Dynamics Pg 1

Does the principle of conservation of momentum applies where two colliding objects lose kinetic energy as a result of sticking to one another at the point of collision ? Explain your answer with reference to Newton’s third law.

Yes. By Newton’s 3rd law, when two objects collide and stick to one another, the forces that the objects exert on each other form an action-reaction pair. These two forces are not external forces

on the system of the two objects. ∴ the principle of conservation of momentum can apply.

A sub-atomic particle of mass 0.113u collides head-on and elastically with a stationary neutron.The neutron moves off with a speed of 3.8106 ms1. By using both the law of conservation of momentum and energy, it can be deduced that the initial speed and final speed of the sub-atomic particle are 1.89107 ms1 and 1.50107 ms1. Suggest why the equation used in the law of conservation of momentum and energy would not necessarily apply if the subatomic particle had a much greater initial speed.

If the initial speed is much higher, then we must take in the relativistic effects for speeds approaching speed of light. Hence both laws may not apply.

Explain why the interaction of gas molecules with each other and with walls of the containing vessel must, on average, be elastic.

If the interaction were not elastic on average, there will be time when the K.E of the molecules decrease to 0 after many collisions. The molecules would stop colliding with the walls of the container. This is impossible because the pressure exerted by a gas does not fall to 0 spontaneously, unless some work is done on it.

Explain why very light particles, such as electrons, or massive particles, such as uranium nuclei, are unsuitable for slowing down neutrons.

When the neutron collides with an electron, it would retain almost all its kinetic energy and hence its velocity. When it collides with a massive uranium atom, it would rebound with most of its speed.In either way, the speed and kinetic energy of neutron is not reduced much. Note that fast neutrons are also absorbed by uranium and therefore lost.

Explain why the terminal velocity of falling spheres of the same density increases with mass.

Spheres with larger masses have greater weights. Hence the drag force has to be larger in order to balance the weight. Larger drag force occur at greater velocity. Hence, spheres of the larger masses have higher terminal velocities.

Dynamics Pg 2

Two bodies of masses A (mass m) and B (mass 3m) are in contact and placed on a frictionless surface. A horizontal force, F, of 12N is applied to body A which in turn pushes body B. What is the force that body A acts on body B ?

Apply Newton’s 2nd law to the combined system A & B : F = (m + 3m) a a = m4F

For body A, F – FBA = ma F – FBA = m (m4F ) = (

4F )

FBA = 43 F = 4

3 (12) = 9N.

Or For body B, FBA = 3ma = m (m4F ) = 4

3 F = 43 (12) = 9N.

By Newton’s 3rd law, FBA = FAB = 9N.

What if the horizontal force F is acting on body B instead of body A ?

Apply Newton’s 2nd law to the combined system A & B :

F = (m + 3m) a a = m4F

For body B, F – FAB = 3ma = 3m (m4F ) = 4

3 F = 43

FBA = 41 F = 4

1 (12) = 3N.

Or, For body A, FBA = ma = m (m4F ) = 4

1 F

= 41 (12) = 3N.

By Newton’s 3rd law, FAB = FBA = 3N.Note that FAB has reduce from 9N to 3N when F apply in opposite direction. A spring balance carrying a mass of 20.0 kg in a lift registered 250 N.What was the acceleration of the lift ?Calculate the balance readings during (a) free fall, (b) motion at constant velocity.

ma = F – mg F = m (g + a) 250 = 20 (9.81 + a) a = 2.69 ms2

(a) O N (b) const. v a = 0 ∴ F = m (g + 0) = mg = 20 (9.81) = 196.2N

Dynamics Pg 3

An 80 kg man is parachuting and experiencing a downward acceleration of 2.5 ms2 . The mass of the parachute is 5.0 kg.(a) What is the force exerted on the parachute by the air ?(b) What is the force exerted by the man on the parachute ?

(a) Let M be the mass of man, i.e. M = 80 kg. Let m be the mass of parachute, i.e. m = 5.0 kg. (M + m) a = (M + m) g – R R = (M + m)(g – a) = (80 + 5)(9.81 – 2.5 ) = 621.35N

(b) Ma = Mg – R R = M (g – a) = 80 (9.81 – 2.5) = 584.8N

A 10 kg monkey is climbing a massless rope attached to a 15 kg mass over a (frictionless) tree Limb.(a) Explain quantitatively how the monkey can climb up the rope so that he can raise the 15 kg mass off the ground.(b) If, after the mass has been raised off the ground, the monkey stops climbing and holds on to the rope, what will its acceleration and the tension on the rope now be ? (Take g = 10ms2 )

(a) mg + ma > Mg 100 + (10) a > 150 a > 5ms2

(b) (M + m) a = 150 – 100 (10 + 15) a = 50 a = 2 ms2

(c) Consider the 15 kg mass, Ma = 150 T T = 150 – (15)(2) = 120 N Or Consider the monkey, ma = T 100 T = (10)(2) + 100 = 120 N

A chain consisting of 5 links, each of mass 1.00 ×101 kg, is lifted vertically with a constant acceleration of 2.50 ms2 , by external force F. Taking g as 10 ms2 , find (a) the forces acting between adjacent links,(b) the lifting force F exerted on the top of the chain,(c) the net force acting on each link.

(a) ma = F – mg F = m (a + g)

For L1 : F = (4 ×1.00 ×101 )(2.5 + 10) = 5 N

L2 : F = (3 ×1.00 ×101 )(2.5 + 10) = 3.75 N

L3 : F = (2 ×1.00 ×101 )(2.5 + 10) = 2.50 N

L4 : F = (1 ×1.00 ×101 )(2.5 + 10) = 1.25 N

(b) F = m (a + g) = (5 ×1.00 ×101 )(2.5 + 10) = 6.25 N

(c) Net force = ma = (1.00 ×101 )(2.50) = 0.25 N

Dynamics Pg 4

A motorist travelling at 13 ms1 approaches traffic lights, which turn red when he is 25 m away from the stop line. His reaction time (i.e. the interval between seeing the red light and applying the brakes) is 0.7s and the condition of the road and his tyres is such that the car cannot slow down at a rate of more than 4.5 ms2. If he brakes fully, how far from the stop line will he stop, and on which side of it ?

Distance travelled during the reaction time = (13)(0.7) = 9.1m

During the application of brake, v2 = u2 + 2as 2 2v u2a =

2 20 132( 4.5)

= 18.8 m

Total distance = 9.1 + 18.8 = 27.9 m

He will stop over the stop line by 27.9 – 25 = 2.9 m

A motorist whose car will not start seeks a tow from a second motorist. The towing vehicles accelerates slowly from rest with the tow-rope slack; when the rope becomes tight, the towed car starts to accelerate and the towing vehicle moves with a constant speed of 0.8 ms1 until the towed car, of mass 1000 kg, achieves the same speed.(a) What is the change of momentum of the towed car ?(b) The average tension in the rope may be assumed to be 4000 N during the acceleration of the towed car. How far does the towed car travel from rest before it reaches the speed of 0.8 ms1 (c) How far does the towing vehicle travel in this time ?(d) The breaking strain of the tow-rope is 0.04. What must be the minimum unstretched length of the rope if the tow is accomplished without breakage of the rope ?

(a) Change of momentum of the towed car = (1000)(0.8) – 0 = 800 Ns

(b) Acceleration, a = mF =

10004000 = 4 ms2.

v2 = u2 + 2as s = 2 2v u2a =

2 2( 0.8) 02(4)

= 0.08 m

(c) v = u + at t = a

uv = 0.8 0

4

= 0.2s

During t = 0.2s, the towing vehicle moves with constant velocity of 0.8 ms1

∴ Distance travelled by towing car = (0.8)(0.2) = 0.16 m

(d) Distance stretched by the tow-rope = 0.16 – 0.08 = 0.08 m

ℓ = strain

extension = 04.008.0 = 2 m

∴ The minimum unstretched length of the rope must be 2 m.

Dynamics Pg 5

A cyclist travel down an inclined road without pedalling. The angle that the road makes with the horizontal is 6.8º. The cyclist and cycle have a combined weight of 760 N.(a) Show that the component of the weight of the cyclist and cycle down the slope 90 N. (b) The variation with time t of the velocity v of the cyclist down the slope is illustrate in the diagram. The cyclist reaches a constant velocity after 30s. Given that the initial acceleration of the cyclist is 0.725 ms2. (i) Calculate the accelerating force acting on the cycle and cyclist

at time t = 0. (ii) Hence determine the resistive force acting on the cycle and cyclist at t = 0. (iii) State the magnitude of resistive force at t = 30s.(c) The cycle is serviced to reduce friction, state and explain what change will occur in the maximum velocity of the cycle if the journey is repeated.(d) After descends, the cyclist travels along a horizontal road at a speed of 7.0 ms1. When the brake are applied, he takes 3.5s to come to rest. Calculate the force opposing motion during the time that the brakes are applied. Is the brakes able to halt the cycle on the inclined road ?

(a) Component of weight down slope, Wd = Wsin 6.8º = 760 sin 6.8º = 90 N

(b) (i) F = ma = ( W9.81

)(0.725 ) = 56.2 N

(ii) F = Wd – R

R = Wd – F = 90 – 56.2 = 33.8 N

(iii) At t = 30s, velocity = const. F = 0 .

∴ Wd = 0 + R R = 90 N

(c) Friction is reduced resistive force R take more than 30s to reach max. 90N.

F = Wd – R with Wd constant. accelerating force F acting on the cycle is longer than 30s the cycle will accelerate longer than 30s max. velocity is higher.

(d) a = a

UV = 0 73.5

= –2 ms2 Opposing force F = ma = ( 7609.81

)(2) = 155 N

On the inclined road, the max. resultant force, Fmax = Wd – Rmin Fmax = 90 – 0 = 90 N

The braking force, 155N > 90N The brakes is able to halt the cycle.

Dynamics Pg 6

A space research rocket stands vertically on its launching pad. Prior to ignition, the mass of the rocket and its fuel is 1.9 ×103 kg. On ignition, gas is ejected from the rocket at a speed of 2.5 ×103 ms1 relative to the rocket, and fuel is consumed at a constant rate of 7.4 kgs–1 .Find the thrust of the rocket and hence explain why there is an interval between ignition and lift-off. Take g = 10 ms1 .

v = 2.5 × 103 ms1 , dtdm = 7.4 kgs1

dtdP =

dt)mv(d

= vdtdm = (7.4)(2.5 × 103 ) = 18.5 × 103 N

Thrust of rocket = 18.5 × 103 N

Weight of rocket and its fuel = (1.9 × 103 ) (10) = 19 × 103 N

Initially, thrust < weight of rocket and its fuel. As the fuel is burnt away, the weight of rocket drops. When the thrust is greater than the weight of rocket, the rocket will take off. This accounts for the time interval.

Dynamics Pg 7

(a) Explain (i) how the thrust developed by rocket motors depends on the principle of conservation of momentum, (ii) why it is that a rocket motor can function in a vacuum, (iii) why the resultant force on the rocket in a vacuum is larger than in air.

(b) A rocket has mass 1.2 × 104 kg (including fuel) and an acceleration of 3.5 ms2 just after a vertical take off. In the question, neglect air resistance. (i) Draw a diagram showing the forces acting on the rocket. (ii) What is the thrust of the rocket motors ? (iii) After 1 minute the rocket is still travelling vertically and the thrust exerted by the motors has not changed, but the acceleration has increased to 18.7ms–2. What is the rate of consumption of fuel ?(c) Why can variation in the value of the acceleration of free fall be neglected in making these calculations ?

(a)(i) Momentum of the rocket = momentum of exhaust gas, MV = – mu where negative sign means opposite direction. M > m, but the gas is jetted at very high velocities so that enough momentum is produced to give the rocket some velocity V. The rate of momentum MV produced (i.e.thrust) is proportional to the rate of mu produced by the rocket motors. (ii) In vacuum, the rocket causes its own oxygen for combustion in its motors. (iii) Air resistance reduces the thrust of the rocket. In a vacuum, this resistance is absent and the full thrust of the motors is available.

(b)(i) W : rocket weight T : engine thrust reaction A : force of exhaust gas

W & T act on rocket

(ii) ma = T mg T = m (a + g) = (1.2 × 106 )(3.5 + 9.8) = 1.6 × 107 N

(iii) T = m(a + g) m = ga

T

= 71.6 10

18.7 9.8

= 5.61 ×105 kg (i.e. mass left)

Change in mass, ∆m = (12.0 – 5.6) × 105 = 6.4 × 105 kg consume in 1 min. Rate of fuel consumed = 6.4 × 105

÷ 60 = 10600 kgs1

(c) Average acceleration, a = 21 (3.5 + 18.7) = 11.1 ms2

Distance, S = ut + 21 at2 = 0 + 2

1 (11.1)(60)2 = 19980m 20km.

20 km << RE (= 6400 km) i.e. height is very small compared to the Earth’s radius change in g is very small negligible.

Dynamics Pg 8

(a) A toy rocket consists of a plastic bottle which is partially filled with water. The space above the water contains compressed air. At one instant during the flight of the rocket, water of density ρ is forced through the nozzle of radius r at speed v relative to the nozzle. Determine, in terms of ρ, r and v, (i) the mass of water ejected per unit time from the nozzle, (ii) the rate of change of momentum of the water. Hence show that the accelerating force F acting on the rocket is given by the expression F = π r2 ρ v2 mg , where m is the mass of the rocket and its contents at the instant considered.

(b) The toy manufacturer recommends that the rocket should contain about 550cm3 of water before take off. If the initial air pressure is 1.6 ×105 Pa, all of this water will be expelled and the pressure is just reduced to atmospheric pressure as the last of the water is expelled. However, on one flight, the initial volume of water was 750 cm3 but the initial air pressure in the rocket was still 1.6 ×105

Pa. State, without calculation but with a reason, the effect on this increased volume of water on (i) the initial thrust, (ii) the initial resultant accelerating force, (iii) the final mass of the rocket and its contents, (iv) the maximum height reached.

(a)(i) Volume of water ejected per unit time, dtdV =

dt)Ax(d

= Adtdx = Av.

Mass of water ejected per unit time, dtdm =

dt)V(d

= ρdtdV = ρ(Av) = π r2ρv

(ii) dtdP =

dt)mv(d

= vdtdm = π r2 ρ v2

By Newton’s 3rd law, upthrust on rocket = rate of change of momentum of water π r2 ρ v2

∴ accelerating force F = upthrust – mg = π r2 ρ v2 – mg (Shown)

(b) (i) Initial pressure difference is same rate of change of momentum of water is same initial upthrust is same.

(ii) From (i), initial uptrust is same but mass is larger. F = uptrust – mg. accelerating force is smaller. (iii) Not all the water expelled out when the internal and external pressure on air is equal final mass is larger. (iv) Larger mass at the peak & from (ii) accelerating force is small max. height reached is lower

Dynamics Pg 9

When a body moves through a fluid, a retarding force due to turbulence may be experienced.In the case a sphere of radius r moving with speed v through a stationary fluid of density which is

at rest, this force is given by F = kρr2 v2 where k is a constant.(a) By relating the retarding force to the transfer of momentum between the sphere and the fluid, explain why F is proportional to ρr2 v2. (b) When spherical raindrops fall through still air, all but the smallest experience a retarding force given by the equation above. It is found that drops of a given radius approach the ground with an approximately constant speed, which is independent of the height of the cloud in which they were formed. Explain this observation by reference to Newton’s laws. Find an expression for this terminal speed v1 in terms of the constant k, the radius r of the drop, its density w , the density A of the air and the acceleration of free fall g.

(Neglect the buoyancy of air) (c) The terminal speed of a raindrop of radius 1 mm is approximately 7 ms1. In freak storms, hailstones with radii as large as 20 mm may fall. Estimate the speed with which such stones strike the ground. (Take density of water = 1 × 103 kgm1 & density of ice = 9 × 102 kgm3 )

(a) When the body moves through the fluid, it displaces fluid in the shape of cylinder.

The mass displaced per sec, dtdm =

time)density)(volume(

= time

)length)(density)(area(

= (area)(density)(velocity) = r2 ρv

Force, F = momentum change per sec = dtdP =

dt)mv(d

= dtdm · v = ( r2 ρv)(v) = ρr2 v2

F ∝ ρr2 v2

(b) By Newton’s 1st law, the particle move at constant speed unless an external force acts on it. Since buoyancy of air is negligible, it will fall at constant speed (or terminal velocity) when air resistance equal to gravity force on it.

∴ F = rw2 2

1v A & F = mg = ( 3

4 rw3

w ) g rw2 2

1v A = 3

4 rw3

w g

1v 2 = w w

A

4r g

3

1v = w w

A

4 r g

3

---------------------(1)

(c) Similarly, icev = ice ice

A

4 r g

3

---------------------(2)

(2) ÷ (1) : ice

1

vv

= ice ice

w w

r

r

= 3 2

3 3(20 10 )(9 10 )

(1 10 )(1 10 )

= 18

icev = 18 1v = 18 ( 7 ) = 29.7 ms1

Dynamics Pg 10

A stream of sand falls at a constant rate from a height h on the pan of a direct reading balance.After a time T, a total mass M has fallen and the delivery of sand ceases.(a) Draw a sketch graph to show the reading m of the balance as a function of time t, giving the values of m at t = 0, t = T and t = 2T.(b) Explain the form of the graph.(c) How would the graph differ if the sand were allowed to fall from a height 2h, a mass M again being delivered in time T ? Trace the transformations of energy during such a process.

(a)

v2 = u2 + 2gh = 0 + 2gh v = 2gh

Change of momentum of sand = M 2gh 0 = M 2gh

∴ Force exerted = change of momentum

time taken =

M 2ghT

.

(b)(i) Time interval OA represents the time for the sand to reach the pan of the balance. (ii) AB represents the time when the sand reaches the pan at a constant rate.

(iii) BC represents the rate of change of momentum when the sand hits the pan. The additional

force, T

gh2M is acting on the pan as long as the sand continues to fall. Once the supply of

sand is stopped, the additional force will cease acting on the pan.

(iv) CD represents the damped oscillations of the pan when all the sand has fallen. (v) The final force at E is the total weight Mg of the sand.

(c) If the height is double, the additional force at BC would be T

gh4M.

The P.E is converted to K.E of the falling sand as the sand falls. This K.E changes to sound and heat on impact with the pan. In addition, some energy is used as work to compress the spring of the balance.

Dynamics Pg 11

A conveyor belt is used to transfer luggage at an airport. It consists of a horizontal endless belt running over driving rollers, moving at a constant speed of 1.5 ms1. To keep the belt moving when it is transporting luggage requires a greater driving force than for an empty belt. On average, the rate at which baggage is placed on one end of the belt and lifted off at the other end is 20 kg per second. Why is an additional driving force required and what is its value ?

Before the luggage is placed onto the belt, it has no horizontal speed. At the end of the belt, it has a speed of 1.5 ms1. Rate of change of momentum is involved for a load of 20 kg in a second. This additional driving force is required to provide this change of momentum.

F = dt

)mv(d = v

dtdm = (1.5)(20) = 30N.

A conveyer belt travelling at a speed of a speed of 3.0 ms1 and at an angle of 20º to the horizontal has 18 kg of sugar dropped on to it each second as shown in the diagram. Assume that the sugar has negligible speed before reaching the belt, calculate (a) the momentum gained in each second by the sugar, (b) the force which the belt must exert on the sugar to accelerate it to speed of belt,

(c) the work done per second by the belt on the sugar in exerting this force. (d) the potential energy gained in each second by the sugar which is on the belt.

(a) P = m (v – u) = 18 (3.0 – 0) = 54 kgms1

(b) F = tP + mg sin θ =

0.14.5 + (18)(9.8) sin 20º = 114.3 N

(c) Work done per second = Power = Fv = (114.3)(3.0) = 343 W

(d) dtdh = v sin θ = (3.0) sin 20º = 1.03 ms1

∆ P.E. per second = mgdtdh = (18)(9.8)(1.03) = 181 W

A machine gun fires 50.0g bullets at a speed of 1000 ms1. The gunner, holding the machine gun in his hands, can exert an average force of 180 N against the gun. Determine the maximum number of bullets he can fire per minute.

<F> ∆ t = N∆P where N is the no. of bullets he can fire.

(180)(60) = N (mVi – mVf ) = N [ (50 103 )(1000) – 0 ] N = 216

Dynamics Pg 12

(a) A projectile of mass 3.2 ×102 kg is fired from a cylindrical

barrel of cross-sectional area 2.8 ×104 m2 by means of compressed gas. The variation with time t of the excess pressure p of the gas in the barrel above atmospheric pressure is shown in the diagram. Given that the maximum pressure is 55M Pa at t = 1 ms. Calculate (i) the maximum force which the gas exerts on the projectile, (ii) the acceleration of the projectile which would result from the

force calculate in (i).(b) Given that the area under the graph is 5.5 ×104 Pas, estimate the total change of momentum due to the compressed gas which is experienced by the projectile.(c) The speed of the projectile changes from 0 to 270 ms1 as it leaves the barrel. What is the change in momentum of the projectile ?(d) Compare your answers to (b) and (c) and comment.

(a)(i) m = 3.2 × 102 kg , A = 2.8 × 104 m2 , Pmax = 55 × 106 Nm2

Fmax = (Pmax)(A) = (55 × 106 )(2.8 × 104 ) = 15.4 kN

(ii) a = mF =

215400

3.2 10 = 4.81 × 105 ms2

(b) Total change of momentum = ∫ Fdt [ ∫ Fdt = ∫ 1

2

v

v dt)mv(d

dt = m 1v – m 2v ]

∫ Fdt = ∫ PA dt = A ∫ Pdt = A × Area under (P – t) graph

= (2.8 × 104 )(5.5 × 104 ) = 15.4 kgms1

(c) Change of momentum = m fv – m iv = (3.2 × 102 )(270 – 0) = 8.64 kgms1.

(d) 4.15

64.8 × 100% = 56% i.e. the momentum gain by projectile is 56% that produced by the

compressed gas. The balance 44 % is used to overcome the frictional force between the inner wall of the barrel and the projectile when firing.

Some makes of car have, as a safety feature, regions at the front and rear which are designed to collapse on impact, but the shell of the passenger compartment is of rigid construction. Give a brief physical explanation of how this design may help to protect passengers from serious injury in the event of a collision.

If the car of mass m, is travelling at a velocity v, before impact and comes to rest after impact, then the change in momentum is m (v – 0) = mv.

By Newton’s 2nd law, the force involved in the impact is F = t

mv where t is the time taken during

the impact. The collapsible design would allow the impact to take a longer time, thereby reducing the force F. The reduced force, F might not be able to break the shell of the passenger compartment which is rigid. Thus, it works as a safety feature.

Dynamics Pg 13

Calculate the change in momentum suffered when a 70 kg person lands on firm ground after jumping from a height of 5.0 m. Estimate the average force exerted on the person’s leg if the landing is (a) stiff-legged, (b) with bent knees. In the former case, assume the centre of mass of the body moves 1.0 cm during impact and in the second case, 50 cm.

(a) v2 = u2 + 2as v2 = 0 + 2 (9.81)(5.0) v = 9.9 ms1. Since a = g is a constant, we can use kinematics equation.

S = 21 (u + v) t (1.0 × 102 ) = 2

1 (0 + 9.9) t t = 2.0 × 103 s.

< F > ∆ t = m∆v < F > (2.0 × 103 ) = (70)(9.9) < F > = 3.47 × 105 N (upwards)

(b) S = 21 (u + v) t 5.0 ×102 = 2

1 (0 + 9.9) t t = 0.10s.

< F > (0.10) = (70)(9.9) < F > = 6.93 × 103 N (upwards)

A steady wind of 50 ms1 blows against a rigid wall, which is in a plane perpendicular to the wind direction. Estimate the pressure exerted by the wind on the wall. Density of air is 1.25 kgm3. State any assumption you made in your calculations.

F = v

dtdm = ρv

dtdV ----------------------(1) since m = ρV, ρ : density , V: volume

But V = Ax, ∴ dtdV =

dtd (Ax) = A

dtdx = Av ---------------------(2)

Subst. (2) into (1) : F = ρv (Av) = Aρv2 AF = ρv2 P = ρv2

∴ P = (1.25)(50)2 = 3125 Nm2.Assumptions : 1. Wind does not rebound. 2. v = 0 when hit the wall. 3. By Newton’s 3rd law, F (wall to wind) = F (wind to wall)

A bird of mass 0.50 kg hovers by beating its wings of effective area 0.30 m2. (a) What is the upwards force of the air on the bird ?(b) What is the downward force of the bird on the air as it beats its wings ?(c) Estimate the velocity imparted to the air (density of 1.3 kgm3 ) by the beating of the wings.(d) Which of Newton’s laws is applied in each of (a), (b) and (c) ?

(a) Upthrust = mg = (0.5)(9.81) = 4.905 N(b) The bird beats its wings not moving Downward force = upthrust = 4.905 N

(c) P = ρv2 AF = ρv2

3.0905.4

= 1.3 v2 v = 3.55 ms1

(d) Newton’s 1st & 2nd laws are applied in (a), 3rd law is applied in (b).

Dynamics Pg 14

Use the Newton’s 2nd and 3rd laws to show that the momentum of a system of two colliding bodies remains constant, provided that no external forces act.

Before impact After impact Two bodies A and B of masses m1 & m2. Before impact, their velocities are U1 & U2. After impact, their velocities become V1 & V2. By Newton’s 2nd law,

Force on A , FA = takentimemomentuminchange

= 1 1 1 1m V m Ut

. Force on B , FB = 2 2 2 2m V m Ut

FA = – FB 1 1 1 1m V m Ut

= – ( 2 2 2 2m V m Ut

) m1U1 + m2U2 = m1V1 + m2V2

∴ Total momentum before impact = Total momentum after impact.

A sphere of mass m travelling in a straight line with speed u collides head-on with a second sphere of same mass which is not stationary but has speed U2 . The speed u of the incoming sphere is greater than U2 .The incoming sphere of kinetic energy W. The

graph shows how WE

, the fractional energy lost by the

incoming sphere, depends on the ratio 2UU

.

(a) What happens to the kinetic energy lost by the incoming sphere ?(b) Use the graph to suggest why paraffin wax, which has a high number density of protons, is a good absorber of high speed

neutrons.

(a) The K.E. lost by the incoming sphere is transferred to the second sphere which causes the latter to travel at travel at higher speed.(b) Since the mass of a neutron is almost the same as a proton, we can consider their collision to be same as above situation. As the paraffin wax has high density of protons, the probability of a neutron collides with protons is very high. In addition, if a high speed neutron (U) collides with a

proton which is virtually fixed in the wax (speed U2 ), then 2UU

0.

From the graph, 2UU

0 WE

1 the neutron loses almost all its K.E

to the proton and comes to rest. Hence, this wax is a good absorber of high speed neutrons.

Dynamics Pg 15

The following data concern a tennis ball at a given instant just before it is struck by a tennis racket : horizontal momentum of tennis ball = 2.4 Ns, kinetic energy of tennis ball = 45 J.(a) Calculate the mass and the velocity of the tennis ball.(b) When the racket hits the ball it strikes it with a constant force of 60 N in a direction opposite to its momentum, bringing it to rest momentarily. Calculate (i) the time the tennis ball takes to stop, (ii) the distance the tennis ball travels while stopping.(c) The force of 60 N then continues to act on the tennis ball for a further 0.060s. Calculate (i) the new momentum of the ball, (ii) the new velocity of the ball.(d) Calculate the increase in kinetic energy of the ball for the whole time that the force is applied to it and hence deduce the mean power being delivered to the ball while it is in contact with the racket.(e) Suggest why, in practice, it is impossible for a constant force to be applied to the ball.

(a) mv = 2.4 -------------------(1) ; 21 mv2 = 45 -----------------------(2)

(2) ÷ (1) : 21 v =

4.245 v = 37.5 ms1.

Subst.into (1) : m (37.5) = 2.4 m = 0.064 kg

(b) Deceleration, a = mF =

064.060 = 937.5 ms2

(i) v = u + at t = a

uv = 37.5 0

937.5

= 0.04 s

(ii) s = ut + 21 at2 = 0 + 2

1 (937.5)(0.04)2 = 0.75 m

(c)(i) New momentum = F t = (60)(0.06) = 3.6 Ns

(ii) New velocity, v = F tm

= 064.06.3 = 56.25 ms1

(d)(i) New K.E = 21

mv2 = 21 (0.064)(56.25)2 = 101.25 J

Increase in K.E = 101.25 – 45 = 56.25 J

Hence, mean power = takentime

E.Kinchange = 56.25

(0.04 0.06)= 563W

(e) The stringing on the racket is not rigid, but stretches upon impact with the ball. The stretching leads a change in the tension of the stringing. Hence the force applied to the ball cannot be constant in practice. Furthermore, the racket is usually in motion when the ball is hit.

Dynamics Pg 16

(a) A particle A of mass M moving with velocity u in the direction shown in the diagram collides head-on with a particle B of mass m which originally at rest. The collision is perfectly elastic. After collision, A and B move off with velocities V and v. (i) Write down equations which summarize the application of the principles of conservation of energy and momentum to this collision.

(ii) What is the ratio mM

such that all the kinetic energy of A is

transferred to B during the collision. (i.e. V = 0) ? (b) Two identical steel spheres suspended so that they are free to move in a vertical plane. The separation to he pairs of suspension threads is equal to the diameter of a sphere. Sphere X is displace to the right and then released. With reference to your answer to (a), discuss the subsequent motion of the spheres.

(a)(i) Conservation of momentum : Mu = MV + mv -------------------------(1)

Conservation of K.E : 21 Mu2 = 2

1 MV2 + 21 mv2 -----------------------(2)

(ii) When V = 0 (i.e. total transfer),

Eqn (1) : Mu = mv mM

= vu ---------------------------------(3)

Eqn (2) : Mu2 = mv2 mM

= 2

2u

v-------------------------------(4)

Subst. (3) into (4) : mM

= 2m

M

Mm

1

Mm = 0

Mm = 0 (rejected) ,

Mm = 1

∴Mm = 1 Total transfer only only occurs when A & B are equal in mass.

(b) When sphere X is released, it moves towards the other sphere (let it be Y). Upon colliding with sphere Y, sphere X transfer all its K.E to it according to the reasoning in (a) since they have the same mass. Sphere X then stay still while sphere Y start to move to the left. After swinging to its peak, sphere Y swings back to the right and collides with sphere X transferring all its K.E to the latter. It then stay still while Sphere X starts to move and repeats the whole motion again. Its moves like a pendulum but with each sphere performing half of each oscillation. Due to small energy loss at each collision and friction with air, the motion loses amplitude and eventually stop.

Dynamics Pg 17

(a) A neutron of mass m and velocity u collides elastically head-on with a stationary carbon atom of mass M. The velocities of the neutron and the carbon atom after the collision are v and V respectively. (i) Write both equations which represent the conservation of momentum and energy. (ii) Given that eliminating M and m from these equations results in the equation (V – v) = u, find an expression for v in terms of m, M and u.(b) In a nuclear reactor, carbon atoms are used t slow down neutrons. (i) assume that m, the mass of a neutron, is 1.0Mu, and M, the mass of a carbon atom, is 12Mu, what fraction of its kinetic energy does a neutron retain after an elastic head-on collision with a carbon atom ? (ii) How many such head-on collisions would be needed to reduce the kinetic energy of the neutron to one millionth of its original value ? Discuss qualitatively the effect on your two answers of not restricting the problem to head-on collisions only.

(a)(i) Conservation of momentum : mu = mv + MV -------------------------(1)

Conservation of K.E : 21 mu2 = 2

1 mv2 + 21 MV2 ------------------------(2)

(ii) V – v = u V = u + v & subst. into (1) :

mu = mv + M (v + u) mu = mv + Mv + Mu v =

MmMm u

(b)(i) K.E of neutron after collision = 21 muv2 = 2

1 mu

2m Mm M

u2 = 2

1 mu

2u u

u u

m 12 mm 12m

u2

= 21 mu

21113

u2 = 211

13

[ 21 muu2 ] = 121

169[ 2

1 muu2 ]

∴ 169121 of fraction retain.

(ii) 2n11

13

6

1

10 2n lg

1311 lg

61

10 n 41.4 n 42.

If the collision takes place at an angle, we can resolve the velocity of the neutron into 2 components. One along the line joining the centres of the neutron and the carbon atom.

The other perpendicular to the line. The parallel component will be reduced by a factor of 1311

while the perpendicular component will be conserved. The sum of the two components which gives the resultant velocity will be larger than that of head-on collision. Therefore, larger fraction of the K.E will be preserved by the neutron and more collisions are collisions are needed to reduce its K.E.

(ii) They have the same velocity at Ot because body A

cannot compress body B further. Before Ot , A is faster than B, so both compress.

After Ot , A is slower than B, so both separate.

At Ot , V1 = V2 from (1), V1 = u – V1

V1 = V2 = 2u

Dynamics Pg 18

(a) A body A of mass m with velocity u makes a perfectly elastic head-on collision with an identical body B which is initially at rest. Describe in words the motion of the bodies after the collision.(b) The elastic collision mentioned in (a) is one in which the bodies become temporarily compressed and remain in contact for a short time. (i) On the same axes of velocity against time, sketch labeled graphs of the velocity of A and the velocity of B. The time axis should extend from a time before the bodies come into contact to a time after they separate : mark on this the time Ct at which they first touch, the time to at which

they suffer maximum compression, and the time St at which they separate.

(ii) Explain why bodies have the same velocity at Ot , ( the time of maximum compression).

What is this velocity ?(iii) Hence find, in term of m and u, the total energy of the bodies at to and again at a time after they have completely separated. Account for the difference between these energies.

(a) Conservation of momentum : mu = mV1 + mV2 u = V1 + V2 V1 = u – V2 --------------------(1)

Conservation of K.E : 21 mu2 = 2

1 mV12 + 2

1 mV22

u2 = V12 + V2

2 -----------------(2) Subst. (1) into (2) : u2 = (u – V2)2 + V2

2 u2 = u2 – 2uV2 + V22 + V2

2

2V2

2 – 2uV2 = 0 V2 (V2 – u) = 0 V2 = 0 (NA) , V2 = u.

Subst. V2 = u into (1) : V1 = u – u = 0. ∴ V1 = 0 & V2 = u The body A reduces its speed and then come to rest whereas body B moves with a velocity of u.

(b)(i)

(iii)Total K.E at Ot = 21 mV1

2 + 21 mV2

2 = 21 m

2 2u u2 2

= 21 m

2u2

= 41 mu2

Total K.E at separation = 0 + 21 mu2 = 2

1 mu2. ∴ Difference in K.E = 21 mu2 – 4

1 mu2 = 41 mu2

The loss in K.E during compression from Ct to St goes into elastic compression energy of both

masses. This energy is returned to translational K.E after separation.

Dynamics Pg 19

A body of mass m makes a head-on, perfectly elastic collision with a body of mass M, initially at rest.

Show that 0EE =

Mm

2Mm

4

1 where E0 is the original kinetic energy of the mass m and

E is the kinetic energy it loses in the collision.

Let u be the initial velocity of body with mass m.Let V1 & V2 be the velocity of the body with mass m and M respectively after collision.Conservation of momentum : mu = mV1 + MV2 -------------------------(1)

Conservation of K.E : 21 mu2 = 2

1 mV12 + 2

1 MV22------------------------(2)

E = 21 mu2 – 2

1 mV12 = 2

1 m (u2 – V12 ).

0EE =

2 2112

212

m( u V )

mu

=

2 21

2u V

u

= 1 –

21V

u

We need to find V1 , u by eliminate V2.

From (1) : V2 = Mm (u – V1) ------------------------(3) Subst. (3) into (2) :

21 mu2 = 2

1 mV12 + 2

1 M2

1m (u V )M

m (u2 – V1

2 ) = 2m

M(u – V1)2

M (u + V1)(u – V1) = m (u – V1)2 M (u + V1) = m (u – V1)

(M + m) V1 = (m – M) u 1Vu

= mMMm

----------------------(4)

Subst. (4) into 0EE :

0EE = 1 –

2

2(m M)

(M m)

= 2 2

2(M m) (m M)

(M m)

= 2 2 2 2

2M 2Mm m m 2mM M

(M m)

= 2

4 mM

(M m) =

Mm

2Mm

4

1 (Shown)

When two strong magnets are held stationary with the north pole of one pushed against the north pole of the other. On letting go, the magnets spring apart. It is apparent that the kinetic energy of the magnets has increased. Explain how the law of conservation of momentum applies in this case.

When the north poles are pushed together, energy is stored in the magnetic field and is then converted into K.E when they are release. Both will move apart in the opposite directions.Total linear momentum before the release = 0Total linear momentum after the release = 1 1m v – 2 2m v (–ve : opposite)

∴ 1 1m v – 2 2m v = 0 1 1m v = 2 2m v

Dynamics Pg 20

A stationary radon nucleus may decay spontaneously into a polonium nucleus and an α-particle

as shown 2 2 2

8 6Rn 2 1 8

8 4Po +

4

2He.

The rest masses of these nuclei are 2 2 2

8 6Rn , 222.0176u ,

2 1 8

8 4Po , 218.0090u ,

4

2He, 4.0026u,

and it may be assumed that no ray is emitted.(a) Calculate the total kinetic energy of the decay products.(b) Explain how the principle of conservation of momentum applies to this decay and calculate the speed of the α-particle.

(a) Mass loss in the decay = mass of (Rn – Po – He) = (222.0176 – 218.0090 – 4.0026) u

= 0.0060 u where 1u = 1.66 ×1027 kg

K.E = 2mc = (0.0060)(1.66 × 1027 )(3 × 108 )2 = 8.96 ×1013 J.

(b) Since the Rn nucleus is stationary at the beginning with initial momentum zero, therefore the momentum of Po and He atoms must be equal and opposite so to add up to zero.

Let M be the mass & u be the speed of Po. Let m be the mass & v be the speed of He.

Conservation of momentum : Mu – mv = 0 u = Mmv

∴ Total K.E = 21 [ Mu2 + mv2 ] = 2

12

2mvM mvM

= 21

22 2m v mv

M

= 2

1 v22m m

M

8.96 × 1013 = 21 v2

2274 4 1.66 10

218

v =1.63 × 107 ms1 or 0.05c.

In a gas, a hydrogen molecule, mass 2.00u and velocity 1.88 × 103 ms1 ,collides elastically and head on with an oxygen molecules, mass 32.0u and velocity 405 ms1. In qualitative terms, what can be stated about the subsequent motion as a result of knowing that the collision is head-on ? Hence determine the velocity of separation of the two molecules after the collision.

“Head-on” implies that velocity of approach of the two molecules is equal to their velocity of separation and the motion takes place in a straight line.Velocity separation = velocity of approach

2 1v v = 1 2u u = 1.88 × 103 – (– 405) = 2285 ms1

Dynamics Pg 21

A proton travelling with speed 8.2 ×105 ms1

collides elastically with a stationary proton in a hydrogen target. One of the protons is observed to be scattered at a 60º angle. At what angle will the second proton be observed ? What will be the velocity of the two protons after the collision ?

For glancing elastic collision between 2 equal masses,θ + = 90º = 90º – 60º = 30º.

1v = u cos = 8.2 × 105 cos 60º = 4.1 × 105 ms1

2v = u sin = 8.2 × 105 sin 60º = 7.1 × 105 ms1

A radioactive nucleus, initially at rest, decays by emitting an electron and a neutrino at right angles to

one another. The momentum of the electron is 1.2 × 1022 kgms–1 and that of the neutrino is

6.4 × 1023 kgms1 . The mass of the residual nucleus is 5.8 × 1026 kg.(a) Find the direction and magnitude of the momentum of the recoiling nucleus.(b) What is the kinetics energy of the recoiling nucleus ?

(a) By conservation of momentum,

Pr2 = Pe

2 + Pn2 = (1.2 ×1022 )2 + (6.4 ×1023 )2

Pr = 1.36 × 1022 kgms1

tan θ = n

e

Pp

= 23

226.4 10

1.2 10

θ = 28º

∴ Direction is 180º – 28º = 152º from the electron track & [ 180º – (90º – 28º ) ] = 118º from the neutrino track.

(b) K.E = 21 mv2 = 2

12(mv)

m= 2

12

rPm

= 21

22 2

26(1.36 10 )

5.8 10

= 1.59 × 1019 J

Dynamics Pg 22

Obtain a relation between the linear momentum p of a body of mass m and its kinetic energy T.Linear momentum is a vector quantity. How do you take this into account when solving problems involving the conservation of momentum ?

A106

44Ru nucleus, originally at rest, undergoes radioactive decay by

emitting a β-particle of momentum 5.5 ×1024 kgms1 . It is found

that the residual nucleus recoils with momentum 8.4 ×1024 kgms1 at an angle of 120º to the direction of emission of the β-particle. A neutrino (a neutral particle of negligible mass, but finite momentum) is also emitted, and moves in the same plane as the β-particle and the residual nucleus.

(a) Write down the atomic number and the mass number of the residual nucleus. Calculate its kinetic energy.

(b) Calculate the magnitude and direction of the vector sum of the momentum Pn of the nucleus and the momentum P of the β-particle.

(c) Deduce the magnitude and direction of the momentum of the neutrino.

Linear momentum p = mv; K.E, T = 21 mv2 =

2 2m v2m

= 2(mv)

2m=

2P2m

P2 = 2mT p = mT2Since linear momentum is a vector quantity, the resultant momentum of the system must remain unchanged if the conservation of momentum applies.

(a) Atomic no. = 45, Mass no. = 106, T = 2P

2m=

24 2

27(8.4 10 )

2(106 1.66 10 )

= 2.0 × 1022 J

(b) From the vector diagram,

P2 = Pn2 + P

2 – 2Pn P cos 60º

= [8.42 + 5.52 – 2 (8.4)(5.5) cos 60º ] ( 2410 )2

= 7.39 × 2410 kgms1.

nPsin

= 60sin

P

sin θ = nPP

sin 60º =39.740.8 sin 60º = 0.9844 θ = 80º to the horizontal.

(c) By law of conservation of momentum, the momentum of neutrino must be equal and opposite to

the resultant of Pn and P .

∴ momentum of neutrino = 7.39 × 1024 kgms1 at 80º to the horizontal.