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Que L1--1A person in space, at a distance R from center of earth, is attracted by gravitational force of 400 N. How far the person should be from center of earth for gravitational force to be 100 N?
F1 = 400N
F2 = 100N
R
X ?
F = G × M1× m2
r2 F 1r2
As G M1 & m2 are not changing
400
100
X2
R2
4 R2X2
2 RX
A truck weighing 6000 kg rams in a car weighing 800 kg. The truck was moving at speed of 15m/sec & car was at rest. If car & truck moved together after collision, what would be final speed of combine car & truck
Que L1--2
6000 kg 800 kg
15m/sec
X m/sec(6000+800)
Equating momentum before & after collision
6000 × 15 = X ( 6000 + 800)
X = 13.2 m/s
A particle moving in straight line covers half
distance with speed 3m/sec.The other half
distance is covered in two equal time intervals
with speed of 4.5m/s & 7.5m/s respectively.
Find average speed of particle
L1 --Q3
V = 3
V = 4.5 V = 7.5
V = velocity m/sec, x = is total distance meters
t1 t2 t2
x/2
.t1= (x / 2) / 3 = x / 6
x/2
x /2 = 4.5 t2+7.5 t2 = 12 t2
t2 = (x /2) /12 = x /24Average velocity
xt1+ t2+t3
=x
x24
x24
x6 ++
=
Total distanceTotal time
=
x6x24
=4xx= = 4m/sec
A ball is weighing 200gms is dropped from 70
m from a tower. Another ball weighing 600gms
is dropped from 80m from same tower exactly
after one second.. Find out separation between
two balls 3secs
( g= 10m/sec2 )
L1-- Q4
600gm
s1
s2
First ball travels for 3secSeconds ball travels for 2 sec
S1= ut +1/2 gt2 =1/2 ×(-10) × 32 = 45
.g = 10m/sec2
S2= ut +1/2 gt2 =1/2 ×(-10) × 22 = 20
Substraction = (S1-S2) +10 = 35m
10
Distance traveled do not depend on weight of ball
A machine guns fired a bullet of mass of 40 gms with a velocity of 1200m/s .Theman holding the gun can exert a max force of 144N on the gun. How manymaximum bullets he can fire per sec.( Ans: = 3 )
Que L1--5
ActionOn bullet
Reactionon man
1200m/sec
Max force = 144 N
• Let n = No. of bullets fired /sec• Initial bullet velocity = 0• Final bullet velocity = 1200 m/sec• Change in momentum / bullet
= (40/1000) ×1200 = 48 kg m/sec• Rate of change in bullet momentum
= n × 48 kgm /sec2 = n×48 – N• Reaction force on man = n×48 – N• 144 N = n×48 , n=3
40 gms
A spring balance is reading 60N in air when a block is suspended to it. This reading changes to 40N when block is submerged in water. Find density of the block. .( Take g =10m/sec2.). ( Ans = 3 gm/cc )
Que : L1-- 8
60 N
.mg = 60N
mg
Buoyancy
.m of block = 60/10 = 6kg --------------- 1 Reduction in spring balance reading (60 – 40) N = Buoyancy force = Volume × Water density × g
40 N
Formula Buoyancy force = weight of liquid displaced= Volume of water displaced × Density of liquid × g= Volume of block × Density of liquid × g
Vcc
60 N
.mg = 60N
mg
Buoyancy
M = 60/10 × 1000 = 6Kg = 6000 gms
40 N
Formula Buoyancy force = weight of liquid displaced=Volume of water displaced × Density of liquid
Vcc
20 kg m /sec2 = V m3 × 1000 kg/m3 × 10 m/sec2
V = 0.002 m3 = 2000 ccDensity of block = m/V = 6000/2000= 3 gm/cc.
Que: L1--9On a unknown planet a ball is dropped from height of 2.5 meter. While rebounding from ground it looses 20 % velocity. Find out height to which it will rebound.( Ans = 1.6 meters )
Va0.8 Va
hb
2.5 mmkg
b
a
PE at ‘c’ = KE at ‘a’ downwardm g 2.5 = ½ m Va2
2.5 g = ½ Va2
cAt point ‘a’ upward velocity will be 0.8 Va as 20 velocity is lost while rebounding
PE at ‘b’ = KE at ‘a’ up ward m g hb = ½ m × (0.8Va)2
g hb = ½ × V2a × 0.8× 0.8
g hb = 2.5g × 0.8× 0.8
hb = 2.5 × 0.8× 0.8 =1.6 m
A bullet having weight 25 gm enters a 7.5cm thick card board at speed 50m/sec It pierces through cardboard & comes out with speed 40m/sec. Find average force exerted by board on bullet
L1 – Q10
Click to view animated diadram
S =7.5 cm
50 m/sec V =40 m/sec
25 gms
Deceleration ‘a’ m /sec2 due to
Cardboard resistance force
U= F
V2 = u2 + 2as40×40 =50× 50 +2a × 7.5/100.a = - (2500-1600) ×100 / 15.a = -6000 m/sec2
F= m × a = (25/1000) × 6000 = 150N
Que L:1--11A car starting from rest with acceleration ‘a’ cm/ sec sq ,covers distance of Xmeters in first 2 secs & Y meters in next two secs. Find relation between X & Y ( Ans Y = 3X is required relation )
2sec 2sec
x y
a m/sec2
Distance = ½ a t2
x = ½ a 22 = 2a
x + y = ½ a 4 2 = 8a y = 6a
y = 3x is required relation
A bullet moving with velocity 100 m/sec can just penetrate two planks of equal thickness. How many number of plates it can penetrate if velocity of bullet is doubled
L1-- Q12
2plates
100 m/sec
25 gms
Deceleration ‘a’ m /sec2 due to
Cardboard resistance force
U= FV=0
S
S1
200 m/sec
25 gms
Deceleration ‘a’ m /sec2 due to
Cardboard resistance force
U= V=0FN plates
V2 = u2 + 2as0 = u2 + 2asU2 ∞ S
100 × 100200 × 200
=SS1
S1= 4S= 4 ×2 plates = 8 plates
L1 –13 :A car traveling with speed of 60km/hr can be stopped within 20 meter distance. Find out stopping distance if speed of the car is doubled.
60 km/hr 20m
120 km/hr X=?
A
BB’
KE at A = work done by break force Fb during motion A to B. ½ mV2 = Fb x Distance.As Fb & m are constant . Breaking distance V2
Hence if V is doubled , d = 4 x 20 = 80 m
Two masses Ma & Mb are hanging over a smooth light pulley by a string .If acceleration of system is g/8 , find ratio of the masses. Mb > Ma , Ans : Ma/Mb = 9/7 )
Que L1-- 14
ma mb
ma<mb
g/8
1.Net downward force = (mb –ma ) g
2.Moving mass = (mb + ma )
Accn = F/M
(mb –ma )(mb + ma )
=18
7mb=9 ma
(mb –ma ) g(mb + ma )
=g8
79
=mamb
L1 – 16: Bullets of 30 g each hit a plate at rate 200 bullet/sec with velocity 50m/sec. Each bullet reflects back with velocity 30m/sec. Find average force on the plate
u = 50m/sec
V = - 30m/sec
Change of movement per bullet= m( - v – -u ) = – 0.03 x 80Change in movement per sec = - 200 x 0.03 x 80 = 480 NAs per Newton's law this = force on plate = 480 N