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NCERT/CBSE PHYSICS CLASS 11 textbook http://www.TutorBreeze.com Contact for Online Tutoring in Physics, Math, Chemistry, English Solutions/Answers to NCERT/CBSE PHYSICS Class 11(Class XI)textbook Exercise and Additional exercise CHAPTER THREE MOTION IN A STRAIGHT LINE EXERCISES 3.12 A ball is dropped from a height of 90 m on a floor. At each collision with the floor,the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Q. 3.12Initial velocity, u = 0 Highest from which ball is dropped, h = 90 m Loss of its speed after every collision = velocity of hitting – its 10% acceleration = g = 9.8m/s2 Sign of acceleration = positive During down ward motion:position = positive velocity = positive acceleration = positive (d)
initial velocity of ball = 9.4 m/s initial velocity of ball = 0 acceleration = -- 9.8 m/s2
So using formula
v 2 − u2 = 2as 2
= O2 − ( 29.4 ) = 2 × ( −9.8 ) s s = +44.1 Maximum height of the ball = 44.1 m. Using formula V = u + at 0 = 29.4 − 9.8 × t t = 3s. ©TutorBreeze.com Please do not copy the answer given here Write to us for help in understanding the solution
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Ball takes the same time to come back, total time = 6 s. If V1 be time taken in first hitting the floor, then using equations V 2 − u2 = 2as----( i ) 1 s = ut + at 2 ----- ( ii ) 2
We get v12 − 0 = 2gt or
and
v1 = 2gh
or
= 2 × 9.8 × 90
h=
1 2 gt1 2
t1 =
or
2g h
2 × 90 30 = = 4.28s 9.8 7
= 42 m/s If v2 be the velocity of its first rise i.e., first rebounce, v2 = v1 ----10% of v1 = 42 – 4.2 = 37.8 m/s If h1, be the height reached after first rebounce then at h = h1 V3 = 0, V2 = U ∴ From equations (i) we get 2
0 − V2 =
( 37.8 ) 2
2
V32 37.8 × 9.8 × 9.8 = = 72.9 m 2 × 9 .8
The ball rises with V2 and returns back to ground with V3 (say), so net displacement is zero i.e., s = 0 Then from (i), we get
V32 − V22 = 2g × 0 V3 = V2 = 37.8 m/s or If t2 be the time elapsed between two successive hits, then using eq., (ii), we get ©TutorBreeze.com Please do not copy the answer given here Write to us for help in understanding the solution
NCERT/CBSE PHYSICS CLASS 11 textbook http://www.TutorBreeze.com
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1 ( −g ) t 22 2 9 .8 2 0 = 37.8 t 2 − t2 2 4.9t 2 = 37.8 0 = V2 t 2 +
t2 =
37.8 54 = = 7.714 s 4 .9 7
Let V4 = velocity of rise after second hit. V3 = --10% of V3 = 37.8 –3.78 = 34.02 m/s And t2 = 7.714 s i.e., the time of flight Also we know that time of ascent = time of descent = t´ (Let) t2 =
7.714 = 3.857 = 3.8 s 2
Now
t1 + t 2 = 4.28 + 7.714 = 11.994 = 12
Graph is
Thus the ball rises only once after fall from 90 m with 37.8 m/s in the time interval from t = 0 to 12 s.
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