Tiêu đề Notes_Objective_03.Kinematics.pdf ✅

www.thinkiit.in Page 1 thinkIIT OBJECTIVE - I 1. At motor car is going due north at a speed of 50 km/h. It makes a 90o left turn without changing the speed. The change in the velocity of the car is about (A) 50 km/h towards west (B*) 70 km/h towards south-west (C) 70 km/h towards north-west (D) zero Sol. B Change in velocity of the car = Vf – Vi =     2 2 50  50 towards south-west = 50 2 km/h = 70 km/h towards south-west 2. Fig. shows the displacement time graph of a particle moving on the X-axis. (A) the particle is continuously going in positive x direction (B) the particle is at rest (C) the velocity increases up to a time to , and then becomes constant. (D*) the particle moves at a constant velocity up to a time to , and then stops. Sol. D Slope of "x-t" graph gives velocity Here (t = 0 to t = t0 ) slope is constant so we can say that velocity is constant. Than aftrer to, slope is zero. So velocity is zero. So velocity is zero. Þ velocity zero means particle stops. 3. A particle has a velocity u towards east at t = 0. Its acceleration is towards west and is constant, Let xA and xB be the magnitude of displacements in the first 10 seconds and the next 10 seconds. (A) xA < xB (B) xA = xB (C) xA > xB (D*) the information is insufficient to decide the relation of xA with xB . Sol. D W ¬ ® E ¬ ® a= constant u V= u + at Some time after velocity is zero O = u – at t = u/a then after particle is return back for this information, we not decided that particle is turn back before 10 sec or after 10
www.thinkiit.in Page 2 thinkIIT sec. 4. A person travelling on a straight line moves with a uniform velocity u1 for some time and with uniform velocity u2 for the next equal time. The average velocity u is given by (A*)     1 2 2 (B) u = 1 2 (C) 2 1 1  2 2   (D) 1 1 1  2 2   Sol. A S1 = u1 t S2 = u2 t Total displacement = S Þ S1 + S2 S = u1 t + u2 t = (u1 + u2 ) t Total time = t + t = 2t average velocity = Total time taken Total displacement =   2t u u t 1  2 = 2 u1  u2 5. A person travelling on a straight line moves with a uniform velocity u1 for a distance x and with a uniform velocity u2 for the next equal distance. The average velocity u is given by (A)     1 2 2 (B) u = 1 2 (C*) 2 2 2 1 1      (D) 2 2 1 1 1      Sol. Total displacement = 2x Total time taken = t1 + t2 = 1 v2x v x  average velocity = Total time taken Total displacement = 1 v2x v x 2x  v = 1 v21 v 1 2  Þ v 2 = 1 v21 v 1  6. A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is (A) a upward (B) (g-a) upward (C) (g-a) downward (D*) g downward Sol. D Stone is released from an elevator The acceleration of the stone after the release is aSe = -g = g downward 7. A person standing neat the edge of the top of a building throws two balls A and B. The ball A is thrown vertically downward with the same speed. The ball A hits the ground with a speed uA and the ball B hits the ground with a speed uB . We have : (A) u A > u B (B) u A < u B (*C) u A = u B (D) the relation between A and B depends on height of the building above the ground. Sol. C After come back at level 1 velocity of the ball 'A' is same and equal to ball 'B'. v 2 = u2 + 2as Ball A + Ball B travels the same height so we can say that
www.thinkiit.in Page 3 thinkIIT uB = uA 8. In a projectile motion the velocity (A) is always perpendicular to the acceleration (B) is never perpendicular to the acceleration (C*) is perpendicular to the acceleration for one instant only (D) is perpendicular to the acceleration for two instants. Sol. C In a projectile motion at maximum height, y component of the velocity is zero. Only instant velocity is perpendicular to the acceleration. 9. Two bullets are fired simultaneously, horizontally and with different speed from the same place. Which bullet will hit the ground first? (A) the faster one (B) the slower one (C*) both will reach simultaneously (D) depend on the masses. Sol. C Both bullet will hit the ground simultaneously because downward acceleration of the both bullet is same & equal to g. t fired at same place. S = ut + 1/2 at2 Sy = uy + 1/2 ay + t2 Sy = O + 1/2 gt2 t = g 2Sy . 10. The range of a projectile fired at an angle of 15o is 50 m. If it is fired with the same speed at an angle of 45o , its range will be (A) 25 m (B) 37 m (C) 50 m (D*) 100 m Sol. D R = g u sin 2 2  50 = g u sin 30o 2 .......... (1) g u 2 = 100 .......... (2) Given Þ fired with the same speed atom angle of 45o R' = g u sin 2 45o 2  = g u 2 sin90o {Q sin90o = 1}