Rugby physics

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