Q.1: How does the force of gravitation between two objects change when the distance between them is reduced to half?
β
Answer:
Gravitational force between two objects:
Fβ1/d^2
Let original distance = d
New distance = d/2
New force:
F^'β1/((d/2)^2 )
=1/(d^2/4)
=4Γ1/d^2
So,
F^'=4F
π When distance is reduced to half, the gravitational force becomes 4 times.
β
Answer:
Gravitational force between two objects:
Fβ1/d^2
Let original distance = d
New distance = d/2
New force:
F^'β1/((d/2)^2 )
=1/(d^2/4)
=4Γ1/d^2
So,
F^'=4F
π When distance is reduced to half, the gravitational force becomes 4 times.
β
Answer:
Gravitational force between two objects:
Fβ1/d^2
Let original distance = d
New distance = d/2
New force:
F^'β1/((d/2)^2 )
=1/(d^2/4)
=4Γ1/d^2
So,
F^'=4F
π When distance is reduced to half, the gravitational force becomes 4 times.
Q.2: Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
β
Answer:
Gravitational force is more on a heavy body because its mass is more.
But acceleration:
a=F/m
For gravitational force: F=mg
βa=mg/m=g
π Acceleration g is the same for all objects near Earth.
So, both heavy and light objects fall with the same acceleration g, hence they do not fall faster or slower (ignoring air resistance).
β
Answer:
Gravitational force is more on a heavy body because its mass is more.
But acceleration:
a=F/m
For gravitational force: F=mg
βa=mg/m=g
π Acceleration g is the same for all objects near Earth.
So, both heavy and light objects fall with the same acceleration g, hence they do not fall faster or slower (ignoring air resistance).
β
Answer:
Gravitational force is more on a heavy body because its mass is more.
But acceleration:
a=F/m
For gravitational force: F=mg
βa=mg/m=g
π Acceleration g is the same for all objects near Earth.
So, both heavy and light objects fall with the same acceleration g, hence they do not fall faster or slower (ignoring air resistance).
Q.3: What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of earth = 6Γ10^24kg, radius = 6.4Γ10^6m)
β
Answer:
We use:
F=G Mm/R^2
Given:
M=6Γ10^24kg
m=1kg
R=6.4Γ10^6m
G=6.67Γ10^(-11) "β" γ"N m" γ^2 γ"kg" γ^(-2)
Step 1: Write formula
F=G Mm/R^2
Step 2: Substitute
F=6.67Γ10^(-11)Γ(6Γ10^24Γ1)/((6.4Γ10^6 )^2 )
Step 3: Square the radius
(6.4Γ10^6 )^2=γ6.4γ^2Γ10^12=40.96Γ10^12
So,
R^2=4.096Γ10^13
Step 4: Simplify fraction
(6Γ10^24)/(4.096Γ10^13 )β1.46Γ10^11
So,
Fβ6.67Γ10^(-11)Γ1.46Γ10^11
Step 5: Multiply
6.67Γ1.46β9.7β9.8
Fβ9.8"βN"
π Gravitational force β 9.8 N.
(This is same as the weight of a 1 kg mass.)
β
Answer:
We use:
F=G Mm/R^2
Given:
M=6Γ10^24kg
m=1kg
R=6.4Γ10^6m
G=6.67Γ10^(-11) "β" γ"N m" γ^2 γ"kg" γ^(-2)
Step 1: Write formula
F=G Mm/R^2
Step 2: Substitute
F=6.67Γ10^(-11)Γ(6Γ10^24Γ1)/((6.4Γ10^6 )^2 )
Step 3: Square the radius
(6.4Γ10^6 )^2=γ6.4γ^2Γ10^12=40.96Γ10^12
So,
R^2=4.096Γ10^13
Step 4: Simplify fraction
(6Γ10^24)/(4.096Γ10^13 )β1.46Γ10^11
So,
Fβ6.67Γ10^(-11)Γ1.46Γ10^11
Step 5: Multiply
6.67Γ1.46β9.7β9.8
Fβ9.8"βN"
π Gravitational force β 9.8 N.
(This is same as the weight of a 1 kg mass.)
β
Answer:
We use:
F=G Mm/R^2
Given:
M=6Γ10^24kg
m=1kg
R=6.4Γ10^6m
G=6.67Γ10^(-11) "β" γ"N m" γ^2 γ"kg" γ^(-2)
Step 1: Write formula
F=G Mm/R^2
Step 2: Substitute
F=6.67Γ10^(-11)Γ(6Γ10^24Γ1)/((6.4Γ10^6 )^2 )
Step 3: Square the radius
(6.4Γ10^6 )^2=γ6.4γ^2Γ10^12=40.96Γ10^12
So,
R^2=4.096Γ10^13
Step 4: Simplify fraction
(6Γ10^24)/(4.096Γ10^13 )β1.46Γ10^11
So,
Fβ6.67Γ10^(-11)Γ1.46Γ10^11
Step 5: Multiply
6.67Γ1.46β9.7β9.8
Fβ9.8"βN"
π Gravitational force β 9.8 N.
(This is same as the weight of a 1 kg mass.)
Q.4: The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force greater, smaller, or the same as the force with which the moon attracts the earth? Why?
β
Answer:
According to Newtonβs Third Law of Motion, every action has an equal and opposite reaction.
So the force with which Earth attracts Moon = force with which Moon attracts Earth.
π They attract each other with equal forces, but in opposite directions.
β
Answer:
According to Newtonβs Third Law of Motion, every action has an equal and opposite reaction.
So the force with which Earth attracts Moon = force with which Moon attracts Earth.
π They attract each other with equal forces, but in opposite directions.
β
Answer:
According to Newtonβs Third Law of Motion, every action has an equal and opposite reaction.
So the force with which Earth attracts Moon = force with which Moon attracts Earth.
π They attract each other with equal forces, but in opposite directions.
Q.5: If the moon attracts the earth, why does the earth not move towards the moon?
β
Answer:
The force on Earth and Moon is equal.
But acceleration:
a=F/m
Mass of Earth is huge compared to Moon.
So acceleration of Earth:
a_"earth" =F/M_"earth" is extremely small.
π Earth does move slightly, but the motion is so small that we do not notice it.
β
Answer:
The force on Earth and Moon is equal.
But acceleration:
a=F/m
Mass of Earth is huge compared to Moon.
So acceleration of Earth:
a_"earth" =F/M_"earth" is extremely small.
π Earth does move slightly, but the motion is so small that we do not notice it.
β
Answer:
The force on Earth and Moon is equal.
But acceleration:
a=F/m
Mass of Earth is huge compared to Moon.
So acceleration of Earth:
a_"earth" =F/M_"earth" is extremely small.
π Earth does move slightly, but the motion is so small that we do not notice it.
Q.6: What happens to the force between two objects, if (i) the mass of one object is doubled? (ii) the distance between the objects is doubled and tripled? (iii) the masses of both objects are doubled?
β
Answer:
Gravitational force:
F=G (M_1 M_2)/d^2
________________________________________
(i) Mass of one object doubled
New mass = 2M_1
F^'=G (2M_1 M_2)/d^2 =2F
π Force becomes double.
________________________________________
(ii) Distance doubled and tripled
Distance doubled:
New distance = 2d
F^'=G (M_1 M_2)/((2d)^2 )=G (M_1 M_2)/(4d^2 )=F/4
π Force becomes one-fourth.
Distance tripled:
New distance = 3d
F^'=(M_1 M_2)/((3d)^2 )=(M_1 M_2)/(9d^2 )=F/9
π Force becomes one-ninth.
________________________________________
(iii) Both masses doubled
New masses = 2M_1and 2M_2
F^'=G ((2M_1)(2M_2))/d^2 =G (4M_1 M_2)/d^2 =4F
π Force becomes 4 times.
β
Answer:
Gravitational force:
F=G (M_1 M_2)/d^2
________________________________________
(i) Mass of one object doubled
New mass = 2M_1
F^'=G (2M_1 M_2)/d^2 =2F
π Force becomes double.
________________________________________
(ii) Distance doubled and tripled
Distance doubled:
New distance = 2d
F^'=G (M_1 M_2)/((2d)^2 )=G (M_1 M_2)/(4d^2 )=F/4
π Force becomes one-fourth.
Distance tripled:
New distance = 3d
F^'=(M_1 M_2)/((3d)^2 )=(M_1 M_2)/(9d^2 )=F/9
π Force becomes one-ninth.
________________________________________
(iii) Both masses doubled
New masses = 2M_1and 2M_2
F^'=G ((2M_1)(2M_2))/d^2 =G (4M_1 M_2)/d^2 =4F
π Force becomes 4 times.
β
Answer:
Gravitational force:
F=G (M_1 M_2)/d^2
________________________________________
(i) Mass of one object doubled
New mass = 2M_1
F^'=G (2M_1 M_2)/d^2 =2F
π Force becomes double.
________________________________________
(ii) Distance doubled and tripled
Distance doubled:
New distance = 2d
F^'=G (M_1 M_2)/((2d)^2 )=G (M_1 M_2)/(4d^2 )=F/4
π Force becomes one-fourth.
Distance tripled:
New distance = 3d
F^'=(M_1 M_2)/((3d)^2 )=(M_1 M_2)/(9d^2 )=F/9
π Force becomes one-ninth.
________________________________________
(iii) Both masses doubled
New masses = 2M_1and 2M_2
F^'=G ((2M_1)(2M_2))/d^2 =G (4M_1 M_2)/d^2 =4F
π Force becomes 4 times.
Q.7: What is the importance of universal law of gravitation?
β
Answer:
The universal law of gravitation explains:
Motion of planets around the Sun
Motion of Moon around Earth
Tides in seas and oceans
Falling of objects towards Earth
Gravitational force between any two objects in the universe
π It unifies many natural phenomena under one single law.
β
Answer:
The universal law of gravitation explains:
Motion of planets around the Sun
Motion of Moon around Earth
Tides in seas and oceans
Falling of objects towards Earth
Gravitational force between any two objects in the universe
π It unifies many natural phenomena under one single law.
β
Answer:
The universal law of gravitation explains:
Motion of planets around the Sun
Motion of Moon around Earth
Tides in seas and oceans
Falling of objects towards Earth
Gravitational force between any two objects in the universe
π It unifies many natural phenomena under one single law.
Q.8: What is the acceleration of free fall?
β
Answer:
The acceleration of free fall is the acceleration produced in a body when it falls freely under gravity alone.
It is denoted by g
Near Earthβs surface, g β 9.8 m/sΒ², downward.
π It is called acceleration due to gravity.
β
Answer:
The acceleration of free fall is the acceleration produced in a body when it falls freely under gravity alone.
It is denoted by g
Near Earthβs surface, g β 9.8 m/sΒ², downward.
π It is called acceleration due to gravity.
β
Answer:
The acceleration of free fall is the acceleration produced in a body when it falls freely under gravity alone.
It is denoted by g
Near Earthβs surface, g β 9.8 m/sΒ², downward.
π It is called acceleration due to gravity.
Q.9: What do we call the gravitational force between the earth and an object?
β
Answer:
The gravitational force between the Earth and an object is called the weight of the object.
β
Answer:
The gravitational force between the Earth and an object is called the weight of the object.
β
Answer:
The gravitational force between the Earth and an object is called the weight of the object.
Q.10: Amit buys a few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? (Hint: g is greater at poles than equator)
β
Answer:
At the poles, value of g is greater, so weight W=mgis more.
At the equator, g is less, so weight of the same mass is less.
So the same mass of gold will show a slightly lesser weight at the equator.
π The friend may say the gold is less in weight if measured at the equator.
(Though the mass of gold is unchanged.)
β
Answer:
At the poles, value of g is greater, so weight W=mgis more.
At the equator, g is less, so weight of the same mass is less.
So the same mass of gold will show a slightly lesser weight at the equator.
π The friend may say the gold is less in weight if measured at the equator.
(Though the mass of gold is unchanged.)
β
Answer:
At the poles, value of g is greater, so weight W=mgis more.
At the equator, g is less, so weight of the same mass is less.
So the same mass of gold will show a slightly lesser weight at the equator.
π The friend may say the gold is less in weight if measured at the equator.
(Though the mass of gold is unchanged.)
Q.11: Why will a sheet of paper fall slower than one that is crumpled into a ball?
β
Answer:
A flat sheet of paper has a larger surface area.
It experiences more air resistance (air friction).
A crumpled ball has less area, so less air resistance.
π Due to greater air resistance, the flat sheet falls slower than the crumpled ball.
β
Answer:
A flat sheet of paper has a larger surface area.
It experiences more air resistance (air friction).
A crumpled ball has less area, so less air resistance.
π Due to greater air resistance, the flat sheet falls slower than the crumpled ball.
β
Answer:
A flat sheet of paper has a larger surface area.
It experiences more air resistance (air friction).
A crumpled ball has less area, so less air resistance.
π Due to greater air resistance, the flat sheet falls slower than the crumpled ball.
Q.12: Gravitational force on the surface of the moon is only 1/6as strong as on Earth. What is the weight in newtons of a 10 kg object on the moon and on the earth?
β
Answer:
On Earth:
m=10"βkg",g=9.8"β" γ"m/s" γ^2
W_E=mg=10Γ9.8=98"βN"
On Moon:
g_"moon" =1/6 g_"earth" =9.8/6β1.63"β" γ"m/s" γ^2
W_M=mg_"moon" =10Γ1.63β16.3"βN"
π Weight on Earth = 98 N, Weight on Moon β 16.3 N
β
Answer:
On Earth:
m=10"βkg",g=9.8"β" γ"m/s" γ^2
W_E=mg=10Γ9.8=98"βN"
On Moon:
g_"moon" =1/6 g_"earth" =9.8/6β1.63"β" γ"m/s" γ^2
W_M=mg_"moon" =10Γ1.63β16.3"βN"
π Weight on Earth = 98 N, Weight on Moon β 16.3 N
β
Answer:
On Earth:
m=10"βkg",g=9.8"β" γ"m/s" γ^2
W_E=mg=10Γ9.8=98"βN"
On Moon:
g_"moon" =1/6 g_"earth" =9.8/6β1.63"β" γ"m/s" γ^2
W_M=mg_"moon" =10Γ1.63β16.3"βN"
π Weight on Earth = 98 N, Weight on Moon β 16.3 N
Q.13: A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (i) the maximum height to which it rises, (ii) the total time it takes to return to the surface of the earth. (Take g=9.8"β" γ"m/s" γ^2)
β
Answer:
Given:
u=49"βm/s",v=0"β"("at top"),g=9.8"β" γ"m/s" γ^2
________________________________________
(i) Maximum height h
Use:
v^2=u^2-2gh
0=49^2-2Γ9.8Γh
2Γ9.8Γh=49^2
19.6h=2401
h=2401/19.6=122.5"βm"
π Maximum height = 122.5 m
________________________________________
(ii) Total time of flight
Time to reach top:
v=u-gt
0=49-9.8t
β9.8t=49
βt=49/9.8=5"βs"
Time to go up = 5 s
Time to come down = 5 s
π Total time = 5 + 5 = 10 s
β
Answer:
Given:
u=49"βm/s",v=0"β"("at top"),g=9.8"β" γ"m/s" γ^2
________________________________________
(i) Maximum height h
Use:
v^2=u^2-2gh
0=49^2-2Γ9.8Γh
2Γ9.8Γh=49^2
19.6h=2401
h=2401/19.6=122.5"βm"
π Maximum height = 122.5 m
________________________________________
(ii) Total time of flight
Time to reach top:
v=u-gt
0=49-9.8t
β9.8t=49
βt=49/9.8=5"βs"
Time to go up = 5 s
Time to come down = 5 s
π Total time = 5 + 5 = 10 s
β
Answer:
Given:
u=49"βm/s",v=0"β"("at top"),g=9.8"β" γ"m/s" γ^2
________________________________________
(i) Maximum height h
Use:
v^2=u^2-2gh
0=49^2-2Γ9.8Γh
2Γ9.8Γh=49^2
19.6h=2401
h=2401/19.6=122.5"βm"
π Maximum height = 122.5 m
________________________________________
(ii) Total time of flight
Time to reach top:
v=u-gt
0=49-9.8t
β9.8t=49
βt=49/9.8=5"βs"
Time to go up = 5 s
Time to come down = 5 s
π Total time = 5 + 5 = 10 s
Q.14: A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground. (Take g=9.8"β" γ"m/s" γ^2)
β
Answer:
Given:
u=0(released from rest), h=19.6"βm",g=9.8"β" γ"m/s" γ^2
Use:
v^2=u^2+2gh
v^2=0+2Γ9.8Γ19.6
v^2=19.6Γ19.6=384.16
v=β384.16=19.6"βm/s"
π Final velocity = 19.6 m/s (downwards)
β
Answer:
Given:
u=0(released from rest), h=19.6"βm",g=9.8"β" γ"m/s" γ^2
Use:
v^2=u^2+2gh
v^2=0+2Γ9.8Γ19.6
v^2=19.6Γ19.6=384.16
v=β384.16=19.6"βm/s"
π Final velocity = 19.6 m/s (downwards)
β
Answer:
Given:
u=0(released from rest), h=19.6"βm",g=9.8"β" γ"m/s" γ^2
Use:
v^2=u^2+2gh
v^2=0+2Γ9.8Γ19.6
v^2=19.6Γ19.6=384.16
v=β384.16=19.6"βm/s"
π Final velocity = 19.6 m/s (downwards)
Q.15: A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g=10"β" γ"m/s" γ^2, find (i) the maximum height reached, (ii) net displacement, (iii) total distance covered.
β
Answer:
Given:
u=40"βm/s",g=10"β" γ"m/s" γ^2,v=0at top.
________________________________________
(i) Maximum height
Use:
v^2=u^2-2gh
0=40^2-2Γ10Γh
2Γ10Γh=1600
β20h=1600
βh=80"βm"
π Maximum height = 80 m
________________________________________
(ii) Net displacement
Stone returns to starting point.
So initial position = final position.
π Net displacement = 0
________________________________________
(iii) Total distance
Upward distance = 80 m
Downward distance = 80 m
Total distance: = 80 + 80=160 βm
π Total distance = 160 m
β
Answer:
Given:
u=40"βm/s",g=10"β" γ"m/s" γ^2,v=0at top.
________________________________________
(i) Maximum height
Use:
v^2=u^2-2gh
0=40^2-2Γ10Γh
2Γ10Γh=1600
β20h=1600
βh=80"βm"
π Maximum height = 80 m
________________________________________
(ii) Net displacement
Stone returns to starting point.
So initial position = final position.
π Net displacement = 0
________________________________________
(iii) Total distance
Upward distance = 80 m
Downward distance = 80 m
Total distance: = 80 + 80=160 βm
π Total distance = 160 m
β
Answer:
Given:
u=40"βm/s",g=10"β" γ"m/s" γ^2,v=0at top.
________________________________________
(i) Maximum height
Use:
v^2=u^2-2gh
0=40^2-2Γ10Γh
2Γ10Γh=1600
β20h=1600
βh=80"βm"
π Maximum height = 80 m
________________________________________
(ii) Net displacement
Stone returns to starting point.
So initial position = final position.
π Net displacement = 0
________________________________________
(iii) Total distance
Upward distance = 80 m
Downward distance = 80 m
Total distance: = 80 + 80=160 βm
π Total distance = 160 m
Q.16: Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6Γγ10γ^24kg and of the Sun = 2Γγ10γ^30kg. The average distance between the two is 1.5Γγ10γ^11m.
β
Answer:
Use:
F=G (M_1 M_2)/d^2
Given:
M_1=6Γ10^24kg (Earth)
M_2=2Γ10^30kg (Sun)
d=1.5Γ10^11m
G=6.67Γ10^(-11)
Step 1: Substitute
F=6.67Γ10^(-11)Γ(6Γ10^24Γ2Γ10^30)/((1.5Γ10^11 )^2 )
Step 2: Multiply masses
6Γ2=12
10^24Γ10^30=10^54
So numerator: 12Γ10^54
Step 3: Square distance
=(1.5Γ10^11 )^2
=γ1.5γ^2Γ10^22
=2.25Γ10^22
Step 4: Fraction
=(12Γ10^54)/(2.25Γ10^22 )
=12/2.25Γ10^32β5.33Γ10^32
So:
F=6.67Γ10^(-11)Γ5.33Γ10^32
Step 5: Multiply
6.67Γ5.33β35.6
10^(-11)Γ10^32=10^21
So:
Fβ35.6Γ10^21
=3.56Γ10^22 "βN"
π Gravitational force β 3.56Γ10^22N
β
Answer:
Use:
F=G (M_1 M_2)/d^2
Given:
M_1=6Γ10^24kg (Earth)
M_2=2Γ10^30kg (Sun)
d=1.5Γ10^11m
G=6.67Γ10^(-11)
Step 1: Substitute
F=6.67Γ10^(-11)Γ(6Γ10^24Γ2Γ10^30)/((1.5Γ10^11 )^2 )
Step 2: Multiply masses
6Γ2=12
10^24Γ10^30=10^54
So numerator: 12Γ10^54
Step 3: Square distance
=(1.5Γ10^11 )^2
=γ1.5γ^2Γ10^22
=2.25Γ10^22
Step 4: Fraction
=(12Γ10^54)/(2.25Γ10^22 )
=12/2.25Γ10^32β5.33Γ10^32
So:
F=6.67Γ10^(-11)Γ5.33Γ10^32
Step 5: Multiply
6.67Γ5.33β35.6
10^(-11)Γ10^32=10^21
So:
Fβ35.6Γ10^21
=3.56Γ10^22 "βN"
π Gravitational force β 3.56Γ10^22N
β
Answer:
Use:
F=G (M_1 M_2)/d^2
Given:
M_1=6Γ10^24kg (Earth)
M_2=2Γ10^30kg (Sun)
d=1.5Γ10^11m
G=6.67Γ10^(-11)
Step 1: Substitute
F=6.67Γ10^(-11)Γ(6Γ10^24Γ2Γ10^30)/((1.5Γ10^11 )^2 )
Step 2: Multiply masses
6Γ2=12
10^24Γ10^30=10^54
So numerator: 12Γ10^54
Step 3: Square distance
=(1.5Γ10^11 )^2
=γ1.5γ^2Γ10^22
=2.25Γ10^22
Step 4: Fraction
=(12Γ10^54)/(2.25Γ10^22 )
=12/2.25Γ10^32β5.33Γ10^32
So:
F=6.67Γ10^(-11)Γ5.33Γ10^32
Step 5: Multiply
6.67Γ5.33β35.6
10^(-11)Γ10^32=10^21
So:
Fβ35.6Γ10^21
=3.56Γ10^22 "βN"
π Gravitational force β 3.56Γ10^22N
Q.17: A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet. (Take g=9.8"β" γ"m/s" γ^2)
β
Answer:
Let they meet after t seconds.
Stone A (falling from top):
Initial velocity u_1=0
Distance fallen in time t:
s_1=1/2 gt^2
=1/2Γ9.8Γt^2
=4.9t^2
Stone B (thrown upward from ground):
Initial velocity u_2=25m/s
Distance from ground in time t:
s_2=u_2 t-1/2 gt^2
=25t-4.9t^2
They meet when the sum of these distances = 100 m:
s_1+s_2=100
4.9t^2+(25t-4.9t^2 )=100
25t=100
βt=4"βs"
Where do they meet?
Height fallen from top:
s_1=4.9Γ4^2
=4.9Γ16
=78.4"βm"
So height above ground:
100-78.4=21.6"βm "
π They meet after 4 s, at 21.6 m above the ground (or 78.4 m below the top).
β
Answer:
Let they meet after t seconds.
Stone A (falling from top):
Initial velocity u_1=0
Distance fallen in time t:
s_1=1/2 gt^2
=1/2Γ9.8Γt^2
=4.9t^2
Stone B (thrown upward from ground):
Initial velocity u_2=25m/s
Distance from ground in time t:
s_2=u_2 t-1/2 gt^2
=25t-4.9t^2
They meet when the sum of these distances = 100 m:
s_1+s_2=100
4.9t^2+(25t-4.9t^2 )=100
25t=100
βt=4"βs"
Where do they meet?
Height fallen from top:
s_1=4.9Γ4^2
=4.9Γ16
=78.4"βm"
So height above ground:
100-78.4=21.6"βm "
π They meet after 4 s, at 21.6 m above the ground (or 78.4 m below the top).
β
Answer:
Let they meet after t seconds.
Stone A (falling from top):
Initial velocity u_1=0
Distance fallen in time t:
s_1=1/2 gt^2
=1/2Γ9.8Γt^2
=4.9t^2
Stone B (thrown upward from ground):
Initial velocity u_2=25m/s
Distance from ground in time t:
s_2=u_2 t-1/2 gt^2
=25t-4.9t^2
They meet when the sum of these distances = 100 m:
s_1+s_2=100
4.9t^2+(25t-4.9t^2 )=100
25t=100
βt=4"βs"
Where do they meet?
Height fallen from top:
s_1=4.9Γ4^2
=4.9Γ16
=78.4"βm"
So height above ground:
100-78.4=21.6"βm "
π They meet after 4 s, at 21.6 m above the ground (or 78.4 m below the top).
Q.18: A ball thrown up vertically returns to the thrower after 6 s. Find (a) the velocity with which it was thrown up, (b) the maximum height it reaches, and (c) its position after 4 s. (Take g=9.8"β" γ"m/s" γ^2)
β
Answer:
Total time = 6 s
Time to go up = 3 s
Time to come down = 3 s
________________________________________
(a) Velocity with which it was thrown up
Use:
v=u-gt
At top, v=0:
0=u-9.8Γ3
βu=9.8Γ3
=29.4"βm/s"
π Initial velocity = 29.4 m/s upward
________________________________________
(b) Maximum height
Use:
h=ut-1/2 gt^2
h=29.4Γ3-1/2Γ9.8Γ9
h=88.2-44.1
=44.1"βm"
π Maximum height = 44.1 m
________________________________________
(c) Position after 4 s
In 3 s, it reaches top.
In next 1 s, it starts coming down.
Distance fallen in 1 s from top:
s=1/2 gt^2
=1/2Γ9.8Γ1^2
=4.9"βm"
So height above ground after 4 s:
=44.1-4.9
=39.2"βm"
π After 4 s, the ball is 39.2 m above the ground, coming down.
β
Answer:
Total time = 6 s
Time to go up = 3 s
Time to come down = 3 s
________________________________________
(a) Velocity with which it was thrown up
Use:
v=u-gt
At top, v=0:
0=u-9.8Γ3
βu=9.8Γ3
=29.4"βm/s"
π Initial velocity = 29.4 m/s upward
________________________________________
(b) Maximum height
Use:
h=ut-1/2 gt^2
h=29.4Γ3-1/2Γ9.8Γ9
h=88.2-44.1
=44.1"βm"
π Maximum height = 44.1 m
________________________________________
(c) Position after 4 s
In 3 s, it reaches top.
In next 1 s, it starts coming down.
Distance fallen in 1 s from top:
s=1/2 gt^2
=1/2Γ9.8Γ1^2
=4.9"βm"
So height above ground after 4 s:
=44.1-4.9
=39.2"βm"
π After 4 s, the ball is 39.2 m above the ground, coming down.
β
Answer:
Total time = 6 s
Time to go up = 3 s
Time to come down = 3 s
________________________________________
(a) Velocity with which it was thrown up
Use:
v=u-gt
At top, v=0:
0=u-9.8Γ3
βu=9.8Γ3
=29.4"βm/s"
π Initial velocity = 29.4 m/s upward
________________________________________
(b) Maximum height
Use:
h=ut-1/2 gt^2
h=29.4Γ3-1/2Γ9.8Γ9
h=88.2-44.1
=44.1"βm"
π Maximum height = 44.1 m
________________________________________
(c) Position after 4 s
In 3 s, it reaches top.
In next 1 s, it starts coming down.
Distance fallen in 1 s from top:
s=1/2 gt^2
=1/2Γ9.8Γ1^2
=4.9"βm"
So height above ground after 4 s:
=44.1-4.9
=39.2"βm"
π After 4 s, the ball is 39.2 m above the ground, coming down.
Q.19: In what direction does the buoyant force on an object immersed in a liquid act?
β
Answer:
Buoyant force always acts upwards,
π Opposite to the direction of gravity.
β
Answer:
Buoyant force always acts upwards,
π Opposite to the direction of gravity.
β
Answer:
Buoyant force always acts upwards,
π Opposite to the direction of gravity.
Q.20: Why does a block of plastic released under water come up to the surface of water?
β
Answer:
Plastic has less density than water.
Upward buoyant force on plastic is greater than its weight.
So the net force is upwards and it rises to the surface.
π Thatβs why a plastic block comes up and floats.
β
Answer:
Plastic has less density than water.
Upward buoyant force on plastic is greater than its weight.
So the net force is upwards and it rises to the surface.
π Thatβs why a plastic block comes up and floats.
β
Answer:
Plastic has less density than water.
Upward buoyant force on plastic is greater than its weight.
So the net force is upwards and it rises to the surface.
π Thatβs why a plastic block comes up and floats.
Q.21: The volume of 50 g of a substance is 20 cmΒ³. If the density of water is 1 g/cmΒ³, will the substance float or sink?
β
Answer:
Density:
Ο="mass" /"volume" =50/20=2.5"β" γ"g/cm" γ^3
Density of water = 1 g/cmΒ³
2.5 g/cmΒ³ > 1 g/cmΒ³
π The substance is denser than water, so it will sink.
β
Answer:
Density:
Ο="mass" /"volume" =50/20=2.5"β" γ"g/cm" γ^3
Density of water = 1 g/cmΒ³
2.5 g/cmΒ³ > 1 g/cmΒ³
π The substance is denser than water, so it will sink.
β
Answer:
Density:
Ο="mass" /"volume" =50/20=2.5"β" γ"g/cm" γ^3
Density of water = 1 g/cmΒ³
2.5 g/cmΒ³ > 1 g/cmΒ³
π The substance is denser than water, so it will sink.
Q.22: The volume of a 500 g sealed packet is 350 cmΒ³. Will the packet float or sink in water if the density of water is 1 g/cmΒ³? What will be the mass of the water displaced by this packet?
β
Answer:
First, find density of packet:
Ο="mass" /"volume" =500/350β1.43"β" γ"g/cm" γ^3
Compare with water: 1.43 > 1
π Packet is denser than water β it will sink.
Mass of water displaced:
When fully submerged, volume of water displaced = volume of packet = 350 cmΒ³
Density of water = 1 g/cmΒ³
"Mass of water displaced"=ΟV=1Γ350=350"βg"
π Water displaced = 350 g
β
Answer:
First, find density of packet:
Ο="mass" /"volume" =500/350β1.43"β" γ"g/cm" γ^3
Compare with water: 1.43 > 1
π Packet is denser than water β it will sink.
Mass of water displaced:
When fully submerged, volume of water displaced = volume of packet = 350 cmΒ³
Density of water = 1 g/cmΒ³
"Mass of water displaced"=ΟV=1Γ350=350"βg"
π Water displaced = 350 g
β
Answer:
First, find density of packet:
Ο="mass" /"volume" =500/350β1.43"β" γ"g/cm" γ^3
Compare with water: 1.43 > 1
π Packet is denser than water β it will sink.
Mass of water displaced:
When fully submerged, volume of water displaced = volume of packet = 350 cmΒ³
Density of water = 1 g/cmΒ³
"Mass of water displaced"=ΟV=1Γ350=350"βg"
π Water displaced = 350 g