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