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WorksheetsRugby physics
Total questions: 25
Worksheet time: 28mins
The bar is 80cm between the masses. The Masses on the ends are both 40kg but Savea has his right hand 10cm from the right mass and his left hand is 20cm from the left mass. The mass of the bar is negligible. What are the 2 conditions for equilibrium?
the total force on the left = total force on the right
the total force up = total force down
total torque clockwise = total anticlockwise torque
total torque clockwise = total anticlockwise torque and the total force up = total force down
total torque up = total torque down and the total force clockwise = total force anticlockwise
The masses are 80 cm apart. The right mass (40kg) is 10cm from the right hand and the left hand is 20 cm from the left mass (40Kg). Which hand provided more force?
left hand
right hand
both the same
not enough information provided
The masses (both 40 Kg) are 80 cm apart. The right mass is 10cm from the right hand and the left hand is 20 cm from the left mass. What is the clockwise torque around Savea's left hand from the right hand mass?
T = F x r = 40 N x 0.8 m
T = F x r = 400 N x 0.6m
T = F x r = 40 N x 0.6 m
T = F x r = 400 N x 0.8 m
T = F x r = 40 N x 0.2m
What is the force from the right hand holding up the bar around the left hand?
F = ΣT/r = (24 Nm + 8Nm) / 0.5
F = ΣT/r = (24Nm - 40 Nm) /0.5
F = ΣT/r = (24Nm -12Nm) / 0.4
F = ΣT/r = (24Nm -8Nm) / 0.4
F = ΣT/r = (24Nm - 8Nm) / 0.5
Richie Mo'unga (95kg) is moving at 3 m/s and is tackled by Makazole Mapimpi (85kg) at 4 m/s. What speed do they both move forward at?
v= p/m = (95 x 3) + (85 x 4) / (95+85)
v= p/m = (85 x 3) + (95 x 4) / (95+85)
v= p/m = (95 x 3) + (85 x 4) / (95)
v= p/m = (95 x 3) + (85 x 4) / (85)
Patchell drops the ball (500g) so it hits his foot at 3 m/s. The ball contacts for 0.1 s and leaves his foot at 10 m/. What was the change in momentum of the ball?
change in p = (500 x 3) - (500 x 10)
change in p = (500 x 10) - (500 x 3)
change in p = (0.5 x 10) + (0.5 x 3)
change in p = (0.5 x 10) - (0.5 x 3)
Patchell drops the ball (500g) so it hits his foot at 3 m/s. The ball contacts for 0.1 s and leaves his foot at 10 m/s. What was the force of his foot on the ball?
F = (change in p)/t = 8.5 / 0.1
F = ma = 0.5 x 10
F = (change in p)/t = 3.5 / 0.1
F = ma = 0.5 x 9.8
Patchell "follows" through with his kick. He does this because
he cant stop his foot
he wants to increase the change in momentum, by making the force impact for longer
he wants to stop the ball from accelerating
Sonny Bill side steps to change direction. If he is running forward but wants to go left, his foot must apply a force that causes the ground to apply a force on him that is directed ...
left
backward
at an angle between backwards and left
right
Sonny Bill side steps to change direction. If he is running forward at 3 m/s but goes left at 3 m/s. His ...
velocity has not changed
speed has not changed and has no acceleration
velocity has not changed and has no acceleration
velocity has changed and he has acceleration
Bauden kicks the penalty at 12 m/s, so that the angle with the ground is 40 degrees. What is the vertical component of the initial velocity?
12 m/s
12sin40
12cos40
12tan40
Bauden kicks the penalty at 12 m/s, so that the angle with the ground is 40 degrees. How long does it take to reach the highest point of its path?
t = (vf - vi)/a
t = (vf - vi)/2a
t = vf2 - Vi2/2d
t = d/v
Bauden kicks the penalty at 12 m/s, so that the angle with the ground is 40 degrees. How far does it travel horizontally?
d = (vf + vi)t/2
d = (vf2 - vi2) /2a
d = ½at2
d = vh x t
Bauden kicks the penalty at 12 m/s, so that the angle with the ground is 40 degrees. How high does get?
d = (vf + vi)t/2
d = vh x t
d = (vf2 - vi2)2a
d = ½at2
Coles throws into the line out at 8 m/s, at an angle of 60 degrees to the horizontal. What is the horizontal component of the initial velocity?
vh = 8tan60
vh = 8 m/s
vh = 8cos60
vh = 8sin60
Coles (2 m tall) throws into the line out at 8 m/s, at an angle of 60 degrees to the horizontal. Sam Whitelock (who is also 2m tall) is 5 metres back from the throw in. How long does it take for the ball to be above his head?
t = d/vh
t = (vf - vi)/a
t = 2d/(vf + vi)
t = √(2d/a)
Coles throws into the line out at 8 m/s, at an angle of 60 degrees to the horizontal. How high must Sam Whitelock jump if he stands 5m from the throw in?
d = vvt ½at2
d = (vf + vi)t/2
d = vv x t
d = (vf2 - vi2)/2a
The match ball is inflated so that 10 N of force compresses it by 1 cm. What is the spring constant for the ball?
k = Fx = 10 x 1
k = x/F = 0.1/10
k = F/x = 10/0.1
k = F/x = 10/1
The ball is compressed 2cm from a bounce, what height is the rebound?
h = mg/(½kx2)
h = (½kx2)/mg
h = mg - (½kx2)
h = (½mv2)/mg
What assumptions are you making, to determine the height the ball reaches on the rebound? What law is used to determine the height of the rebound?
Assume that gravity is constant. Law of conservation of momentum
Assume the ball is normally size and mass. Law of conservation of mass
Assume that gravity is 9.8 m/s2. Law of conservation of speed
Assume no energy is lost as heat due to air resistance (or friction due to the stretching of the rubber). Law of conservation of energy
How does headgear protect a player in a scrum?
it is soft and light so it does not hurt when being worn to reduce the force of gravity on the head
it is hard so it cant be crushed and absorbs energy instead
it compresses so that the time of collision increases while the change in momentum of the player remains the same.This reduces the force since F = Δ p / Δ t
it compresses so that the time of collision decreases while the change in momentum remains the same. This reduces the force since F = Δ p / Δ t
Spidercam is suspended by 4 wires above the stadium. What statements are true?
The sum of Forces on the spidercam = zero
The sum of tension Forces = weight force
The vector sum of Forces = zero
There are 4 forces acting on the spidercam
The spidercam moves right. Which statements are true?
The tension forces remain the same and the weight forces remain the same
The tension in the wires on the right side are greater than the tension forces in the wires on the left side. The weight force is the same.
The tension in the wires on the right are less than the tension forces in wires on the left side. The weight force is the same.
The tension in the wires on the right are greater than the tension forces in the wires on the left side. The weight force is the more than before.
This is a VECTOR DIAGRAM of the spidercam. Which of the following statements are true?
The red vector is the weight vector. The blue vectors are the tension forces added nose to tail. The resultant is zero
The red vector is the resultant vector. The blue vectors are the tension forces added nose to tail
The red vector is the weight vector. The blue vectors are the tension forces added on the left and subtracted on the right
The red vector is the resultant vector. The blue vectors are the tension forces added on the left and subtracted on the right
The masses (both 40 Kg) are 80 cm apart. The right mass is 10cm from the right hand and the left hand is 20 cm from the left mass. What is the anticlockwise torque around Savea's left hand from the left hand mass?
T = F x r = 40 N x 0.8 m
T = F x r = 400 N x 0.6m
T = F x r = 40 N x 0.6 m
T = F x r = 400 N x 0.8 m
T = F x r = 40 N x 0.2m
