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25 Cards in this Set
- Front
- Back
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Explain why the following statement is incorrect:
‘If you don’t push something it won’t move.’ Use an example to illustrate your answer. |
Objects move without a resultant force on them; they change their motion when there is a resultant force on them. If they are already moving they carry on moving if there is no resultant force. An example could be a skater moving in a straight line. |
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Use a labelled diagram to describe all of the forces that would be acting on a cyclist riding along a straight road at a steady speed.
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The diagram should show two larger arrows of the same length pointing in opposite directions vertically, the downward one labelled weight and the upward one labelled reaction. There should be two smaller arrows in opposite directions and of equal length horizontally, the one in the same direction as the bike movement labelled push and the other could be friction or air resistance. |
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Explain how three forces, all of which are larger than 40 N, can produce a resultant force of less than 10 N. |
Forces can cancel each other out, and so for example two forces to the right of 50 N each could be opposed by a 95 Newton force in the opposite direction, giving a resultant force of 5 N to the right. |
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Explain the difference between length and extension for a spring.
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The length of a spring is the overall length of a spring, including any extension. The extension is the amount by which the spring’s length increases. |
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Use the data in the Table 1 to produce a graph of the results. Describe why this graph supports the suggestion that the force applied is proportional to the extension of the spring. From the data work out the spring constant in N/m. |
Marks for suitable axes with labels, accurate point-plotting and best-fit line. The extension is proportional to the force given that this graph is of length not extension. If the graph was re-plotted for extension against force then the line would be straight and through the origin. F = k x. Rearranging, k= F/x. Therefore k = 1/0.05 = 20 N/m |
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A stiffer spring of the same length was tested using the same forces. Add a line to your graph that shows the relationship between force and extension for this new spring. A weaker spring of the same initial length was also tested. It was found that, after the experiment, its length when the force was removed was 7.2 cm. Explain why this might have happened. |
The line would be a straight line starting at 6 cm and steeper than the original.
The spring had been overstretched and started to deform. It stopped behaving elastically. |
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The following is an extract from a live radio commentary during the launch of a space vehicle. ‘This massive man-made machine is slowly inching its way off the launch pad. All 500 tonnes of it are balanced on a pillar of flame, and now it starts to gather speed and climb into the night sky.’ Use the information in this passage to fully describe the forces involved with the take-off of this vehicle. |
The weight of the spacecraft will be 5000 kN when it is resting on the launch pad. The launch pad will react with an equal and opposite force. As the spacecraft lifts off, it must have an upwards force of more than its weight and it accelerates upwards. The upwards force is a reaction to the force of the hot gases being pushed out by the engines. |
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When somebody dives off the high board in a swimming pool, they move upwards to start with and then plunge into the water. The high board is further off the surface of the water than the depth of the water. Use your understanding of forces to describe fully the motion of the diver from the time that they leave the board to the moment they reach their deepest point inthe water. |
The diver jumps upwards, but immediately starts to accelerate downwards because their weight is no longer balanced by the reaction force of the diving board pushing upwards. After the initial upward acceleration of the jump they slow down, stop, and begin to move downwards with increasing speed. When they hit the water they slow down quickly, because the force of water resistance on them is much larger than their weight. They then quickly come to a halt. |
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Three guns are used to launch fireworks into the sky at a display.
The first applies a force of 40 N to a firework that has a mass of 200 g. The second throws a 500 g firework with a force of 80 N and the third gives an acceleration of 180 m/s^2 to a 300 g firework. (a) Which firework has the greatest force on it? (b) Which firework has the greatest acceleration |
(a) F = ma. Rearranging, a= F/m.
For first gun a= 40/0.2 = 200 m/s^2. For second gun a= 80/0.5 = 160 m/s^2. For third gun F= ma= 0.3 × 180 = 54 N. Second firework: fired with a force of 80 N. (b) First firework: has acceleration of 200 m/s^2. |
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The Figure shows a distance–time graph fora car. Calculate its speed at a, b and c. |
Speed at a = 0 m/s speed at b = 3 m/s at c = 6 m/s. 0 |
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Use the data in the Table to draw a graph of velocity against time for a radio-controlled car. |
Suitable graph drawn from data: Suitable axes with labels. Accurate plotting of points. Line to fit points. |
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(a) Calculate the distance travelled by the car. (b) Calculate the maximum acceleration of the car. (c) If the mass of the car is 2 kg, what is the greatest accelerating force on the car? |
(a) 39 m.
(b) –3 m/s^2. (c) –6 N. |
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A driver in a car sees a deer run out into the road and puts on the brakes.Unfortunately the car hits the deer. Explain why it is much more likely that the deer will be severely injured if the car was travelling at 40 miles per hour than if it was travelling at 30 miles per hour. |
First, the stopping distance is larger because the thinking distance is greater, so the car is closer to the deer before it starts to slow down. It is also slowing from a higher speed. These two factors together mean that the car will hit the deer at a higher speed. The deer may also be dragged further before the car stops. The higher speed of impact means that a greater force will be exerted on the deer’s body, causing more injury. |
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A car is travelling at 5 m/s when a second car pulls out into the road. The first car decelerates at 2 m/s^2. Draw a velocity–time graph to show the motion of the first car as it stops. From the graph work out the distance the car travels before it stops. |
Suitable axes with labels. Accurate plotting of points. Line to fit points. It travels 6.25 m. |
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A car with a mass of 800 kg brakes with a force of 2000 N. Calculate its deceleration. |
F = ma. Rearranging, a = F/m. Therefore a = 2000/800 = 2.5 m/s^2. |
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A tractor is pulling a plough through soil with a force of 1500 N. The tractor drags the plough a distance of 3.2 km during the day. The tractor has a mass of 3 tonnes and the plough weighs 500 kg. How much work does the tractor do to pull the plough? |
Work done W = F d = 1500 N × 3200 m = 4 800 000 J. |
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As it moves along, a car engine not only pushes the car forward, but also generates electricity to recharge the battery and run the lights. Two students discussing this issue disagree. One student thinks that the car is moving anyway so the lights do not take any fuel to run. The other student insists that it must usemore fuel to have the lights on than to have them off . Choose one student’sargument to support and explain which student is correct. |
You cannot create or destroy energy, so turning the lights on does use extra fuel. The engine has to push a bit harder to drive the belts that drive the generator. |
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Below is an excerpt from a magazine article about regenerative braking used in trains in Switzerland. Read the article and then explain why this system would be unlikely to have many applications in this country.
‘In Switzerland, many of the train lines in the more mountainous areas use overhead electric power lines and regenerative braking. This reduces power consumption on the railway. In some places, one train engine going up a mountain can be powered by the descent of two other trains. The electric motors that are used to push the trains up the mountains turn into generators when they come down again, feeding electricity back into the overhead power lines.’ |
There are very few places in this country where this system would be useful. Most of our train lines are built on flat land and so there are very few chances to make use of regenerative braking. |
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It takes a parent 10 seconds to push a pushchair up a slope. The parent pushes with a force of 80 N and the slope is 5 metres long. (a) How much work has the parent done? (b) Whatwas their power output? |
(a) W = F d = 80 N × 5 m = 400 J (b) P = E/t = 400 J/10 s = 40 W : |
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A power station’s power output is 500 MW. If the power station was only 40% efficient, how much energy would it need, in the form of fuel, to operate for 1 hour? |
E = P × t. So energy output in 1 hour = 500 000 000 W × 3600 s = 1 800 000 000 000 J. Efficiency = 40%. Energy needed for 1 hour = 100 × 1 800 000 000 000/40 = 4 500 000 000 000 J = 4.5 × 10^12 J. |
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If a 100 g ball is thrown up into the air at 5 m/s, and assuming that no energy is lost, calculate the following: (a) The kinetic energy that the ball has initially. (b) The height it will reach before it stops. (c) The speed it will be travelling at when it comes down again. (d) Inreality, as the ball comes down, it will move a little more slowly that when it was thrown up, and it will not get quite as high as you have calculated. Explainwhy this is so. |
(a) Ek = ½ mv^2 = 0.5 ×0.1 × (52) = 1.25 J (b) Ep = mgh. Rearranging, h = Ep/mg = 1.25 J/(0.1 ×10) = 1.25 m. (c) 5 m/s. (d) Resistive forces will remove some of the energy, so it will not get as high, or be travelling as fast as these calculations suggest. |
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In an experiment using a friction-compensated runway and an 800 g trolley, students measured the velocity of the trolley as it was pulled down the slope by different forces. (a) Describe what is meant by a friction-compensated slope. (b) Calculate the acceleration of the trolley if a force of 0.4 N pulled it down the slope. Show your working. (c) If the force of 0.4 N continued for 3 s, what would be the change in the velocity of the trolley? Show your working. (d) Using the answer from part (c), calculate the increase of momentum of the trolley after 3 s. Show your working. (e) The trolley is stopped by a brick placed at the end of the runway. To protect the trolley, the teacher asked the pupils to place a piece of sponge in front of the brick. Use your understanding of the change in momentum of colliding masses to explain how the sponge protects the trolley. |
(a) Gravity pulling the trolley down the slope compensates for the frictional forces that would slow the trolley down. The result is that if the trolley is released down the runway, it will move at a constant speed if there is no resultant force on it. (b) Use a = F/m.(1 mark) a = 0.4/0.8 = 0.5 m/s^2. (c) Use a= (v– u)/t. Rearranging, v – u = a × t. Change in velocity = 0.5 m/s^2 × 3 s = 1.5 m/s, or 3 times the answer to part (b). (d) Use momentum change = mass × velocity change. Momentum change =0.8 kg × 1.5 m/s = 1.2 kg m/s, or 0.8 times the answer to part (c). (e) To change the momentum of a mass quickly, a large force is needed. The sponge increases the time it takes to slow the trolley down so the force on the trolley is less. This protects the trolley. |
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A student trying to model the way that a crumple zone works used a steel tube strapped to an egg. The egg is dropped from a height of 20 cm on to the bench. Another student used a tube made of thin cardboard. Explain why one student ended up with a broken egg and the other ended up with a crumpled tube. |
The steel tube is stiff and transfers the force of the impact straight to the egg, so it breaks. The cardboard tube is less stiff and crumples on impact. This spreads the force of the impact over a longer period, reducing the maximum force on the egg. The egg therefore does not break. |
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Explain how the ideas in the previous question relate to the design of crumple zones in cars that are made from steel sections. |
The forces in a car crash are much larger, so specially designed steel sections do bend during a crash. The designers must not make the crumple zone too stiff, otherwise it will not crumple in a crash and the passengers will be injured just like the egg. |
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A 400 g toy plane moving at 4 m/s crashes into a wall and is damaged. (a) How much momentum did the plane have before the crash? (b) How much momentum did it have after the crash? (c) How much kinetic energy did it havebefore the crash? (d) How much kinetic energy did it have afterthe crash? (e) Describe where the momentum and kineticenergy have gone. |
(a) p = m × v = 0.4 kg ×4 m/s = 1.6 kg m/s (b) 0 (c) Ek = ½ mv^2 = ½ × 0.4 kg ×(42) m/s = 3.2 J (d) 0 (e) The momentum is conserved, as the wall will move backwards ever so slightly. The energy is conserved and will be transformed into heat and sound. |
