1. a) 7.75 m/s.
b) 0.00775 km/s.
c) 27.9 km/h.
2. a) 500 seconds and
b) 8⅓ minutes.
3. 39 735 360 000 000 kilometres (roughly 40 trillion
... [Show More] kilometres).
4. a) Graph (b) (the steepest straight line).
b) Graph (d) (change in gradient from positive, moving away from the start, to negative moving
back).
c) Graph (c) (the gradient is continuously increasing).
d) Graph (a) (straight line, less steep than graph (b)).
5. a) 25 m/s.
b) 3.125 m/s2.
6. a) 3.5 m/s2.
b) At a constant velocity of 14 m/s.
c) 140 m (area under graph = area of triangle + area of rectangle (2 × 14 + 8 × 14) m).
d) 7 m/s.
7. a) b) The sum or resultant of balanced
forces is zero as they are equal
in size, but opposite in direction.
8. a) b) When the string is cut the tension
force will no longer act on the
balloon.
c) The forces will no longer be balanced; there will be a resultant force upwards which will cause
the balloon to accelerate vertically upwards.
d) As soon as the balloon starts to move it will be subjected to a (viscous) drag force opposing its
upward motion through the air. This force will increase with the speed of the balloon, so the
balloon will accelerate quickly at first, then more and more slowly until the three forces balance
once again. The balloon will then rise at a steady terminal speed.
Physics Revision Guide
2
9. An unbalanced force can cause an object to accelerate (speed up), decelerate (slow down) or change
the direction in which it is moving.
10. a) Between A and C (the straight part of the graph) Hooke’s Law is obeyed. At C Hooke’s Law
ceases to be obeyed.
b) The elastic limit or the limit of proportionality.
11. a) 54 000 000 N (b) The drag force due to its movement through the water. (c) It must be greater
than the drag force by 54 000 000 N to cause the acceleration.
12. a) 8500 N.
b) 1.9 m/s2. (If the LEM weighed 7500 N on the Moon, its mass = weight ÷ Moon gravity, approx
4490 kg).
c) As the rocket expels the products of burning fuel, the mass of the rocket decreases; if the thrust
is constant then the acceleration will increase.
13. 0.008 kg (or 8 grams).
14. a) Car C. It has the largest area under the graph line.
b) Car A. The horizontal section of the graph shows no braking (constant velocity) because the
driver has yet to react to the hazard – in the graph for car A this is the shortest. The reaction time
for the other two graphs could be because the driver was tired, under the influence of alcohol or
other drugs, was distracted by use of a mobile phone or because visibility was poor (any two).
c) If you cannot see the car in front you cannot ensure you have left enough braking distance and
you are likely to take longer to realise that there was an emergency stop required. The faster you
are going and the longer you take to react (graph C) the greater the distance needed to come to a
halt.
d) The slope of the braking part of the graph would be less steep because you could not brake as
hard without skidding. This would mean that the total stopping distance (thinking distance +
braking distance) would be greater still.
15. a) The hammer and feather hit the Moon’s surface at the same instant. It showed that in the
absence of an atmosphere, all objects fall with the same acceleration.
b) The hammer would hit the ground after a shorter time than it did on the Moon, because the
Earth’s gravity is about six times stronger than the Moon’s. The feather would also accelerate
more quickly, but the effect of air resistance would mean it soon reached a terminal velocity and
therefore took noticeably longer to reach the ground.
16. a) The acceleration is virtually constant because the resultant force on the object is virtually
constant.
b) When the speed of the object increases, the drag force due to air resistance increases. Eventually
this becomes significant and the resultant force becomes noticeably smaller – so too does the
acceleration.
c) The terminal velocity.
d) Zero. The forces on the object are balanced. [Show Less]