Title 04. MOtion in a plane Medium Ans.pdf ✅

1. (b) Here, Velocity of water flowing in river, 1 vr 3mms− = Velocity of swimmer in still water, 1 vs 4ms− = Form figure, The resultant velocity of the swimmer is 2 2 2 1 r 2 v vs v (4) (3) 25 5 ms− = + = + = 2. (d) Observed speed, v (10) (6) 136 2 2 = + = 1 11ms− = 3. (a) It is an example of projectile motions. Therefore, the trajectory of the bomb is parabola. 4. (c) At the highest point, velocity is acting horizontally and accelerations (= accelerations due to gravity) is acting vertically downwards. Therefore, at the highest point the angle between velocity and accelerations is 90 . 5. (b) Let u be the velocity of projections, then 100m g u R 2 max = = Or u 100g or u 100g 2 = = For upward throw of the ball, we have, u = 100g,v = 0,a = −g As v u 2as or 0 100g 2( g)s 2 2 − = − = − 50m 2g 100g s = − −  = 6. (a) Horizontal velocity of a projectile is not affected by gravity. 7. (c) Time taken by the bomb to fall through a height of 490 m 10s 9.8 2 490 g 2h t =  = = Distance at which the bomb strikes the ground = horizontal velocity × time = 360kmh 10s 1  − h 1km 3600 10 360km h 1 =  = − 8. (c) For a projectile launched with velocity u at an angle , the horizontal range is given by g u sin 2 R 2  = (i) For1 = 45 −  g u cos2 g u sin(90 2 ) R 2 2 1  =  −  = (ii) For  = +  0 2 45 1 R R R g u cos2 g u sin(90 2 ) R 1 2 1 2 2 2 =  =  =  +  = 9. (b) Here, 1 u 56ms− = Let  be the angle of projections with the horizontal to have maximum range, with maximum height = 40 m Maximum height, 2g u sin H 2 2  = 2 9.8 (56) sin 40 2 2   = or 4 1 (56) 2 9.8 40 sin 2 2 =    = 2 1 Sin =  =        = − 30 2 1 sin 1 10. (a) The horizontal range is the same when angle of projections is or(90 − ). 11. (d) 2g u sin H 2 2 1  = ..... (i) 2g u cos 2g u sin (90 ) H 2 2 2 2 2  =  −  = ..... (ii) Divide (i) by (ii), we get 1 tan H H 2 2 1  = 12. (c) Maximum height, 2g u sin H 2 2  = ..... (i) Horizontal range, g 2u sin cos g u sin2 R 2 2   =  = ..(ii) Divide (i) by (ii), we get 4 tan R H  = or        =  = − R 4H or tan R 4H tan 1 13. (a) Time to reach maximum height m = t Time to reach back to ground m = t Total time of flight, f m m m T = t + t = 2t
14. (d) In the presence of air resistance, the range, maximum height, speed at which the projectile strikes the ground will decrease whereas the angle at which the projectile strikes the ground will increase. 15. (d) Let u1 and u2 be the initial velocities of the two particles and 1 and 2 be their angles of projections with the horizontal. The velocities of the two particles after time t are j ˆ i (u sin gt) ˆ v (u cos ) 1 1 1 1 1 =  +  − And j ˆ i (u sin gt) ˆ v (u cos ) 2 2 2 2 2 =  +  − Their relative velocity is 1 2 v12 = v − v  j ˆ i (u sin u sin ) ˆ (u cos u cos ) = 1 1 − 2 2 + 1 1 − 2 2 Which is a constant. So the path followed by one, as seen by the other is straight line, making a constant angle with the horizontal. 16. (d) Centripetal accelerations, R v a 2 c = Where v is the speed of an object and R is the radius of the circle. It is always directed towards the centre of the circle. Since v and R are constant for a given uniform circular motions, therefore the magnitude of centripetal accelerations is also constant. However, the direction of centripetal accelerations changes continuously. Therefore, a centripetal acceleration is not a constant vector. 17. (b) Here, v = 27 km 1 1 ms 18 5 h 27 − − =  1 1 ms 7.5ms 2 15 v − − = = R = 80m Centripetal accelerations, r v a 2 x = 2 1 2 c 0.7ms 80m (7.5ms ) a − − = = Tangential acceleration, 2 a1 0.5ms− = Magnitude of the net accelerations is 2 2 2 2 t 2 a (ac ) (a ) (0.7) (0.5) 0.86ms− = + = + = 18. (c) According to cosine formula 2 2 2 2 2r r r x cos60 + −  = 2 2 2 2r cos60 = 2r − x x 2r 2r cos60 2r [1 cos60 ] 2 2 2 2 = −  = −  2 2 2 = 2r [2sin 30] = r x = r Displacement AB = x = r 19. (b) In a uniform circular motion, the accelerations is directions towards the centre while velocity is acting tangentially. 20. (d) We known that scalar product is cyclic, (B A),A = (AB).A (A A).B = 0.B = 0 21. (a) B) A ˆ ˆ B).( ˆ A. ˆ B ( A ˆ ˆ 2 − = − B ˆ B. B ˆ A ˆ ˆ A. A ˆ ˆ A. ˆ = − − + B 1 ˆ A. B ˆ ˆ A. ˆ =1− − + = 2 − 2cos = 2(1− cos) 2 4sin 2 2 2sin2 2   =       = 2 B 2sin A ˆ ˆ  − = 22. (b) Given j ˆ i B ˆ ˆ j, ˆ i A ˆ = + = − A (1) (1) 2 2 2  = + = and B (1) ( 1) 2 2 2 − + − = Let  be angle between the vector A  and B . Then according to definition of scalar product (or dot product) A.B = A B cos Or 0 ( 2)( 2) j) ˆ i ˆ j).( ˆ i ˆ ( A B A.B cos = + −  =  = =  − cos (0) 90 1 23. (d) (a) The statement is false, It is because energy (scalar quantity) is not conserved during inelastic collision. (b) The statement is false, It is because the temperature (scalar quantity) can be negative. (c) The statement is false. It is because gravitational potential energy (scalar quantity) vary from point to point to point in space. (d) The statement is true because the value of scalar does not change with orientation of axes.
24. (b) As per figure, in j, ˆ i b ˆ u = a + both a and b are Positive while in j, ˆ i q ˆ v = p + p is positive and q is negative. Thus a, b and p are positive and q is negative. 25. (a) Give : k ˆ j 5 ˆ i 2t ˆ r 3t 2 = + + Velocity, 2 1 jms ˆ i 4t ˆ k) 3 ˆ j 5 ˆ i 2t ˆ (3t dt d dt dr v − = = + + = +   At t = 1 s, 1 jms ˆ i 4 ˆ v 3 − = +  Let  be angle which the directions of v makes with the x- axis, Then  =        = =  = − 53 3 4 or tan 3 4 v v tan 1 x y 26. (c) The path followed by the motorist is a regular hexagon ABCDEF of side length 500m as shown in the figure. Let the motorist starts from A and taken the third turn at D. Therefore, the magnitude of this displacement is AD = AG + GD = 50 m + 500 m = 1000 m 27. (b) The relation (b) is true, other are false because relations (a), (c) and (d) hold only for uniform accelerated motions. 28. (a) Here, 1 u 720kmh − = 1 1 ms 200ms 18 5 720 − − =  = H = 1.5km =1.51000m =1500m Time taken by the bomb to attack the target 300s 10 3s 10ms 2 1500m g 2H t 1 = =  = = − R u t 200ms 10 3s 2000 3m 1 =  =  = − From Figure, Angle of sight (w.r.t. horizontal)        = − T H tan 1          =       =       = − − − 4 3 tan 4 3 3 tan 2000 3 1500 tan 1 1 1  =  =  − tan (0.43)or 23 1 29. (c) Displacement of each girl = PQ Magnitude of displacement for each girl = PQ = 2 200m = 4000m For girl C, displacement = actual length of path 30. (c) The positions of the particle is given by 2 0 0 at 2 1 r = r + v t + Where, r0 is the position vector at t = 0 v0 is the velocity at t = 0 Here, 1 2 0 0 jms ˆ i 2 ˆ ims ,a 3 ˆ r 0,v 5 − − = = = + j ˆ i 1t ˆ j)t (5t 1.5t ) ˆ i 2 ˆ (3 2 1 i ˆ r 5t 2 2 2  = + + = + + Compare it with j, ˆ i y ˆ r = x + we get 2 2 x = 5t +1.5t ,y =1t Given : x = 84 m 2 84 = 5t −1.5t 1.5t 5t 84 0 2 + − = 3t 10t 168 0 2 + − = On solving, we get t = 6s At t = 6s, y = (1) (6) 36m 2 = 2 0 0 at 2 1 r = r + v t + 31. (c) Here, Velocity of swimmer in still water, 1 vs 10ms− = Velocity of water flowing in river, 1 vr 5ms− = From figure, 2 1 10 5 v v sin s r  = = =  =        = − 30 2 1 sin 1 west of north Alternative solutions 2 2 2 1 r 2 v vs v (10) (5) 5 3ms− = − = − = 3 1 5 3 5 v v tan r  = = =
Or =           = − 30 3 1 tan 1 west of north 32. (b) In figures, vr represent the velocity of rain and vb , the velocity of the bicycle, the woman s riding. To protect herself from rain, the woman should hold her umbrella in the directions or velocity of rain with respect to the bicycle, vrb .        = = =  = − 5 2 or tan 5 2 20 12 v v tan 1 r b Therefore, the woman should hold her umbrella at an angle of      −  5 2 tan 1 with the vertical towards the west. 33. (a) Assume north to be i ˆ directions and vertically upwards to be j ˆ directions Let the velocity of rain be j ˆ i b ˆ v = a +  In the first case, g v  = velocity of girl = i ˆ 5 j ˆ i b ˆ i (a 5) ˆ j) 5 ˆ i b ˆ vgr = vr − vg = (a + − = − + Since rain appears to fall vertically downwards.  a – 5 = 0 or a = 5 In the second case i ˆ vg =10  j ˆ i b ˆ i (a 10) ˆ j) 10 ˆ i b ˆ i (a ˆ vgr = vr −10 = + − = − + j ˆ i b ˆ = −5 + Since rain appears to fall at 45 with the vertical  b = -5  The velocity of the rain is j ˆ i 5 ˆ vr = 5 − Speed of the rain is 2 2 1 vr (5) ( 5) 5 2ms− = + − = 34. (a) Velocity of object A relative to that of B is vAB = vB − vA vAB = −vBA And v AB = vBA 35. (a) Let u be the velocity of projection of the ball. The ball will cover maximum horizontal distance when angle of projections with horizontal , Then g u R 2 max = Here, Rmax =100m 100m g u 2  = .....(1) As v u 2as 2 2 − = Here, v = 0 (At highest point velocity is zero) a = - g, s = H  50m 2 100 2g u H 2 = = = 36. (c) Take the x – axis along the incline and y – axis perpendicular to the plane. us = ucos uy = usin a x = −gsin a y = −gcos When the particle lands at P its y coordinate becomes zero.  2 y y a t 2 1 0 = u t + 2 gcos t 2 1 0 = usint −  Or   = g cos 2u sin t ..... (i) For motions along inclined plane 2 gcos t 2 1 x = usint −  2 gsin t 2 1 L = u cost =  Substituting the value of t form Eq. (i) we get           −            =  gcos 2usin gsin 2 1 gcos 2usin L u cos    −    = 2 2 2 2 gcos 2u sin sin gcos 2u sin cos