  = g 2u cos = . So ratio of time of flight tan  2 1 = T T . 16. (d) Range  horizontal component of velocity. Graph 4 shows maximum range, so football possess maximum horizontal velocity in this case. 17. (b) g u R sin 2 2 =  R  1 / g Moon Earth Earth Moon g g R R = = 6       g Moon = g Earth 6 1  R Moon = 6 REarth = 6R 18. (c) For maximum horizontal Range  = 45 From R = 4 H cot = 4H[As  = 45o , for maximum range.] Speed of the particle will be minimum at the highest point of parabola. So the co-ordinate of the highest point will be (R/2, R/4) 19. (a) Velocity at the highest point is given by 2 cos u u  = (given)   = 45o Horizontal range g u g u g u R 2 2 o 2 sin 2 sin(2 45 ) =  = =  20. (b) The maximum area will be equal to area of the circle with radius equal to the maximum range of projectile Maximum area ( ) 2 max 2 r =  R 2 2         = g u  2 4 g u =  [As r R u / g 2 = max = for  = 450 ] 21. (b) Initial velocity ( i J)m /s ˆ 8 ˆ = 6 + (given) Magnitude of velocity of projection 2 2 u = ux + uy 2 2 = 6 + 8 = 10 m/s Angle of projection 6 8 tan = = x y u u  3 4 =  5 4 sin = and 5 3 cos = Now horizontal range g u R sin 2 2 = 10 5 3 5 4 (10) 2 2    = = 9.6 meter 22. (c) g u sin 2 R 2  = sin 2 2  R  u                 = 1 2 2 1 2 1 2 sin 2 sin 2   u u R R 1 2 2 1 8 sin 30 2 sin 90 R u u R R o o =                = 23. (a) If the velocity of projection is u then at the highest point body posses only u cos 2 cos u u  = (given)  = 60 Now g u R sin(2 60 ) 2   = g u 2 2 3 = 24. (d) m r F 45o Y X R/2 R/4
We know g u u R 2 x y = 9.8 2  9.8  19.6 = = 39.2 m Where u x = horizontal component of initial velocity, uy = vertical component of initial velocity. 25. (a) We know R = 4 H cot 2H = 4H cot  2 1 cot = ; 5 2 sin = ; 5 1 cos = As R = 2H given  g u . 2.sin . cos Range 2 = g u 5 1 . 5 2 2 2 = g u 5 4 2 = 26. (d) For equal ranges body should be projected with angle  or (90 −) o from the horizontal. And for these angles : g u h 2 sin 2 2 1  = and g u h 2 cos 2 2 2  = by multiplication of both height : 2 2 2 2 1 2 4 sin cos g u h h   = 2 2 sin 2 16 1         = g u   2 16h1 h2 = R  R = 4 h1h2 27. (c) Horizontal distance travelled by grasshopper will be maximum for  = 45 m g u R 1.6 2 max = =  u = 4 m / s. Horizontal component of velocity of grasshopper u cos = 4 cos 45 = 2 2 m /s Total distance covered by it in10 sec. S = u cos  t = 2 2 10 = 20 2 m 28. (c) Angle of projection a b v v x 1 y 1 tan tan − −  = =  a b tan  = ...(i) From formula R = 4H cot = 2H  2 1 cot =  tan  = 2 ...(ii) [As R = 2H given] From equation (i) and (ii) b = 2a 29. (c) sec g u T 2 2 sin = =  (given) u sin = 10 Now 5 . 2 10 (10) 2 sin 2 2 2 m g u H =  = =  30. (d) When two stones are projected with same velocity then for complementary angles  and (90o – ) Ratio of maximum heights :  2 2 1 = tan H H 3 3 tan 2 = =   3 3 1 2 H Y H = = 31. (c) g u H 2 sin 2 2  = 2 H  u [As  = constant] If initial velocity of a projectile be doubled then H will becomes 4 times. 32. (b) Time of flight for the ball thrown by Pankaj g u T 1 1 2 = Time of flight for the ball thrown by Sudhir g u T o 2 sin(90 ) 2 2 − = According to problem T1 = T2  g u g 2u1 2 2 cos =  u1 = u2 cos Height of the ball thrown by Pankaj g u H 2 2 1 1 = Height of the ball thrown by Sudhir g u H o 2 sin (90 ) 2 2 2 2 − = g u 2 cos 2 2 2  = 16m Short Trick : Maximum height H  T 2 2 2 1 2 1         = T T H H  1 2 1 = H H (As T1 = T2) m u1 2m u2 90o –   1.6 m