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Question 215: [Dynamics > Conservation of Linear momentum]
Smith and Jones are skating on ice (assumed frictionless) so that they are moving with equal speeds s in the same straight line. Smith is skating backwards facing Jones. Smith throws a ball to Jones at time t₁ and receives it back at time t₂. Assuming that the time of flight of the ball is that the time of flight of the ball is negligible, which one of the sketches below gives the correct speed-time relationship for the two skaters?

Reference: Past Exam Paper – N76 / II / 3



Solution 215:
Answer: A.
First, let’s try to visualize the situation. Smith is in front of Jones, facing Jones. In this position, by skating backwards, Smith is actually moving in the same direction as Jones who is skating forwards towards Smith. Since they are both skating at the same speed s, the distance between them is constant.
The ball is initially with Smith. Smith will throw the ball to Jones.  Then, Jones will throw the ball back to Smith. Their positions are as described above. The time of flight of the ball is negligible, that is, it is assumed as the ball is thrown by one of them, the other one catches it at the same instant.  

The choices are speed-time graphs. It is known that apart from speed and time, 2 other quantities can be obtained from a speed-time graph. These are the distance (area under graph) and the acceleration (gradient). Since the distance is not being considered here, it is hinted that the acceleration will be an important quantity that will help us solve the problem.

Note that even though in the graph, before t₁, the speeds seem different; they are actually the same as mentioned in the question. This is only to help us to differentiate between the graphs of the 2 persons.

To throw the ball to Jones (at time t₁), Smith must apply a force on the ball (towards Jones – this direction is opposite to the direction of motion of Smith).  From Newton’s third law, the ball will exert a force equal in magnitude and opposite in direction. Since Smith is facing Jones and is skating backwards, the reaction force of the ball on Smith will be in the SAME direction as the motion of Smith. The force causes an acceleration on Smith. Since the direction of the acceleration and that of motion of Smith is the same, the speed of Smith increases.
At the same instant, Jones catches the ball. To catch the ball, Jones needs to apply a force on the ball (towards Smith – this direction is the same as the direction of motion of Jones) to slow it down to rest. From Newton’s third law, the ball will exert a force equal in magnitude and opposite in direction. Since Jones is moving forwards, the direction of the reaction force exerted by the ball on Jones will be OPPOSITE to the direction of motion of Jones. The force causes an acceleration on Jones. Since the direction of the acceleration and that of motion of Jones are opposite, the speed of Jones decreases.

But the throwing of the ball and the catching of the ball occurs at the same instant. So, the increase in speed of Smith and the decrease in speed of Jones occur at time t₁ itself, not from t₁ to t₂. [D is incorrect] Afterwards, the motion of Smith and Jones occur at constant speed since no forces continuously act on them. [C is incorrect]


At time t₂, Jones throws the ball at Smith again. This requires a forward force (in same direction as the direction of motion of Jones) and thus, the ball exerts a backward reaction force on Jones, in a direction opposite to the direction of motion of Jones. This causes the speed of Jones to further decrease.
At the same time t₂, Smith catches the ball by applying a force on the ball (the direction of the force applied by Smith on the ball is opposite to the direction of motion of Smith). The ball exerts a forward reaction force on Smith which is in the same direction as the direction of motion of Smith. This causes the speed of Smith to increase further at the instant t₂. [A is correct]


Note that the system consists of Smith, Jones and the ball which are all moving with the same speed initially. Each of them will have their own linear momentum. The sum of linear momentum (momentum is a vector) in the system remains constant at all instant.









Question 216: [Dynamics]
Car with front-wheel drive accelerates in direction shown.

Which diagram best shows direction of the total force exerted by the road on front wheels?

Reference: Past Exam Paper – June 2003 Paper 1 Q11 & November 2007 Paper 1 Q11



Solution 216:
Answer: B.
The force due to the weight of the front wheels act downwards on the road. From Newton’s third law, the road will exert a reaction force equal in magnitude and opposite in direction. So, the direction of the reaction force exerted by the road on the car is upwards.

Force = ma. The car is accelerating forwards. This acceleration is caused by the rotation of the wheel. At the POINT of contact with the road, the direction of the motion (acceleration) of the wheel is BACKWARDS - towards the left (This is obvious. For the car to accelerate towards the right, the wheels need to rotate in a clockwise direction). This acceleration of the wheel causes a force (in the same direction as the acceleration of the wheel) on the road. From Newton’s third law, the road exerts a force of the same magnitude in the opposite direction. Thus, this force exerted by the road is towards the RIGHT.

The total force exerted by the road on the front wheels is the vector sum of these 2 forces.









Question 217: [Dynamics > Newton’s laws]
A tennis ball is dropped onto table and bounces back up. Table exerts force F on the ball.
Which graph best shows variation with time t of force F while ball is in contact with the table?

Reference: Past Exam Paper – June 2014 Paper 13 Q10



Solution 217:

Answer: C.

Force is defined as the rate of change of momentum.

Momentum, p = mv

In this case, as the ball makes contact with the table, its momentum changes (because its speed reduces to zero with time). So, it exerts a force on the table. From Newton’s third law; the table exerts a (reaction) force F on the ball, equal in magnitude but in opposite direction.

Let’s analyze how the speed of the ball changes.

At the instant the ball makes contact with the table, its speed is still unchanged {so, the momentum is still unchanged} and the rate of change of momentum [i.e. force] is INITIALLY zero. Therefore, initially, the force is zero.

After that instant, the speed of the ball decreases gradually {speed is changing, so momentum changes. Thus, force F gradually increases} until it becomes zero {at this instant, the change in speed, and hence the change in momentum, is maximum. So, force F is maximum at this instant}.

Then, the speed of the ball increases in the opposite direction {while still in contact with the table – but in this case, the force by the ball is now AWAY from the table (in the opposite direction than before). So, the resultant force (reaction) F from the table gradually decreases} until it (that is, the speed of the ball) reaches a maximum speed in that opposite direction {at this instant, the resultant reaction force F from the table becomes zero}. At the instant the ball is about to leave the table, it has the same maximum speed reaches (it no longer exerts a force on the table since it’s about to separate from the table – so force F is zero).

The table does not exert any force on the ball at the instant they make contact or at the instant they separate. The force must gradually increase and then decrease between these points, so only C could be correct.









Question 218: [Dynamics > Momentum]
June 2007 Paper 1 Q11:
Lorry of mass 20 000 kg is travelling at 20.0 m s^–1. Car of mass 900 kg is travelling at 30.0 m s^–1 towards the lorry.

What is the magnitude of total momentum?
A 209 kN s                  B 373 kN s                  C 427 kN s                  D 1045 kN s


For June 2004 Paper 1 Q11:
Diagram shows a situation just before a head-on collision. A lorry of mass 20 000 kg is travelling at 20.0 m s^–1 towards a car of mass 900 kg travelling at 30.0 m s^–1 towards the lorry.
What is the magnitude of the total momentum?
A 373 kN s                  B 427 kN s                  C 3600 kN s                D 4410 kN s

Reference: Past Exam Paper – June 2004 Paper 1 Q11 & June 2007 Paper 1 Q11



Solution 218:

June 2007 Paper 1 Q11 - Answer: B.

June 2004 Paper 1 Q11 – Answer: A.

Momentum p = mv

Momentum is a vector quantity. So, it has a direction. Let the positive direction be defined as the direction towards the right.

So, the lorry is travelling in the positive direction while the car is travelling in the negative direction.

Total momentum = Momentum of lorry + Momentum of car

Total momentum = 20000(20) + 900(-30) = 373000Ns = 373kNs

Question 219: [Work, Energy, Power]
(c) The minimum flying speed for a bird called a house-martin is 9.0ms^-1. It reaches this speed by falling from its nest before swooping away. Calculate the minimum distance its nest must be above the ground.

(d) A house-martin has a mass of 120g. When it returns to its nest, it is travelling horizontally at P with a speed of 13.0ms^-1 and at a distance 7.5m below its nest. It then glides upwards to the nest, as shown in Fig.

Neglecting air resistance, calculate
(i) the kinetic energy of the house-martin at P
(ii) the total gain in potential energy as it glides upwards to its nest
(iii) its kinetic energy as it reaches its nest
(iv) its speed as it reaches its nest

(e) Discuss qualitatively how air resistance affects, if at all, each of your answers in (d).

Reference: Past Exam Paper – N98 / III / 1 (part)



Solution 219:
(c)
Initial speed at its nest, u = 0
Acceleration of free fall, a = 9.81ms^-2
Final speed, v = 9.0ms^-1

Consider the equation for uniformly accelerated motion: v² = u² + 2as
9.0² = 0² + 2 (9.81) s
Minimum distance, s = 9.0² / (2x9.81) = 4.13m

(d)
(i) Kinetic energy at P = ½ mv² = 0.5 (0.120) (13.0)² = 10.14 = 10.1J

(ii) Gain in potential energy = mgh = 0.120 (9.81) (7.5) = 8.829 = 8.83J

(iii)
From the conservation of energy,
Kinetic energy at nest = Kinetic energy at P – Gain in potential energy
Kinetic energy at nest = 10.14 – 8.829 = 1.31J

(iv)
Kinetic energy at nest = ½ mv² = 1.31J
Speed at nest, v = √[(2x1.31) / 0.120] = 4.67ms^-1    

(e)
Air resistance opposes motion. It tends to decrease the speed of the bird.
Since the speed is initially taken to be 13.0ms^-1 at P, the kinetic energy at P remains unchanged.
The gain in potential energy depends on the gain in height, not the speed. So, this is also unchanged.
Air resistance opposes motion. So, some work needs to be done to move against the air resistance. So, the kinetic energy at it reaches its nest is less.
The speed also decreases since it has less kinetic energy.









Question 220: [Kinematics > Linear motion]
A radio-controlled toy car travels along straight line for a time of 15 s.
Variation with time t of velocity v of the car is shown below.

What is average velocity of the toy car for journey shown by the graph?
A –1.5 m s^–1                B 0.0 m s^–1                  C 4.0 m s^–1                  D 4.5 m s^–1

Reference: Past Exam Paper – June 2014 Paper 12 Q6



Solution 220:

Answer: B.

The question asks for the average VELOCITY, not speed. Velocity is a vector (it has a direction and this needs to be considered) while speed is a scalar.

Average Velocity = Total displacement / Total time

Displacement is given by the area under graph. For positive velocities, the displacement is positive and for negative velocities, the displacement is negative.

Total displacement = (3x10) + (-6x5) = 0

Therefore, the car starts and finishes at the same point.

So, average velocity = 0