Physics Problem Sets Answer Book
1
1 2 3 C
Motion I
7 (a) From 1 January 2009 to 10 January 2009, the watch runs slower than the actual time by 9 minutes. Therefore, when the actual time is 2:00 pm on 10 January 2009, the time shown on the watch should be 1:51 pm on 10 January 2009.
Practice 1.1 (p. 6)
D (a) Possible percentage error 10 −6 = × 100% 24 × 3600
= 1.16 × 10 % 1 (b) = 1 000 000 days 10 −6
–9
It would take 1 000 000 days to be in error by 1 s.
(b) Percentage error 9 = × 100% 9 × 24 × 60
= 6.94 × 10–2%
4
(a) One day
= 24 × 60 × 60 = 86 400 s
Practice 1.2 (p. 15)
1 2 3 4 5
C B D D
(b) One year
= 365 × 86 400 = 31 500 000 s
5
Let t be the period of time recorded by a stop-watch. Percentage error =
0.4 × 100% ≤ 1% t t ≥ 40 s
(a) Total distance she travels 2 × × 10 2 × × 20 2 × × 15 + + = 2 2 2
= 141 m
(b) Magnitude of total displacement
= 10 × 2 + 20 × 2 + 15 × 2 = 90 m Direction: east Her total displacement is 90 m east.
The minimum period of time is 40 s.
6
(a) Percentage error error due to reaction time = × 100% time measured
0.3 = × 100% 10 = 3%
6 7
His total displacement is 0. With the notation in the figure below.
(b) From (a), the percentage error of a short time interval (e.g. 10 s) measured by a stop-watch is very large. Since the time intervals of 110-m hurdles are very short in the Olympic Games, stop-watches are not used to avoid large percentage errors.
Since ZX = ZY = 1 m, α = β = 60°. Therefore, XY = ZX = ZY = 1 m The magnitude of the displacement of the ball is 1 m.
©
8
(a) The distance travelled by the ball will be longer if it takes a curved path.
7
(a) Length of the path
= 0.8 × 120 = 96 m
(b) No matter which path the ball takes, its displacement remains the same.
(b) Length of AB along the dotted line 96 = 30.6 m = (c)
Magnitude of Jack’s average velocity 30.6 × 2 = = 0.51 m s–1 120
Practice 1.3 (p. 23)
1
B Total time 5000 5000 = + = 9821 s 1.4 0.8 5000 + 5000 = 1.02 m
1 2 3 C
Motion I
7 (a) From 1 January 2009 to 10 January 2009, the watch runs slower than the actual time by 9 minutes. Therefore, when the actual time is 2:00 pm on 10 January 2009, the time shown on the watch should be 1:51 pm on 10 January 2009.
Practice 1.1 (p. 6)
D (a) Possible percentage error 10 −6 = × 100% 24 × 3600
= 1.16 × 10 % 1 (b) = 1 000 000 days 10 −6
–9
It would take 1 000 000 days to be in error by 1 s.
(b) Percentage error 9 = × 100% 9 × 24 × 60
= 6.94 × 10–2%
4
(a) One day
= 24 × 60 × 60 = 86 400 s
Practice 1.2 (p. 15)
1 2 3 4 5
C B D D
(b) One year
= 365 × 86 400 = 31 500 000 s
5
Let t be the period of time recorded by a stop-watch. Percentage error =
0.4 × 100% ≤ 1% t t ≥ 40 s
(a) Total distance she travels 2 × × 10 2 × × 20 2 × × 15 + + = 2 2 2
= 141 m
(b) Magnitude of total displacement
= 10 × 2 + 20 × 2 + 15 × 2 = 90 m Direction: east Her total displacement is 90 m east.
The minimum period of time is 40 s.
6
(a) Percentage error error due to reaction time = × 100% time measured
0.3 = × 100% 10 = 3%
6 7
His total displacement is 0. With the notation in the figure below.
(b) From (a), the percentage error of a short time interval (e.g. 10 s) measured by a stop-watch is very large. Since the time intervals of 110-m hurdles are very short in the Olympic Games, stop-watches are not used to avoid large percentage errors.
Since ZX = ZY = 1 m, α = β = 60°. Therefore, XY = ZX = ZY = 1 m The magnitude of the displacement of the ball is 1 m.
©
8
(a) The distance travelled by the ball will be longer if it takes a curved path.
7
(a) Length of the path
= 0.8 × 120 = 96 m
(b) No matter which path the ball takes, its displacement remains the same.
(b) Length of AB along the dotted line 96 = 30.6 m = (c)
Magnitude of Jack’s average velocity 30.6 × 2 = = 0.51 m s–1 120
Practice 1.3 (p. 23)
1
B Total time 5000 5000 = + = 9821 s 1.4 0.8 5000 + 5000 = 1.02 m