An open elevator is ascending with constant speed of 10 m/s. A ball is thrown vertically up by a boy on elevator…

Q: An open elevator is ascending with constant speed of 10 m/s. A ball is thrown vertically up by a boy on elevator when it is at a height of 40 m from the ground. The velocity of projection with respect to the elevator is 30 m/s. The maximum height attained by the ball is
(take g = 10 ms2)

(a) 85 m

(b) 60 m

(c) 120 m

(d) 45 m

Ans:(c)

Sol: h = 40 m ; Initial vel. w.r.t. elevator = 30 m/s

Velocity of elevator = 10 m/s

Velocity w.r.t. ground, u = 30 + 10 = 40 m/s

From v2 = u2 + 2 a S

0= 1600 + 2(-10)h’

h ‘ = (-1600)/(-20) = 80 m

Total height above the ground = 40 + 80 = 120 m

A juggler keeps 5 balls going with one hand, so that any instant, 4 balls are in air and one ball in the hand…

Q: A juggler keeps 5 balls going with one hand, so that any instant, 4 balls are in air and one ball in the hand. If each ball rises to a height of 20 m, then the time for which each ball stays in the hand is (g = 10 m/s2 )

(a) 1.0 s

(b) 1.5 s

(c) 2.0 s

(d) 4.0 s

Ans: (a)

Sol: Here n = 5 , h = 20 m
Time of Flight \displaystyle T = 2\sqrt{\frac{2 h}{g}}
Time for one ball is hand
\displaystyle = \frac{T}{n-1}

\displaystyle = 2\sqrt{\frac{2 h}{g}}/(n-1)

\displaystyle = \frac{2}{n-1} . \sqrt{\frac{2 h}{g}}

\displaystyle = \frac{2}{5-1} . \sqrt{\frac{2 \times 20}{10}}

= 1 sec

A train is moving along a straight path with uniform acceleration. Its engine passes across a pole…

Q: A train is moving along a straight path with uniform acceleration. Its engine passes across a pole with a velocity of 60 km/h and the end (guard’s van) passes across same pole with a velocity of 80 km/h. The middle point of the train will pass across same pole with a velocity:

(a) 70 km/h

(b) 70.7 km/h

(c) 65 km/h

(d) 75 km/h

Ans:(b)

Sol: Let S = length of train,

t = time taken by full train to cross the pole,

a= uniform acc. of train.

From v2 = u2 + 2 a S

S = v2 – u2/2a

S = 802 – 602/2a

S = 1400/a

The middle point of the train is to cover a distance is

S’ = S/2 = 700/a

From v2 = u2 + 2 a S’

v2 = 602 + 2 a (700/a)

v2 =1400 + 3600

\displaystyle v = \sqrt{5000}

v = 70.7 km/h

A body starts from rest and falls vertically from a height of 19.6 metres. If g = 9.8 m/sec2, then …

Q: A body starts from rest and falls vertically from a height of 19.6 metres. If g = 9.8 m/sec2, then the distance travelled by the body in the last 0.1 second of its motion is

(a) 0.049 m

(b) 1.9 m

(c) 17.7 cm

(d) 19.6 m

Ans:(b)

Sol: u = 0,a = 9.8 m/s2,S = 19.6 m

\displaystyle S = u t + \frac{1}{2}a t^2

\displaystyle 19.6 = 0 + \frac{1}{2}(9.8) t^2

t2 = 19.6/4.9 = 4

t = 2 sec

Let us calculate distance travelled in 1.9 s

\displaystyle S = u t + \frac{1}{2}a t^2

\displaystyle S' = 0 + \frac{1}{2}(9.8) (1.9)^2

S’ = 17.69 m

Distance travelled in last 0.1 s

= S- S’ = 19.6-17.69

=1.91 m

A train accelerating uniformly from rest attains a maximum speed of 40 ms-1 in 20 seconds…

Q: A train accelerating uniformly form rest attains a maximum speed of 40 ms-1 in 20 seconds. It travels at this speed for 20 seconds and is brought to rest with uniform retardation in further 40 seconds. What is the average velocity during this period.

(a) 80/3 ms-1

(b) 25 ms-1

(c) 40 ms-1

(d) 30 ms-1

Ans: (b)

Sol: (i) v = u + at1

40 = 0 + a × 20 ,

a = 2 m/s2

From v2= u2 + 2 a S

402 = 0 + 2 × 2 S1

(ii) S2 =v × t1

S2 = 40 × 20

= 800 m

(iii) v = u + at

0 = 40 + a × 40 ,

a = -1 m/s2

From, v2= u2 + 2 a S

0 = 402 + 2(-1)S3

S3 = 800 m

Total distance travelled , S = S1 + S2 + S3

S = 400 + 800 + 800 = 2000 m

Total time taken = 20 + 20 + 40 = 80 s

Average velocity = 2000/80

= 25 m/s