A cannon is shot an angle of 20°. Calculate the Vertical velocity if the cannon has an initial velocity of 40 m/s.
Vy = 40 Sin 20
Vy = 13.7 m/s
If 5 revolutions take 65 seconds. How long is one revolution.
13 seconds.
All planets move around the sun in an elliptical orbit.
Keplers 1st Law
Keplers 2nd Law
Keplers 3rd Law
Keplers 1st Law
Give a potential energy formula;
U = mgh
U = GMm/r
A planet covers equal areas in equal periods of time.
Keplers 1st Law
Keplers 2nd Law
Keplers 3rd Law
Keplers 2nd Law
A cannon is shot an angle of 20° with an initial velocity of 40 m/s.
Given the VERTICAL VELOCITY is 13.7 m/s
and the HORIZONTAL VELOCITY is 37.6 m/s
Calculate the time taken to reach the peak/maximum height.
Vy = Uy + at
0 = 13.7 + 9.8t
t = 1.4 seconds
What is the formula for ORBITAL VELOCITY?
V = 2 (pi) x Radius / Period
V = 2 (pi) x R/ T
A skateboarder moves off a ramp at a velocity of 5m/s at an angle of 30 degrees.
Calculate the X and Y components of the velocity.
Y = 5 Sin 30 Y = 2.5 m/s
X = 5 Cos 30 X = 4.3 m/s
Give the kinetic energy formula.
KE = 1/2 mv2
What is the formula for Centripetal force?
(Fc) = mv2 /r
Calculate the velocity of the cannon as it hits the ground given the intial y velocity was 0 m/s and the final x velocity was 20 m/s. The time taken was 5 seconds.
First find the final y velocity.
Vy = Uy + at
Vy = 0 + 9.8(5)
Vy = 49 m/s
V^2 = 49^2 + 20^2
V = square root (2401 + 400)
V = 53 m/s
Calculate the ORBITAL VELOCITY given the
RADIUS of the ride is 4m and the PERIOD (time taken to complete one revolution) is 20 seconds.
V = 2(pi) x radius / T
V = 2(pi) x 4 / 20
V = 1.26 m/s
Calculate the range for a skater after leaving at a ramp. Given the HORIZONTAL VELOCITY is 4.3 m/s.
Note the time taken was calculated to be 5 seconds.
Range = Horizontal velocity x Time taken
Range = 4.3 x 5
Range = 21.5 m
If the Kinetic energy (1/2 mv2) equals the Potential energy (GMm/r). Use these two formulas to derive the ESCAPE VELOCITY
Escape velocity = 2GM/r
The square of a planets orbital period is proportional to the cube of the radius.
T2 = R3
Keplers 1st Law Keplers 2nd Law Keplers 3rd LawKeplers 3rd Law
Calculate the range for a skater after leaving at a ramp. Given the HORIZONTAL VELOCITY is 2.3 m/s.
Note the time taken was calculated to be 6 seconds.
Range = 13.8 m
Calculate the ANGULAR VELOCITY given the VELOCITY is 2.3 m/s for a ride with a radius of 5m.
Angular velocity = Velocity/Radius
Angular velocity = 2.3/5
= 0.46 rads/sec
How long will the skaterboarder be in the air after moving off a ramp.
A skateboarder moves off a ramp at a velocity of 5m/s at an angle of 30 degrees.
NOTE the X and Y components of the velocity is
Y = 2.5 m/s
X = 4.3 m/s
Vy = Uy + at
0 = 2.5 + 9.8t
t = 0.23 seconds
Calculate the escape velocity for a 500kg rocket that is launched from Earth given
r Earth = 6.38 x 10^6 m
m Earth = 5.97 x 10^24 kg
V esc = √2GM/r
V esc = √(2(6.67 x 10^(-11) )5.97 x 10^24)/ 6.38 x 10^6
V esc = 1.12 x 10^4 m/s
Calculate the Centripetal force (Fc) acting on a 60kg person on a ride with a radius of 10m which is travelling at a velocity of 2m/s.
(Fc) = mv2 /r
= (60)(22) / 10
= 24N
How long will the skaterboarder be in the air after moving off a ramp.
A skateboarder moves off a ramp at a velocity of 5m/s at an angle of 30 degrees.
NOTE the X and Y components of the velocity is
Y = 5.5 m/s
X = 4.2 m/s
Vy = Uy + at
0 = 5.5+ 9.8t
t = 0.56 seconds
Calculate the ANGULAR VELOCITY
Angular velocity = Velocity / Radius
Note the following information is given;
RADIUS of the ride is 3m and the PERIOD and the is 15 seconds
V = 2 (pi) x 3 / 15
V = 1.26 m/s
Angular velocity = 1.26 / 3
Angular velocity = 0.42 rads/sec
A satellite with a mass of 800kg orbits Earth at an altitude of 35 800km.
Calculate the gravitational potential energy.
r Earth = 6.38 x 10^6 m
m Earth = 5.97 x 10^24 kg
radius = r Earth + altitude
= 6.38 x 10^6 + 3.58 x 10^7
= 4.22 x 10^7 m
U = - GMm/r
U = (6.67 x 10^-11)(5.97 x 10^24)(800)/ 4.22 x 10^7
U = - 7.55 x 10^9 J
A satellite with a mass of 1390 kg orbits Earth at an altitude of 357 km.
Calculate the SPEED of the satellite
Given the total energy is
4.11 x 10^10J
m Earth = 5.97 x 10^24 kg
K = ½ mv2
4.11 x 10^10 = ½ 1390 x v2
V2 = 5.91 x 10^7
v = √ 5.91 x 10^7
v = 7690 m/s
Jupiters largest moon has a mass of 1.66 x 10^23 kg and has an orbital radius of 1.07 x 10^6km with an orbital period of 7.15 days.
Use Kepler’s law to calculate the orbital radius of another of Jupiter’s moon which has an orbital period of 3.55 days.
T2 /R3 = T2 /R3
7.152 /1.7 x 10^63 = 3.552 /R3
R3 = (3.55)^2 (1.7 x 10^63) / 7.15^2
R3 = 1.21 x 10^18
R = 1.66 x 10^13 km