Forbidden.
Banned
Space skills practice questions.
Neglect:
- Air resistance
- Friction
- Curvature of the surface of the earth
Use the necessary data and formula provided by the data sheet here
unless values are already given:
www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/physics_formulaesheet.pdf
1.
A projectile is fired with an initial velocity 100 m/s at 30 degrees above the horizontal.
Find:
a) The vertical component of the initial velocity
b) The maximum height
c) The time of flight
d) The range
2.
A golf ball is hit on a flat course and its maximum height
is known to be 9.8 m and its range is 29.4 m.
Find the magnitude and direction of its initial velocity.
3.
A bomber is flying at a constant velocity of 200m/s at an altitude of 500m.
a)
Describe the pilot's observation of the path of the projectile when it
is dropped from the bomber.
b)
How long will it take a bomb dropped from the bomber to reach the ground ?
c)
At hwat distance in front of the target does the bomber have to drop its bomb
to hit the target ?
4.
The mass of the Earth is given as 5.98 x 1024 kg.
a)
From the Law of Energy Conservation, derive an expression and calculate the
escape velocity of a 200kg object.
b)
How would the escape velocity of the object differ if it was projected
at an angle of 45 degrees to the horizontal ?
5.
The radius of the Earth is given as 6380 km and the acceleration due to
gravity close to the Earth's surface is given as 9.8 ms-2
Calculate the mass of the Earth.
IDEAL ANSWERS:
You should only check them after you have attempted the questions.
Note: These answers are highly simplistic, values are immediately substituted in without writing them separately as a list of values and all answers are rounded to one whole number.
1.a)
1.b)
1.c)
1.d)
Δx = uxt
Δx = (100cos30)(10) m
Δx = 866 metres
2.
3.a)
3.b)
4.a)
4.a)
5.
Neglect:
- Air resistance
- Friction
- Curvature of the surface of the earth
Use the necessary data and formula provided by the data sheet here
unless values are already given:
www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/physics_formulaesheet.pdf
1.
A projectile is fired with an initial velocity 100 m/s at 30 degrees above the horizontal.
Find:
a) The vertical component of the initial velocity
b) The maximum height
c) The time of flight
d) The range
2.
A golf ball is hit on a flat course and its maximum height
is known to be 9.8 m and its range is 29.4 m.
Find the magnitude and direction of its initial velocity.
3.
A bomber is flying at a constant velocity of 200m/s at an altitude of 500m.
a)
Describe the pilot's observation of the path of the projectile when it
is dropped from the bomber.
b)
How long will it take a bomb dropped from the bomber to reach the ground ?
c)
At hwat distance in front of the target does the bomber have to drop its bomb
to hit the target ?
4.
The mass of the Earth is given as 5.98 x 1024 kg.
a)
From the Law of Energy Conservation, derive an expression and calculate the
escape velocity of a 200kg object.
b)
How would the escape velocity of the object differ if it was projected
at an angle of 45 degrees to the horizontal ?
5.
The radius of the Earth is given as 6380 km and the acceleration due to
gravity close to the Earth's surface is given as 9.8 ms-2
Calculate the mass of the Earth.
IDEAL ANSWERS:
You should only check them after you have attempted the questions.
Note: These answers are highly simplistic, values are immediately substituted in without writing them separately as a list of values and all answers are rounded to one whole number.
1.a)
u = uysinθ
u = 100sin30
u = 50 m/s
u = 100sin30
u = 50 m/s
1.b)
vy2 = uy2 + 2ayΔy
Let Δy be h for height.
vy2 = uy2 + 2ayh
(0) = (50)2 + 2 (-9.8) h
(0) = 2500 – 19.6h
19.6h = 2500
h = 128 metres
Let Δy be h for height.
vy2 = uy2 + 2ayh
(0) = (50)2 + 2 (-9.8) h
(0) = 2500 – 19.6h
19.6h = 2500
h = 128 metres
1.c)
Δy = uyt + ½ ayt2
Δy is zero when the projectile hits the ground.
(0) = (50)t + ½ (-9.8) t2
0 = 50t – 4.9t2
0 = 50 – 4.9t (Divide both sides by t to remove the power of two)
4.9t = 50
t = 10 seconds
Δy is zero when the projectile hits the ground.
(0) = (50)t + ½ (-9.8) t2
0 = 50t – 4.9t2
0 = 50 – 4.9t (Divide both sides by t to remove the power of two)
4.9t = 50
t = 10 seconds
1.d)
Δx = uxt
Δx = (100cos30)(10) m
Δx = 866 metres
2.
vy2 = uy2 + 2ayΔy
vy2 = 0 at max. height.
(0) = uy2 + 2ayΔy
uy2 = - 2ayΔy
uy2 = - 2 (-9.8) (9.8) m/s
uy2 = 192 m/s
uy = 14 m/s
Δy = uyt + ½ ayt2
Δy is zero when the projectile hits the ground.
(0) = (14)t + ½ (-9.8)t2
0 = 14 - 4.9t
4.9t = 14
t = 3 seconds
Δx = uxt
ux = Δx / t
ux = (29.4) / (3)
ux = 10 m/s
u = √(ux2 + uy2)
u = √[(10)2 + (14)2]
u = √296
u = 17 m/s
tan-1 (uy / ux) = θ
tan-1 [(14)/(10)] = θ
tan-1 1.4 = θ
θ = 54 degrees above the horizontal
vy2 = 0 at max. height.
(0) = uy2 + 2ayΔy
uy2 = - 2ayΔy
uy2 = - 2 (-9.8) (9.8) m/s
uy2 = 192 m/s
uy = 14 m/s
Δy = uyt + ½ ayt2
Δy is zero when the projectile hits the ground.
(0) = (14)t + ½ (-9.8)t2
0 = 14 - 4.9t
4.9t = 14
t = 3 seconds
Δx = uxt
ux = Δx / t
ux = (29.4) / (3)
ux = 10 m/s
u = √(ux2 + uy2)
u = √[(10)2 + (14)2]
u = √296
u = 17 m/s
tan-1 (uy / ux) = θ
tan-1 [(14)/(10)] = θ
tan-1 1.4 = θ
θ = 54 degrees above the horizontal
3.a)
Straight line
3.b)
Δy = uyt + ½ ayt2
Δy will be – 500 metres from its original position when it strikes the ground.
uy is zero as the bomber is only flying horizontally.
(500) = (0)t + ½ (-9.8) t2
– 500 = – 4.9t2
500 = 4.9t2
t2 = 102
t = 10 seconds
Δx = uxt
Δx = (200)(10)
Δx = 2000 metres
Δy will be – 500 metres from its original position when it strikes the ground.
uy is zero as the bomber is only flying horizontally.
(500) = (0)t + ½ (-9.8) t2
– 500 = – 4.9t2
500 = 4.9t2
t2 = 102
t = 10 seconds
Δx = uxt
Δx = (200)(10)
Δx = 2000 metres
4.a)
EK + EP = 0
½mv2 + (– GmM / r) = 0
½mv2 – GmM / r = 0
½mv2 = GmM / r
mv2 = 2GmM / r
v2 = 2GM / r
v = √(2GM / r)
v = √[2(6.67 x 10-11)(5.98 x 1024) / (6,380,000)]
v = 11,182 m/s
½mv2 + (– GmM / r) = 0
½mv2 – GmM / r = 0
½mv2 = GmM / r
mv2 = 2GmM / r
v2 = 2GM / r
v = √(2GM / r)
v = √[2(6.67 x 10-11)(5.98 x 1024) / (6,380,000)]
v = 11,182 m/s
4.a)
The escape velocity of the object is independent of the angle of trajectory.
5.
F = mg
F = GmM / r2
mg = GmM / r2
GM / r2 = g
M / r2 = g/G
M = g r2/G
M = (9.8)(6,380,000) / (6.67 x 10-11)
M = 6 x 1024 kg
F = GmM / r2
mg = GmM / r2
GM / r2 = g
M / r2 = g/G
M = g r2/G
M = (9.8)(6,380,000) / (6.67 x 10-11)
M = 6 x 1024 kg
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