Unit 2 Motion in a Straight Line PDF

UNIT -2 MOTION IN A STRAIGHT LINE 55 v(m/s) AN 10}----. Ane i f ' } > t(s) 10...

UNIT -2 MOTION IN A STRAIGHT LINE 55 v(m/s) AN 10}----. Ane i f ' } > t(s) 10 20 #30 ©6440 * -50 Fig 2.32 V -t graph of example - 47 b. a=0, fromt = 10stot=20s; and fromt = 30stot=40s a is positive fromt=Otot= 10s a is negative fromt = 20s tot = 30 s; and fromt = 40s tot =45 5 Sree Ag SR A A Cee Te Be Eas preys 2.9 Free Ball | oe ea ag oe rea Objects released from rest or thrown upward or downward that move under the influence of gravity alone are said to be in free fall. If you carefully observe the motion of a ball which is dropped from a small height above the surface of the earth, you would sce that the distance covered per unit time increases as the ball descends. This indicates that a free fall is an accelerated motion. The acceleration of a freely falling body is called acceleration due to gravity, denoted by a small letter g. When air resistance is neglected all bodies irrespective of their mass and shape have the same acceleration when in free fall, Near the surface of the earth the gravitational acceleration is equal to 9.8m/s*. This fact will indicate that a free fall is a uniformly accelerated rectilinear motion. Thus, it is governed by the equations of a uniformly accelerated motion, except the acceleration is changed by an acceleration due to gravity g. Equations of a freely falling body 1. ¥ =u+gt 2.16 2..V., # os i7 3 y= (22) 2.18 4 yaut+ > gr’ 2.19 Sv? =u? + 269. 2.20 Scanned with AltaScanner Scanned with Scanned with AltaScanner  UNIT -2._ MOTION IN A STRAIGHT LINE —— oe Solution:- Since the second stone is thrown 1. 0 second later, the time of free fall for the first ball is t; = 3. Os + 1. Os = 4. Os and for the second ball t2 = 3. Os. For both balls i = 20m/s up, and g = 9.8m/s* down. Then their corresponding height is y, =tt, + : Bt, = 20m/s (4. Os) + = 80m - 78.4m= 1. 6mup y, = tit, + ; gt,” = 20m/s (3. 0s) + ; (- 9.8m/s” ) (3. Os) = 60 m - 44.1 m= 15. 9m up Thus the two stones are y2 - y; = 15. 9m up - 1.6 mup = 14. 3 m apart. Example — 54. A ball player catches a ball 3.0 s after throwing it vertically upward (a) with what speed did he throw it, and (b) what height did it rise? Solution:- a) From problem 52 (d), we have seen that the total time the ball remains in air is given by lio = = a — egt,, > (-9.8m/s?) (3.08) 5h Sees i= - ; =. ek tie =.7m/sup b) y= oy , but at the maximum height ¥ =0 0-(14.7m/s) _ = 216, 09m? /s? Ymax = a te ag - =11.025m 2 (- 9.8m/s* ) -19.6 m s° 2.10 Relatiye Velocity in One Dimension «7. ").. In this secuion we will discuss of relative velocity of two objects Moving in a straight-line in the same or in opposite directions. When the velocity of body A is described with respect to another body, body B, which is also moving it is termed as the velocity of A relative to B, written as Vin Scanned with AltaScanner Scanned with Scanned with AltaScanner  UNIT -2 MOTION INA STRAIGHT LINE 57 —— —— Example - 50, The astronaut on the moon drops a spanner. It takes 1.7s to fall a distance of 2.3 m. Determine the acceleration due to gravity on the moon. Given: u=0,t=1.7s, y=2.3m Solution:- I 2 y =ut+ —> £ pt Since u =0 y = —7I g gt 2 9] 2 (2.3m g=— = ( x 4.6m = 1. 59m/s? t (1.7s) 2. 89s” Example —- $1. An object is dropped from a helicopter at a height of 45 m above the ground. With what velocity will the object strike the ground if (a) the helicopter is at rest? (b) the helicopter is moving vertically down with a velocity of 4m/s? (c) the helicopter is moving vertically up with a velocity 4m/s?(Take g = 10m/s? ) Solution:- a) wu =0, y=45m down, g = 10m/s? down v? =a? +2g¥=0+2 (10m/s”) (45m) = 900m?/s? i= 900m? /s? = 30m/s down b) wd = 4m/s down, #= 10m/s* down, ¥ = 45m down V=u? + 2gy = j(4m/sy +2 (10m/s? )(45m) =/916m?/s? Vv = 30.3 m/s down 4 Sass, c) u=4m/s up, g = 10m/s? down, y = 45m down ow Taking up as positive and down as negative V= fu?+2gy = |(4/ms)° +2(-10m/s2) (45m) V = 916 m/s = 30.3 m/s down Why is the answer for question (b) and (c) the same? Scanned with AltaScanner Scanned with Scanned with AltaScanner  56 ALPHA PROBLEM SOLVER SERIES PHYSic. Note that the above equations (eqn 2.16 - 2.20) are vector equations Therefore, we should take care of the directions of each of the vecto, quantities U, Vv, § andy. The direction of i is the direction of the initia) velocity, which in most cases is taken as a reference direction. The direction of g 1s always downward towards the surface of the earth (negative y-axis) Example — 48. Suppose an imaginary speedometer is attached to a rock. What would it read after 3.5s (a) if the rock falls froma position of res} (b) if it is thrown downward: and (c) thrown upward with an initia velocity of Sm/s? Solution:- (a) i = 0, § = 9.8m/s? down, t = 3. 5s ¥ =u+ gt =0+9.8m/s’ (3.5s) down v = 34,3m/s down (b) u = 5m/s down, g = 9.8m/s?_ down, t = 3. 5s =u+gt=Sm/s + 9.8m/s* (3. Ss) = 39. 3m/s downwards