Year 11 Physics Exam Questions Part 2: Moving About
Posted on May 2, 2017 by DJ Kim
Year 11 Physics Exam Questions: Moving About
Assess your depth of knowledge and understanding for the Year 11 Physics module ‘Moving About’. Work through these Moving About exam questions below!
You can refer to the Glossary of Key Words for the meaning of any verbs in the questions.
Detailed answers are provided at the bottom of this blog post.
What is the difference between a vector and a scalar?
What is the result of subtracting 10 m East from 10 m North?
If you travel around a square track, where each side of the square is 50 m long, and come back to your starting position what distance did you travel and what is your displacement?
What is the difference between instantaneous speed and average speed?
If I travel at 40 km/h North for 1 hour, and then at 60 km/h West for 30 minutes, what was the average speed and the average velocity? Express your answer in SI units.
Consider the graph below showing the position of a person at different times.
(a) What was the person’s displacement at t = 6s?
(b) What was the person’s average speed between t = 2 and 7 s?
(c) What was the person’s average velocity between t = 2 and 7s?
A plane can fly at 1000 km/h and heads North at full speed. It encounters a cross-wind of 120 km/h from the west. What will the plane’s velocity be relative to the ground?
A boat that can travel at 5 m/s wishes to cross a 200 m wide river. It needs to travel North to reach a point directly across its starting position. There is a current in the river flowing at 3 m/s from East to West. In what direction should the boat head, and how long will it take it to cross the river?
A woman is driving West at 20 m/s on a still day with no wind. She notices a bird flying South East relative to her. If the speed of the bird relative to the woman is 8 m/s, how fast is the bird flying compared to the ground?
Consider the graph below showing the motion of a car in a straight line.
(a) What distance did the car travel between t = 0 and 8 s?
(b) What was the displacement of the car at the end of the trip?
(c) In which time period did the car slow down?
(d) What was the car’s acceleration between t = 0 and 4 s?
(e) What was the car’s acceleration between t = 4 and 8 s?
(f) If the car’s mass was 1800 kg, what was the average force on the car between t = 4 and 8 s?
Compare mass and weight.
Explain the following situations in terms of Newton’s Laws:
(a) A cyclist has to keep pedalling to move at constant velocity.
(b) A rocket accelerates forwards as the rocket engine ejects the hot exhaust gas backwards.
A car is parked on an inclined plane. Draw a free body diagram showing the forces acting on the car, and explain why there must be a frictional force up the plane.
A Formula 1 car is driving around a circular curve in the track at 30 m/s. If the radius of the curve is 25 m and the mass of the car is 1250 kg, what is the net force acting on the car, and where does this force come from?
A 1 kg ball is held out of a window 10 m above the ground and released.
(a) What was the potential energy of the ball when it was released?
(b) When the ball reaches a height of 5 m of the ground, how fast will it be travelling?
A donkey pulls a cart that is tied behind it with a rope. The rope makes an angle of 20 degrees with the horizontal. If the donkey pulls with a force of 1000 N, and the cart weighs 600 kg and starts from rest, how fast will the cart be moving after being pulled for 2 m?
Explain in terms of work and impulse why a lower speed limit, like in School Zones, is safer.
A car of mass 2000 kg is travelling East at 16 m/s and collides with a 2500 kg van travelling West at 15 m/s.
(a) After the collision the van remains stationary. What is the final velocity of the car?
(b) How would you compare the impulse experienced by the car and the van?
Explain why crumple zones in cars reduce the severity of injuries for the passengers.
A scalar quantity only has a magnitude, whereas a vector quantity has both magnitude and direction.
10 m North – 10 m East = 10 m North + (– 10 m East) = 10 m North + 10 m West = 14.1 m N45°W.
The distance is 200 m and the displacement is 0.
The instantaneous speed is how fast you are travelling at a given moment in time. The average speed is equal to the total distance covered divided by the total time taken for a trip.
Average speed = (total distance)/(total time) = (40 x 1 + 60 x 0.5)/1.5 = 46.7 km/h = 13.0 m/s
To find the average velocity you must first find the displacement.
Displacement = 40 km North + 30 km West = 50 km N37°W
Average velocity = displacement/time = (50 km N37°W)/1.5 = 33.3 km/h N37°W = 9.3 m/s N37°W
(a) 5.5 m
(b) Speed = distance/time = 4/5 = 0.8 m/s
(c) Velocity = displacement/time = -4/5 = -0.8 m/s
As the plane is moving in the air, and then air is moving relative to the ground, the velocity of the plane relative to the ground will be given by the vector addition:
vplane rel. to ground = vplane rel. to air + vair rel. to ground = 1007 km/h N7°E
The boat travels at 5 m/s relative to the water, and the water travels at 3 m/s W compared to the ground. The final velocity required is the velocity of the boat relative to the ground, which is given by the vector addition:vboat rel. to ground = vboat rel. to water + vwater rel. to ground
For the boat must travel North relative to the ground, it must head in a direction of N36.9°E
Its speed relative to the ground will be 4 m/s, so it will take 50 s to cross the river.
The velocity of the bird relative to the ground is given by the following vector addition:
Vbird rel. to ground = vbird rel. to woman + vwoman rel. to ground
The velocity of the bird relative to the ground is 15.4 m/s S68.5°W
(a) Distance is the area under the graph (area below axis treated as positive) = 54 m
(b) Displacement is the area under the graph (area below axis treated as negative) = 30 m
(c) The car slows down when the speed decreases, between t = 5 and 6 s
(d) a = (v-u)/t = (10-0)/4 = 2.5 m/s2
(e) a = (v-u)/t = (-8-10)/4 = -4.5 m/s2d
(f) F = ma = 1800 x -4.5 = -8100 N
Mass a measure of inertia, how much an object resists changes to its state of motion whereas weight is the force with which an object is attracted to the Earth due to the Earth’s gravity, given by w = mg, where m is the mass and g is the gravitational field strength, g = 9.8 N/kg.
(a) The cyclist experiences air resistance and friction, so must keep pedalling to produce a force that cancels out the air resistance and friction. This makes the net force on the cyclist zero and they travel at constant velocity (Newton’s First Law).
(b) As the rocket pushes the exhaust gas backwards, there is an equal and opposite force from the gas pushing the rocket forwards (Newton’s Third Law). Since there is a net external force acting on the rocket, it will accelerate (Newton’s Second Law).
The free body diagram should show:
- The weight force downwards
- The normal reaction force perpendicular to the inclined plane
- A frictional force upwards.
The truck is parked, and hence not accelerating, so the net force on the truck is zero. The weight can be decomposed into a component perpendicular to the slope, which is cancelled out by the normal force. There is also a component of the weight (mg sinθ) which is down the slope. This requires a frictional force up the slope to cancel it out.
F = mv2/r = (1250 x 302)/25 = 45000 N towards the centre of the curve. This centripetal force is provided by the friction between the tyres and the road.
(a) EP = mgh = 1 x 9.8 x 10 = 98 J
(b) At a height of 10 m the ball was stationary so its kinetic energy was zero. The total energy was EK + EP = 0 + 98 = 98 J. As energy is conserved, the total energy will remain at 98 J as the ball falls. At a height of 5 m the potential energy will be 1 x 9.8 x 5 = 49 J, hence the kinetic energy will be 98 – 49 = 49 J.
EK = ½ mv2 = 49 J, so v = 9.9 m/s.
The work the donkey does on the cart is given by W = F x distance x cosθ = 1000 x 2 x cos 20° = 1879 J. This work will increase the cart’s kinetic energy, so EK = ½ mv2 = 1879 J, so v = 2.5 m/s
As a car stops, its kinetic energy and its momentum will both reduce to zero.
A force is required (from the brakes of the car) to slow the car down. This force will have a given value, and is assumed to be a constant.
The force will be applied over some distance, meaning it does work (W = F x r) on the car to reduce its kinetic energy to zero. Also, the force will be applied over some time, meaning the car will experience an impulse (I = F x t), which will reduce its momentum to zero.The initial kinetic energy and the initial momentum will depend on how fast the car was going (EK = ½ mv2, p = mv). The slower the initial speed of the car, the smaller the kinetic energy and the momentum. Since the force is taken to be constant, a smaller kinetic energy and momentum mean a shorter stopping distance and stopping time.Hence, a lower speed limit is safer as it allows the driver to stop more quickly if necessary, so as to avoid a collision.
(a) Take East to be the positive direction.
Initial momentum = mcarucar + mvanuvan = 2000 x 16 + 2500 x (-15) = -5500 kgm/s
Due to conservation of momentum, the final momentum is also -5500 kgm/s.
Final momentum = -5500 kgm/s = mcarvcar + mvanvvan = 2000 x vcar + 2500 x 0vcar = -2.75 m/s = 2.75 m/s West
(b) The forces on the two vehicles are equal and opposite (Newton’s 3rd Law) and they experience these forces for the same amount of time, hence the impulses they experience are equal and opposite.
The crumple zones deform allowing the stronger passenger cell to slow down and stop over a longer time and distance. If it is slowing down over a longer time, the acceleration must be lower, so the net force on the passengers must be lower, and the severity of any injuries is reduced.
If you missed part 1 of this guide you can visit Year 11 Physics Exam Questions Part 1.
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