BROOKHOUSE INTERNATIONAL SCHOOL Force and Motion Practice Questions
1. (a) (a) (i) State State the diffe differen rence ce betwe between en scala scalarr and and vector vector quanti quantitie tiess [1] (ii) Give Give two exam example pless of a vector vector quantity quantity.. [2] (b) Bill travels travels 10km 10km North-east North-east and then then 12km 12km due due East (i) Draw a vector diagram showing s howing Bill’s route. [2] (ii) Calculate, without the use of a scale diagram, Bill’s resultant displacement in components East and North. All working working should be be shown shown clearly. clearly. [3] 2. (a) (a) (i) Comp Complet letee the the vector vector diagr diagram am below below show showing ing the the resul resulta tant nt forc force. e. 30N
Scale 1cm: 5N
120o
20N [1] (ii) label label the the result resultant ant with its magn magnitud itudee [1] (b) The boat shown in the the diagram below leaves leaves the Isle of of Sheppy and arrives arrives at the coast coast at point point A 25 seconds later. It travels in a straight line.
300 Sheppy
400
Calculate (i) the dista distanc ncee trav travell elled ed by by the the boat. boat. [2] (ii) the avera average ge veloc velocity ity towa towards rds point point A. [1] (iii) the component component of the velocity in direction direction X-B [3] 3. A ship ship is pulled pulled at at a constan constantt speed, speed, v, v, of 2.5m 2.5mss-1 by two tugs, A and B. Each tug is connected to the ship by a o cable so that the angle each of the cables makes with the direction of travel is 41 . The ship experiences a drag 2 force given by 8000v newtons.
v = 2.5ms-
Tug A 41o
Ship
o
41
Tug B
(a) Calculate the tension tension in each cable while while travelling travelling at this constant constant speed. speed. [3]
-1
(b) As the tugs attempt attempt to increase the speed speed of the the ship from from 2.5 ms , tug A breaks down, with its cable to the ship becoming slack. (i) Calculate Calculate the speed speed to which which the ship initially decelerates, decelerates, assuming assuming the tension tension in the other cable cable remains cons consta tant nt.. [2] [2] (ii) The ship also veers veers off-course. off-course. Explain Explain why why this happens. happens. [2] (c) The harbour harbour authorities authorities want to move the ship to clear the harbour walls sooner. This requires that the tugs increase increase the tensions in the the ccables ables to the the ship. ship. The maximum safe tension in the cables is 50kN and the tugs o need to maintain a minimum angle of 60 between the cables connecting them to the ship. Calculate the maximu maximum m speed speed at which which the tugs tugs can pull pull the ship. ship. [4] 4. A child throws throws a ball vertically upwards, upwards, then catches catches it again. -1 -2 (a) Taking Taking the the initia initiall velocity velocity to to be 10.0 10.0 ms , g to be 9.8ms and ignoring effects of air resistance, calculate: (i) the total total time time of of fligh flightt of the ball ball [3] (ii) the maximu maximum m heig height ht it reach reaches es [2] (b) Sketch the following following graphs graphs for its complete complete flight, taking taking the upwards upwards direction direction as positive: positive: (i) (i) velo veloci city ty - tim time graph raph [3] [3] (ii) (ii) acce acceler lerati ation on – time time graph graph [2] 5. A dart player player throws a dart horizontally horizontally.. By the time time it reaches reaches the dartboard, dartboard, 3.0m 3.0m away, it it has -2 fallen fallen a height height of of 0.20m. 0.20m. Taking Taking g as 9.8ms , find: 0.20m
3.0m (a) (a) The The time time of flig flight ht [2] [2] (b) (b) The The init initia iall velo veloci city ty [2] [2] (c) The magnitude magnitude and and direction direction of the the velocity velocity as it it is just about to hit the dartboar dartboard. d. [6] 6. A tennis ball is dropped from from a height h and bounces so that the speed speed immediately immediately after each bounce is half the speed just before the bounce. Sketch the following graphs from the time of release until the ball hits the ground for the third time. In each case, take the upward direction as positive. (a) (a) The The vel veloc ocit ity y – time time graph graph [3] (b) (b) The displa displace ceme ment nt – time time graph graph [3] -2
7. A rocket accelerates from rest for 20s with a constant upward acceleration of 10ms . At the the end end of of 20s 20s the fuel fuel is used up and it completes its flight under gravity alone. Assuming that air resistance can be neglected and -2 taking g = 9 8ms , calculate the:
(a) (a) (b) (b) (c) (c) (d) (e) (e)
speed speed reache reached d after after 20s. 20s. [2] heig height ht afte afterr 20s 20s.. [2] [2] maxim maximum um heigh heightt reach reached. ed. [3] speed speed just just before before the the rocke rockett hits hits the ground. ground. [2] total total tim timee take taken n for for the the flig flight. ht. [4]
8. Water flows flows from a fire fire hydrant at at the rate of of 1.2m3 min-1. The water rises vertically from the supply and flows round the bend and into a hose, of internal diameter 7.0cm
7.0cm Fixed pipe
hose, held horizontally
Water flow -3
(a) The density density of water water is 1000k 1000kg g m . Calculate the: (i) mass mass flow flow rate rate of of wate waterr in kg s-1. [2] -1 (ii) speed of the water water as it leaves leaves the nozzle nozzle in ms [3] (b) (i) State State the the change change in in horizon horizontal tal veloc velocity ity of of the water water as it goes goes roun round d the bend in the the pipe. pipe. [1] (ii) Calcula Calculate te the change change in horizon horizontal tal moment momentum um per second. second. [2] (iii) State the force force a fireman must must exert on the hose hose to keep it still when when the water is flowing out out horizon horizontally tally.. [1] (c) The mass mass of the the firema fireman n is 75kg. 75kg. If the water water is sudde suddenly nly turned turned on on when when he is holding holding the hose hose horizontally, horizontally, with what acceleration acceleration will will he start to move backwards backwards if he does not brace himself himself to oppose the force? (Assume that that his hands do not slip on the hose and that no friction friction acts between his feet and the ground.) ground.) [2] 9. (a) Identify three properties of pairs of forces that are linked by Newton’s third law. [3] (b) A person person stands stands on bathroom scales on the ground.
Bathroom scales Ground Draw a free-body force diagram diagram for the person. Identify all forces clearly. clearly. [2] (c) For the situation in (b), state the other force forming a Newton’s third law pair with the reaction force of the scales acting on the person’s feet. [1] (d) A slimming club is situated at the top of a tall building; to motivate motivate its clientele, clientele, the club has installed its -2 own lift lift which contains contains a weighing weighing machine. machine. The lift accelerat accelerates es uniformly uniformly at 1.0ms for 90% of its journey, both going up and coming down. Calculate: -2 (i) The resultant resultant force force required required to accelerate accelerate a person person whose whose mass is 80kg at 1.0ms 1.0ms . [1] (ii) The reading reading (in kg) kg) on the weighing weighing machine machine when when the 80kg person stands stands on it as the the lift accelerates accelerates upwards. [3] (iii) The reading (in kg) kg) of the machine when the same person person stands on it as it accelerates accelerates downwards. downwards. [2]
BROOKHOUSE INTERNATIONAL SCHOOL Force and Motion Motion Practice Questions Questions - Marks
1. (a) (a) (i) (i) A vec vecto torr has has a dir direc ecti tion on,, a scal scalar ar does does not. not. [1] (ii) Any two examples examples such such as force, displacemen displacement, t, velocity, velocity, momentum etc. [2] (b) (i)
12km
10km
45
[2] (ii) Resultant Resultant vector, vector, in terms of of (E, N) components components is: R = (10 cos45, 10 sin45) + (12, 0) = (19 km E, 7.1 km N)
[3]
30N
2. (a)
o
120
20N
magnitude: 26.5 N
[2] (b) (i) d2 = 3002 + 4002 d = 500m
[2] [2] (ii) (ii) v = d/t d/t = 500/2 500/25 5 = 20m 20mss-1
[1] [1] (iii) if angle between between XA XA and XB = component of v in direction XB = v cos cos = 400/500 -1 vxb = 20400/500 = 16ms [3] 3. (a) The ship ship travels travels at a steady speed so the forward force due to the cables cables must have the same same
8000v
T
magnitude as the drag force. Ropes are symmetrical so the magnitude of the tensions in them are equal; Resolve forces in direction of travel: 8000v2 = 2T cos41o ; 2 o 8000 2.5 = 2T cos41 T = 33125N = 33 kN [3] (b) (i)
8000v 41 T Resolving forces in direction of travel: 8000v2 = 33,000 cos41 o ; v = 1.8 ms-1 [2]
(ii) The ship veers veers off-course off-course as Tug B exerts a force force on the ship that that is perpendicular perpendicular to the original original direction direction of travel and is no longer balanced balanced by Tug A; either This provides a turning effect (moment) on the ship. This causes the ship to accelerate in this perpendicular direction [2] (c) Maximum speed speed is attained when the angle angle between the rope to each tug with the direction direction of travel is the same. Otherwise some of the force of the tugs goes to turning the ship; In this case the angle is 30o. The tension in the ropes is 50kN. Resolving in the direction of motion: 2 50,000 cos30o = 8000v 2 ; -1 Maximum speed, v = 3.3 ms -1
[4]
-2
4. (a) (i) s = 0, u = 10.0ms , a = -9.8 -9.8 ms ms , t = ? 2 1 s = ut + 2 at 0 t
= 10.0t = 0 or t =
2
1 2
9.8
10.0 4.9
t
2
= t(10.0 – 4.9t) 4.9t)
= 2.0s (2 s.f.)
[3]
2
(ii) v = u + 2as v = 0, u = 10 ms -1, a = -9.8 -9.8ms ms-2 2 2 9.8s 0 = 10.0 – 2 s
=
10.0
2
2 9.8
= 5.1m [2]
9(b)(i) v/ ms-1 10
0 1.0
2.0
-10
t/ s
[3] (ii) a/ ms-2 10
0 1.0
-10
t/ s
2.0
[2] 5. (a)Vert (a)Vertica icall motion: motion: s = ut ut + 0.20 t
=
=0+ 2 0.2 9.8
1 2
1 2
at2
2
9.8t
= 0.20s [2]
(b) (b) Horiz Horizon ontal tal motio motion: n: ux =
s
3.0
t
0.20
= 15ms-1 [2]
(c) ux
V
vv
Vertical final velocity: vv = uv + at = 0 + 9.8 0.20 Magnitude of total final velocity: V2 = ux2 + vv2 V = -1 -1 V = 15.1 ms (15ms to 2 s.f.) Angle: = tan-1
1.96 15
=
= 1.96 ms 2
15
1.96
2
-1
7.4 below horizontal [6]
6. (a) v
t
General shape correct Ratios correct, 1 : ½ : ¼ Slopes parallel
[3]
(b) s
t
Parabolic curves Reducing in size 1 Ratio of heights, 1 : ¼ : / 16 16
[3] 7. (a)
-2
u = 0, a = 10ms , t = 20s v = u + at = 0 + 10 20 2
= 200ms-1 [2]
2
(b) v = u + 2as s
=
v2
u
2
=
2a 2
200 2
Or: Or: s = ½ at at = ½
0
=
2 10
2
10 20
2000m = 2000m [2]
(c) (c) Furt Furthe herr heig height ht:: s = Total
v2
u
2
2a
=
0 200 2
2 9.8
= 2040m (3 s.f.)
height = 2000 + 2040 = 4040m = 4000m (2 s.f.)
[3] 2
2
(d) v = u + 2as v = 0 2 9.8 4040
-1
= 280 ms (2s.f.) [2]
(e) Time Time to fall back to earth earth is: t =
2s a
=
2 4040 9.8
Time to reach max height after fuel used up is: t = Time
v u a
= 29s (2s.f.) =
0 200 9.8
= 20.4s
for entire flight = 20 + 20.4 + 29 = 69.4s = 69s (2s.f.) [4]
8. (a) (i) 1.2 1000 = 1200 kg min -1 -1 1200/60 = 20 kgs [2] (ii) volume leaving pipe per second = cross-sectional cross-sectional area speed
1.2
0.035
2
v
60 -1
5.20 ms = v
(-1 for no unit or incorrect units) [3]
(b) (i) 5.20 ms
-1
[1] -1
(ii) 20 5.20 = 104 kg ms per second
[2] -1
(iii) (iii) Force Force time = 104 kg ms time = 1 second Force [1] (c) 104 = 75 a 1.39 ms -2 = a [2] 9. (a) (a) equa equall in in mag magni nitu tude de opposite direction act on different bodies [3] (b) weight
reaction of scales on person
[2]
(c) The person’s weight [1] (d) (i) F = ma = 80 1 = 80 N [1] (ii) Total Total force force on on scales scales = person' person'ss weight weight + acceler accelerating ating force force = 80 9.8 + 80 = 864 N Mass reading = force/g = 88.2 kg
[3] (iii) Total force force on scales = 80 80 9.8 – 80 80 = 704 mass reading = 71.8 kg (2 marks for final answer) [2]
