A ball is dropped from 400m high tower. At the same time another ball is thrown up with velocity 50m/sec. At what height they will meet from ground ?

L2--Q2

Ball ‘a’ covers 400 -h meters -(400 – h) = - 1/2 gt2 h = 400 - 1/2 gt2400

h

Ub = 50m/sec

Ua=0

a

b

.a & b meet after t sec

S= ut +1/2 (-g)t2

+ve-ve

.g = 10 m/sec2

.g = 10 m/sec2

Ball ‘b’ covers ‘h’ meters h = ut - 1/2 gt2 h = 50t - 1/2 gt2400

h

Ub = 50m/sec

Ua=0

a

b

.a & b meet after t sec

S= ut +1/2 (-g)t2

+ve-ve

For Ball ‘a’ h = 400 - 1/2 gt2

.g = 10 m/sec2

400 - 1/2 gt2 = 50 t - 1/2 gt2 400 = 50 t , t = 8 sec h = 50 t - 1/2 gt2 .h = 50 x 8 - 1/2 x 10 x 82 = 80m400

h

Ub = 50m/sec

Ua=0

a

b

.a & b meet after t sec

+ve-ve

For Ball ‘a’, h = 400 - 1/2 gt2

For Ball ‘b’ , h = 50t - 1/2 gt2

A car traveling with uniform acceleration covers

200m in first 2 sec & 220 m in next 4 sec .Find

it’s velocity after 7 secs

L2 – Q3

200m2sec

220m4sec

U m/seca m/sec2

S = ut +1/2 at2

200 = 2u + ½ a x 22 200 = 2u + 2 a

100 = u + a ------a

S = ut +1/2 at2

200+ 220 = 6u +1/2 a x 62 420 = 6u + 18 a

70 = u + 3a ------ b

From a – b, 30 = -2a, a = -15m/sec2, u=115m/sec Velocity after 7sec = u+at =115+(-15)x7=10m/sec

A ball of 2 kg weight rolls down a hill 80 m height .It then rolls down a 10m deep valley near the hill & then rolls up from the other edge of the valley to ground level. Find out magnitude of velocity of ball when it leaves ground .Neglect friction .Take g =10m/sec sq

L2-- Q6

80 meters

10 meters

V m/sec

2 kg mass

Ground level

Two skaters A and B of mass 50 Kg & 70 Kg

respectively stand facing each other

6m apart with a rope stretched between them If

they pull the rope how far A & B

will travel when they meet

L2 – Q9

6 meter6 meter

A , Mass 50 Kg B , Mass 70 Kg

6m

6 meter6 meter

A , Mass 50 Kg B , Mass 70 Kg

Pull F Pull F

• A & B Pull rope with force F( belt tension)• Due to reaction of F, Rope pulls A & B by

force F’= F in apposite direction

F’F’

6 meter6 meter

A , Mass 50 Kg B , Mass 70 Kg

Pull F Pull F

F’F’Aa (Accn) Ab

F =m × aF’= 50 AaF’ =70 Ab

AbAa

57

=

6m

6 meter6 meter

A , Mass 50 Kg B , Mass 70 Kg

Pull F Pull F

F’F’Aa = Ba

Da

D = 1/2 at2 as u =0 ,

Db

AbAa

57

=

AbAa

57

= DbDa

=

6m

& Db BaDa Aa

.ta= tb

D a

6 meter6 meter

A , Mass 50 Kg B , Mass 70 Kg

Pull F Pull F

Da Db

Da + D b = 6m

Db + DaDa

=5 +7

7=

127

= 57

DbDa

6 meter6 meter

A , Mass 50 Kg B , Mass 70 Kg

Pull F Pull F

6mDb + Da = 6

6Da

=7

12

Da =12

6 × 7 Da = 3.5m

Now Db + DaDa

=127