* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Year-11-solutions-to-test-on-Newton`s
Roche limit wikipedia , lookup
Negative mass wikipedia , lookup
Coriolis force wikipedia , lookup
Electromagnetism wikipedia , lookup
Newton's law of universal gravitation wikipedia , lookup
Lorentz force wikipedia , lookup
Fictitious force wikipedia , lookup
Centrifugal force wikipedia , lookup
Year 11 set 3 topic test on Newton’s laws solutions 1a • Explain the difference between mass and weight • Mass is the amount of substance in a body and does not change if the object is taken into space [1] • Weight is the pull of gravity on that mass and will be different on the moon [1] 1b • What is the weight of a 150g mass? • You have to change the mass into kg and then multiply by 10 • 0.15 kg x 10 = 1.5N [2] 2a • What does it mean when we say two forces are balanced? • They are the same size and are acting on the object in opposite directions [2] 5N 5N 2b • Explain what Newton’s law says when the forces acting on an object are balanced • When forces are balanced there is no resultant force and so there is no change in velocity (remain at rest or at same speed in same direction) [2] 2c • A parachutist of weight 700N has a constant velocity, called terminal velocity. Draw a simple diagram to show the direction and size of the forces acting on the person. 700N from air friction on parachute [4] 700N from weight 3a • Write Newton’s Second law as an equation • • • • F = ma Where F = resultant force m is the mass in kg a is the acceleration in m/s2 [3] 3b • What does ‘resultant force’ mean • The force left over after you have subtracted forces pulling in opposite directions, or the single force that could replace two forces pulling at an angle to each other [2] 3c • A box of mass 20kg, pulled along a floor with a force of 75N, accelerates at 2m/s2. Calculate the friction force • We use F = ma, but we need to know F • Resultant force F = (pull – friction) • So (75 – friction) = 10 x 2 • Changing sides, 75 - 20 = friction • Friction = 55N [3] 3d • A car is moving from left to right when the breaks are applied. Draw a diagram to show the direction of motion, the 2 horizontal forces, and the direction of acceleration. acceleration Force from breaks Direction of motion, but the car is slowing down Driving force [4] 4 • Two forces are acting on a small object. One is 5N acting to the right, and the other is 6N acting upwards on the page. Use scale drawing or calculation to find the resultant force You could use Pythagoras Use 1cm = 1N for your scale and then find the length of the red line to the nearest mm. 6N It should be about 7.8cm so answer is 7.8N [4] 5N 5a A car is turning a wide corner at a constant speed • Explain how it has a changing velocity • Velocity changes if speed or direction changes, and in this case the car is changing its direction each second so it has an acceleration [2] 5b • What is the name of the force causing this change in velocity? What direction does this force have? • The force is called centripetal force and it acts towards the centre of the circle or the curve [2] 5c • For the example with the car, what is causing this force? • The turning force is from the sideways friction of the tyres on the road, and without it the car would slide or skid straight ahead [2] 6a • How do we calculate the moment of a force? • The moment, or the turning effect of a force is the force x distance between the force and the pivot. The distance has to be the perpendicular distance. [2] Distance to force from pivot 6b • A uniform rod of length 80cm is balanced on a pivot with a mass of 200g attached 5cm from the left end. Draw a diagram clearly showing these details and the position of the pivot. 40 cm 5cm a Weight of 200g mass b We do not know what distances a and b are yet Weight of the rod at the centre of mass [5] 6c • Calculate the mass of the rod in grams • We cannot work this out because the question has not given us the distance b • moment clockwise from rod’s mass is mass of rod x distance b • Moment anticlockwise is 200 x distance a [this question will not be counted] 6d • The 200g mass is now removed and replaced by 100g. Calculate the new position for the pivot in order to balance the rod. • Once more, we cannot calculate this because the question has not given us enough information. The pivot would need to me moved away from the 100g mass to increase the distance a • [this question will not be counted] 7 • Using three drawings, show examples of stable, unstable, and neutral equilibrium. Stable Unstable – the slightest movement will make it fall over neutral [6]