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G484The Newtonian World
MODULE 1:
Newton’s Laws and Momentum
Answer Booklet Lesson 1 questions – Newton’s First Law
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1 Without looking at your notes, define (a) Newton’s First Law, (b) Newton’s Second Law and (c) Newton’s Third Law.
(a) Newton’s first law states that an object will remain stationary or continue at a constant velocity unless acted on by a resultant force. (1) If it is acted on by a resultant force it will either accelerate, decelerate or change direction (1) depending on the direction of the force.(1) … (3)
(b)… Newton’s Second Law says that an object will accelerate quicker the harder you push it. (1) But if you push two different objects with the same force, (1) the heavier one won’t accelerate as much. (1)
OR
F=ma
(each symbol must be defined) (3)………………………………………………… (3)
(c)… When body A exerts a force on body B,(1) then body B exerts on body A a force that is equal, (1) opposite in direction and of the same type (1)…… (3)
2 Define the SI unit of force.
…………………………………………………………………………………………..
The force needed to accelerate an object of 1kg by 1ms-2.……………………..
……………………………………………………………………………………… (2)
3 The figure below shows a ball of 50g resting on the strings of a racket held horizontally. Contact Force
Weight
a) Draw and label the 2 forces acting on it. (2)
b) Calculate the weight of the ball and give the units.
Weight = mass x gravitational field strength
0.05 x 10
Weight = ……0.5……………… unit……N(ewtons)…………. (3)
c) Explain, with reference to Newton’s Laws what we know about the forces acting on the ball if it is at rest.
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
Lesson 2 questions – Momentum
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1. Define linear momentum and state its SI unit.
……… Momentum=mass×velocity [1]
SI unit of momentum is kgms–1. [1]
…………………………………………………………………………………………
…………………………………………………………………………………………(2)
2. A bumper car collides at right-angles with a metal barrier and rebounds at the same speed. A student suggests that the change in momentum of the car is zero.
Explain why the student is wrong.
Momentum is a vector quantity – it has both direction and magnitude. [1]
If the initial momentum of the car is +p, then its final momentum must be –p
(see diagram):
Change in momentum, Δp= final momentum– initial momentum
Δp=–p–p= –2p (the change is not zero) [1]
…………………………………………………………………………………………(2)
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3. Work out the momentum changes in the following cases:
(a) a mass of 3 kg slows from 3 m/s to 1.5 m/s;
Momentum change = mv – mu =
………- 4.5………………. Kgm/s (2)
(b) a mass of 500 g accelerates from 4 m/s to 8 m/s;
………2 ………………. Kgm/s (2)
(c) a ball of 50 g moving at 25 m/s is caught;
………-1.25 ………………. Kgm/s (2)
(d) a force of 4 N acts on an object for 12s;
……48 …………………. Kgm/s (2)
(e) a ball hits a wall at 4 m/s and bounces back at 3.5 m/s. Mass of ball is 2 kg;
………26.25 ………………. Kgm/s (2)
(f) a stone of mass 750 g hits a wall at 20 m/s and bounces back at 15 m/s;
Momentum change = 0.75(15 – (-20)) = 15 kgm/s or 15 Ns
(notice that the direction change has been allowed for by the use of another minus sign)
………15………………. Kgm/s (2)
(g) a ball of mass 1.5 kg hits the ground at 80cm/s and bounces back at 65 cm/s.
……2.175 …………………. Kgm/s (2)
Lesson 3 questions – Newton’s Second Law
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1 Fig 1.1 shows a car of mass 1200kg, pulling a caravan of mass 400kg along a horizontal road.
Fig 1.2
Calculate, for the first 20s of the journey
i) The acceleration of the car
Gradient = 13/20
acceleration = …0.65…..ms-2 (2)
ii) the resultant force acting on the car
Force = ma
1200x0.65
resultant force on the car = 780…N (2)
iii) the resultant force acting on the caravan.
Force = ma
400x0.65
resultant force acting on the caravan = ……260…. N (2)
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2. A boy running at 8 m/s is stopped by a friend in 2s. If the mass of the boy is 40kg what is the force on him during stopping?
40x8 = Fx2 F = 160N
………160………………. N (2)
3. A ball of mass 20 g hits a vertical wall with a horizontal velocity of 12 m/s. If it rebounds with a velocity of 10 m/s in the opposite direction and is in contact with the wall for 0.05 s find:
(a) the change in momentum of the ball;
0.02x22 = 0.44 Ns
…………0.44 ……………. Kgm/s (2)
(b) the force of the wall on the ball.
0.44/0.05 = 8.8 N
…………8.8 ……………. N (2)
4. A parachutist lands on the ground. If his mass is 65 kg and he hits the ground at 4 m/s and stops in 0.5 s find:
(a) his change of momentum;
65x4 = 260 Ns
………260 ………………. Kgm/s (2)
(b) the force of the ground on him.
260/0.5 = 520 N
……… …520 ……………. N (2)
5. A second parachutist with the same mass lands with the same velocity but forgets to bend his knees and stops in 0.01 s. Find his change of momentum and the force of the ground on him.
260/0.01 = 26000 N!
…………260 ……………. Kgm/s (2)
………26000 ………………. N (2)
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6. F=ma is a special case of Newton’s Second Law for a body that has a constant mass but changing velocity.
- Define a change in linear momentum, Δp in terms of Δt and F.
…………………………………………………………………………………………
…………………… Δp = F Δt …………………………………………………………………………………………………………………………………………………………………… (1)
- Define acceleration.
………………a= Δv/ Δt ………………………………………………………………………………… (1)
Show that F=ma is true for a body that has a constant mass but changing velocity using your definitions from (a) and (b).
Since FΔt = Δp = Δ(mv);
And acceleration is the change in velocity (Δv) in a change in time (Δt)
or a = Δv/Δt;
If an object of constant mass has a changing velocity we can write that:
FΔt = m Δv
If we rearrange this to make F the focus of the equation we can get:
F = m Δv/Δt
We have said that Δv/Δt = a and so substituting this in we now get
F = ma which is a special case of Newton’s Law when mass remains constant and velocity changes.
QED
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Lesson 4 questions - Newton’s third law
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1. State Newton’s 3rd Law.
When body A exerts a force on body B, then body B exerts on body A a force that is equal, opposite in direction and of the same type. (2)
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2. When a golf ball is dropped to the pavement, it bounces back up. (a) Is a force needed to make it bounce back up? (b) If so, what exerts the force?
a. ……………Yes…………………………………………………
……………………………………………………………………………………… (1)
b. ………The pavement’s contact force must be greater than the weight of the ball in order for the ball to accelerate vertically.………………………………………………………………..
……………………………………………………………………………………… (2)
3. If you walk along a log floating on a lake, why does the log move in the opposite direction?
……… Newton’s 3rd Law says that if I push the log away with a force it will push back the same amount. The frictional force between your foot and the log is greater than the frictional force between the water and the log and so the log moves in the opposite direction to the dircetion I walk.………………………………… (2)
4. When an object falls freely under the influence of gravity there is a net force mg exerted on it by the Earth. Why doesn’t the Earth move?
…………The weight exeerted on me by the Earth is the same as the weight exerted by me on the Earth. My inertia is less than the Earth’s and so I accelerate much more than the Earth.……………………………………………………………..
……………………………………………………………………………………… (2)
5. A person exerts an upward force of 40 N to hold a bag of shopping. Describe the “reaction” force (Newton’s third law) by stating (a) its magnitude, (b) its direction, (c) on what object it is exerted, and (d) by what object it is exerted.
a) …………40N………………………………………………………
……………………………………………………………………………………… (1)
b) ……upwards……………………………………………………
…………………………………………………………………………………………..
……………………………………………………………………………………… (1)
c) ………Bag of shopping………………………………
……………………………………………………………………………………… (1)
d) …The Earth…………………………………………………………
……………………………………………………………………………………… (1)
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9. According to legend, the Russian mystic Nostradamus's horse once refused to pull a cart after being hit on the head by an apple. The horse reasoned that, according to Newton's third law, any force that he exerted would be balanced by an equal and opposite force. The cart would thus never move, no matter how hard the horse tried.
Explain why this horse is wrong.
…The cart feels the unbalanced force applied by the horse and therefore accelerates if this force is greater than the frictional force of the ground on the cart. On the other hand, the horse feels an equal and opposite force applied by the cart but is able to accelerate forward if the frictional force of the ground on the horse is greater… (3)
Lesson 5 questions – Impulse
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1 This question is about kicking a football.
a) The graph below shows how the force F applied to a ball varies with time t when it is kicked horizontally. The ball is initially at rest.
i) Use the graph to find
1 the maximum force applied to the ball
maximum force = ……48…………… N
2 the time the boot is in contact with the ball.
Time = ……0.25………….s (1) Both answers correct for 1 mark
ii) What is the quantity called that the mean force multiplied by the time of contact describes?
……………………Impulse (accept change in momentum)………………… (1)
iii) How would you use the graph to estimate this quantity delivered to the football?
…………… Estimating area under graph (1)…..
……………………………………………………………………………………… (1)
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iv) Use the graph to estimate this quantity delivered to the football.
Impulse = ………6.5 +/- 1 (1)……………. Ns (1)
b) The mass of the ball is 0.5kg. Use your answer to (a) to calculate
i) the maximum acceleration of the ball
a=F/m (1)
acceleration = ……96 (1)……………….ms-2 (2)
ii) the final speed of the ball
Ft=mv (1)
v=a(ii)/0.5
=2a(ii) (1) ecf
Speed = ……13…………….. ms-1 (2)
iii) the kinetic energy of the ball after the kick.
Ke=1/2 mv2
ke=1/2 x 0.5 x b(ii)
Kinetic energy = ……42………………….. J (2)
c) The ball hits the wall with a speed of 14 ms-1. It rebounds from the wall along its initial path with a speed of 8.0 ms-1. The impact lasts for 0.18s. Calculate the mean force exerted by the ball on the wall.
Ft=mv-mu
Ft=0.5 (8-(-14))
F=11/0.18
Force = ……61.1……………. N (3)
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2 A 0.30 g fly moving at 1.5ms–1 is trapped by a spider’s web. The fly comes to rest in a time of 0.40 s. Calculate the magnitude of:
a) the change in momentum of the fly, (give the units);
Δp= final momentum – initial momentum
Δp=0 – (0.30×10–3×1.5) (the final momentum is zero) [1]
Δp= 4.5×10–4 kgms–1 (magnitude only) [1]
change in momentum = ……4.5×10–4 …………….. units …… kgms–1 …[1]… (3)
b) the average force exerted by the web on the fly.
Newton’s second law: F = Δp / Δt [1]
F = 4.5×10–4/0.4 [1]
F = 1.13x10–3N [1]
Average force = ……1.1x10–3……………… N (3)
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3 a) State Newton’s second law.
…………………The rate of change of momentum of an object is directly proportional to the net force acting on the object and takes place in the direction of this force. ………….
…………………………………………………………………………………… (1)
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b) A ball of mass 320 g hits the ground at right-angles at a speed of 15ms–1and rebounds vertically at a speed of 7.0ms–1. The ball is in contact with the ground for a time of 0.16 s.
i) Show that the change in momentum of the ball is about 7.0kgms–1.
Momentum, p = mass × velocity [1]
Δp = final momentum – initial momentum
Δp = (0.320 × –7.0) – (0.320 × 15) [1]
Δp = 7.04kgms–1 » 7.0kgms–1 (magnitude only)
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ii) Calculate the average force on the ball during impact with the ground.
Δp = 7.0kgms–1, Δt = 0.16 s
F = Δp/ Δt [1]; F=7.0/0.16 [1]
F » 44N [1]
Average force = ………44………..N (3)
Lesson 6 – 8 questions – Conservation of Momentum
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1. State the principle of conservation of momentum and explain what is meant by elastic collisions and inelastic collisions.
……The momentum before a collision or explosion = the momentum after that collision or explosion………………………………………………………………..
…………………………………………………………………………………………..
……In elastic collisions momentum and KE are conserved.………..
…………………………………………………………………………………………..
……In inelastic collisions momentum is conserved but the KE before the collision is greater than afterwards.…………..
……………………………………………………………………………………… (3)
2 In the fission of a uranium 235 nucleus a neutron collides with the nucleus causing it to break up. The particles formed are two smaller nuclei and three neutrons. Is the linear momentum of the system conserved during the collision? Explain your answer.
…………………………………………………………………………………………..
Yes – momentum is conserved in all explosions and collisions as long as all particles involved are taken into account