9 few L2 solutions.ppt
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Transcript of 9 few L2 solutions.ppt
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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
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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
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.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
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.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
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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
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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
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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
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80 meters
10 meters
V m/sec
2 kg mass
Ground level
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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
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6 meter6 meter
A , Mass 50 Kg B , Mass 70 Kg
6m
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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’
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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
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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
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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
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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
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