4.1 - Further Mechanics

1 | Q u a z i N a f i u l I s l a m – w w w . s t u d e n t t e c h . c o . c c

FURTHER MECHANICS 1.1 – MOMENTUM

1. Momentum 2. Collisions 3. Energy in Collisions 4. Collisions in 2D

1.2 – CIRCULAR MOTION

1. Angular Displacement 2. Centripetal Force

MOMENTUM

MOMENTUM

The momentum is a measure of the amount of force that has to be applied over a period of time in order to bring an object to the speed at which it is moving. In simple terms, it’s how hard something hits you. The symbol 𝒑 is used for momentum.

Here are the equations relating to momentum:

𝑭 =𝒑𝒕

𝒑 = 𝒎𝒗

The product of a force applied for a certain time is called impulse.

𝑰(𝒊𝒎𝒑𝒖𝒍𝒔𝒆) = 𝑭𝒕

QUESTIONS ON PAGE 13 – EDEXCEL A2 PHYSICS

1. What is the momentum of a. A plane of mass 2 × 105 Kg flying at a speed of 270𝑚𝑠−1?


b. A horse of mass of 400 Kg carrying a rider of mass 75 Kg, galloping at 10ms−1

2. In the worked example above, it was estimated that a cricket batsman might need to use 80N to stop a ball dead. a. Explain why a batsman hitting a ball so that it travels away at the same speed might need less than 80N force

to play his shot.

It would need 80N to stop the ball entirely. However, if the batsman were to cut the ball i.e. slightly change the direction of the ball, less force would be required to keep the ball travelling away at its original speed

b. Explain why the same batsman playing the same ball might need more than 80N force to play a shot.

If the ball is returned back in the direction of the bowler, this would be a greater momentum change than in the worked example, so more force would be needed.

3. An 800 Kg car is travelling at 33 ms−1. How much force must the brakes apply, if they are to stop the car completely in four seconds?

Figure 1 - We have to view the whole problem as a system. So we need to assume that the horse and its rider is one particle, hence 475 Kg


4. The following questions are related to one another a. In a car crash, a passenger with a mass of 82 Kg is not wearing a seatbelt. The car is travelling at 45 Kmh−1.

What impulse must the car’s airbag provide in order to stop the passenger’s motion?

b. Explain why hitting an airbag will cause less injury than if a passenger hits the dashboard.

The air bag will cushion the passenger, meaning that the impact of the collision i.e. the momentum will be applied over a longer period of time, reducing the force applier to the passenger as F = p

t.

5. A wooden mallet is being used to hammer a tent peg into hard ground. a. The head of the mallet is a cylinder of diameter 0.100m and length 0.196m. The density of the wood is 750

Kgm−3. Show that the mass of the head is approximately 1.2 Kg.


b. The head strikes the tent peg as shown at a speed of 4.20 ms−1 and rebounds at 0.58 ms−1. Calculate the magnitude of its momentum change in the collision.

Figure 2 - Remember that velocity is a vector. We took the direction of the mallet going down as positive and thus the direction of the mallet going up must be negative.


c. The head is in contact with the peg for 0.012s. Estimate the average force exerted on the peg by the head during this period.

d. Give a reason why your value for the force will only be approximate.

It only takes into account the weight of the mallet head. It does not take into account the weight of the handle as well as the weight of the person’s hand, as we are summing that the only particle involved is the mallet head, whereas it is a combination of the head, the handle and the person’s hand. This reduces the mass in the equation and therefore reduces the momentum. As the momentum is being reduced but the contact time is constant, it gives us an approximate of the value that is less.

e. With reference to your calculations above, discuss whether a mallet with a rubber head of the same mass would be more or less effective for hammering in tent pegs.

The mallet with a rubber head would be less effective as the rubber would have a larger contact time as it is softer than wood. As it has the same mass, it would have the same momentum as the wooden head, but the force applied would be less. This is because since the contact time is higher, the force applied would be reduced as 𝐹 = 𝑝

𝑡.


COLLISIONS AND COLLISIONS IN 2D

• The conservation of linear momentum states that in a system with no external forces acting; the initial moment will be equal to the final momentum.

• For there to be no external forces, we would have to do the experiment in outer-space. So, we assume that there are no external forces acting such as air resistance or friction between surfaces.

• There are mainly two types of collisions in terms of energy o Elastic collisions o Inelastic collisions

• Explosions are where the particles are a single particle to begin with, and they break up into two different particles each having the same momentum as the other

• When dealing with collisions in 2D, the momentum of the particles have to be equal both horizontally and vertically.

Here are the equations for collisions:

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑡𝑤𝑜 𝑐𝑜𝑙𝑙𝑖𝑑𝑖𝑛𝑔 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 = 𝐹𝑖𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑝𝑎𝑟𝑡𝑖𝑐𝑒𝑠


1. A movie stuntman with a mass of 90 kg stands on a stationary 1 kg skateboard. An actor shoots the stuntman with a 9mm pistol. The 8-gram bullet leaves the pistol at 358 ms−1 and embeds completely in the stuntman’s bulletproof vest. At what speed will the stuntman roll away?


2. A girl in a stationary boat on a still pond has lost her oars in the water. To get the boat moving again, she throws her rucksack horizontally out of the boat with a speed of 4 ms−1.

Mass of boat = 60 kg; mass of girl = 40 kg; mass of rucksack = 5 kg.

a. How fast will this action make the boat move?


b. If she throws the rucksack by exerting a force on it for 0.2s, how much force does she exert?


3. How can Newton’s third law help to explain the problem suffered by the boy stepping out of the boat in Worked example 2 in the section ‘Explosions’?

The boy is applying a force against the boat and this is a 3rd pair force. The initial momentum was 0. So the final momentum must be 0. So, the boy goes in front by a applying a force against the boat and the boat moves away from the boy by applying a force in the oppose direction to his movement. The two forces are equal and opposite and this they are 3rd pair forces and they are third pair forces because they are formed in reaction to one another.

4. In a stunt for an action movie, the 100 Kg actor jumps from a train which is crossing a river bridge. On the river below, the heroine is tied to a raft floating towards a waterfall at 3 ms−1. The raft and heroine have a total mass of 200 Kg.

a. If the hero times his jumps perfectly so as to land on the raft, and his velocity is 12 ms−1 at an angle of 80° to the river current, what will be the velocity of the raft immediately after he lands? Draw a vector diagram to show the momentum addition. (Ignore any vertical motion.)


b. If the waterfall is 100m downstream, and the hero landed when the raft was 16m from the bank, would they plummet over the fall? (Assume the velocity remains constant after the hero has landed.)

5. The questions below are related: a. Define linear momentum.

Linear momentum is the product of an object’s mass and velocity. p = mv

b. The principle of conservation of linear momentum is a consequence of Newton’s laws of motion. An examination candidate is asked to explain this, using a collision between two trolleys as an example. He gives the following answer, which is correct but incomplete. The lines of his answer are numbered on the left for reference. In which of his arguments below did the candidate use Newton’s second law?

i. During the collision the trolleys push each other. ii. These forces are of the same size but in opposite directions.

iii. As a result, the momentum of one trolley must increase at the same rate as the momentum of the other decreases.

iv. Therefore the total momentum of the two trolleys must remain constant.

In statement (III) – Because Force ∝ acceleration, as the two trolleys decelerate and then accelerate away from each other, as their velocities are changed (the velocity changes from positive to negative if we take that from left to right is positive for the

trolley coming from the left). We need to also remember, F = ∆mv∆t

c. In which line is he using Newton’s third law?

In statement (II) – Because he states that the forces are equal and opposite and are in reaction to one another, hence they are related and thus they are Newton’s Third pair forces.


d. The student is making one important assumption which he has not stated. State this assumption. Explain at what point it comes into the argument.

He has assumed that there are no external forces acting on the two trolleys. If air resistance and drag were taken into consideration, it would reduce the forces applied by the trollies to each other. And this reduction would not be the same for both trolleys as the ground may not be the same.

e. Describe how you could check experimentally that momentum is conserved in a collision between two trolleys. • Have two light gates before each trolley, so as to measure the velocity of the trolleys before and after contact • Measure the mass of the trolleys • Apply forces to the trollies using large force springs. • Calculate momentum before and after collision

ENERGY IN COLLISIONS

• There are two types of collisions o Elastic – where the kinetic energy applied to the object(s) is conserved – a rubber ball bouncing off of a table is

a good example; however this is not completely elastic. o Inelastic – where the kinetic energy applied to the object is not conserved, a meat ball falling splat onto a

table.

Equations that needs to be known:

𝑬𝒌 =𝟏𝟐𝒎𝒗𝟐

𝑬𝒌 =𝒑𝟐

𝟐𝒎

𝝀 =𝒉

�𝟐𝑬𝒌𝒎


1. A bowling ball travelling at 5 ms−1 strikes the only standing pin straight on. The pin flies backward at 7 ms−1. Calculate:

a. the velocity of the bowling ball after the collision


b. the loss of kinetic energy in this collision


2. Calculate the kinetic energy of an alpha particle which has a momentum of 1.08 × 10−19 kgms−1 a. in joules b. in electron volts


c. in MeV

3. Explain why the slingshot orbit of a satellite passing near a planet and then flying away at a different angle as a result of the effect of the planet’s gravity would be an elastic collision.

As no energy would be lost in doing so, it is not coming into contact with anything, not even air particles. Thus, it has to be elastic.


1. An alpha particle moving at 3% of the speed of light collides elastically with a stationary aluminium nucleus (atomic number 13 and mass number 27). If the alpha particle bounces backwards at 0.1% of the speed of light, what is the velocity of the aluminium nucleus after the collision?


2. In a pool shot, the cue ball of mass 0.17 kg travels at 6 ms−1 and hits the stationary black ball in the middle of the table. The black ball, also of mass 0.17 kg, travels away at 45° with a speed of 4.24 ms−1, ending up in the corner pocket. By resolving the components of the black ball’s momentum, find out what happens to the cue ball – On the next page.


2


CIRCULAR MOTION

ANGULAR DISPLACEMENT

Arc Length



1. Convert: a. 4𝜋 radians into degrees b. 36° into radians


2. Vinyl records could be played at one of three speeds. Calculate the angular velocity of each. a. 33 revolutions per minute b. 45 rpm c. 78 rpm


3. What is the angular velocity of an athletics hammer if the athlete spins at a rate of three revolutions per second? 4. A man standing on the equator will be moving due to the rotation of the Earth. What is his angular velocity? What is his

instantaneous velocity?


5. There are 6283 milliradians in a complete circle. The army’s rounding of this to 6400 mils causes an error. a. How far sideways from the target could this rounding cause an artillery shell to be when aimed at a target 20

km away? b. Why does the actual bearing to the target not affect your answer to part a?

Please note, that the answer given in the answer books is wrong. This is the correct way to do it, and I hope that you will see how it is correct.


CENTRIPETAL FORCE

• When an object rotates, there is a constant centripetal force applied to keep it in circular motion. There is a constant change in velocity because there is a change in direction.

• Formulas of importance:

The first formula:

𝑭 =𝒎𝒗𝟐

𝒓

The second formula is derived from the first and another:

𝑣 = 𝑟𝜔

𝐹 =𝑚(𝑟𝜔)2

𝑟

𝐹 =𝑚𝑟2𝜔2

𝑟

𝑭 = 𝒎𝒓𝝎𝟐

The third formula is derived from the third and yet another formula that we have been introduced to before:

𝐹 = 𝑚𝑎

𝑚𝑎 =𝑚𝑣2

𝑟

𝒂 =𝒗𝟐

𝒓


1. A roller coaster has a complete (circular) loop with a radius of 20 m. A 65 kg girl rides the roller coaster and the car travels once round the loop in 1.5 seconds. What centripetal force does the girl experience?


2. A man of mass 75 kg standing on the equator is moving because of the rotation of the Earth. a. What centripetal force is required to keep him moving in this circle? b. How does this compare with his weight? c. How would the reaction force with the ground be different if he went to the South Pole?

(b) The value is much less than his weight.

(c) It would be higher as the centripetal force there is 0. It would be 736, compared to 733 on the equator.

4.1 - Further Mechanics

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