## Four persons K,L,M,N are initially at the four corners of a square of side d. Each person now moves with a uniform speed…

Q: Four persons K,L,M,N are initially at the four corners of a square of side d. Each person now moves with a uniform speed v in such a way that K always moves directly towards L, L directly towards M, M directly towards N and N directly towards K. The four persons will meet at a time …………

Ans: d/v

Sol: Initially distance between two person = d

Relative velocity of approach = v + v cos90o = v

The person will meet after time , t = d/v

## 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