# Physics 9702 Doubts | Help Page 35

__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 t

_{1}and receives it back at time t

_{2}. 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 t

_{1}, 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 t

_{1}), 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 t

_{1}itself, not from t

_{1}to t

_{2}. [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 t

_{2}, 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 t

_{2}, 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 t

_{2}. [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: v

^{2}= u

^{2}+ 2as

9.0

^{2}= 0

^{2}+ 2 (9.81) s

Minimum distance, s = 9.0

^{2}/ (2x9.81) = 4.13m

**(d)**

(i) Kinetic energy at P = ½ mv

^{2}= 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 mv

^{2}= 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

in q217, does the ball not move away from the table as a result of the reaction force(the reaction force for force it exerts on the table when it hits the table) of the table on it?

ReplyDeleteIt does, but this is not instantaneous. There is a time of collision, though this time may be small. The graphs shows how the force varies during this time of collision. Details of this is explained above.

Deleteplease explain question 10 and 16 of june 2014 paper 12

ReplyDeleteGo to

Deletehttp://physics-ref.blogspot.com/2014/10/9702-june-2014-paper-12-worked.html

In q217, why does the ball compress on hitting the ground? I know that the ball has kinetic energy on impact with ground but what force causes it to compress ?

ReplyDeleteAlso u said that the speed of the ball gradually decrease to zero . If this is the case, then the normal reaction force from the ground on the ball should be greater than the weight of the ball so that the resultant force on the ball is upwards causing it to deccelerate but why is the force exerted by ball on ground(newtons third law) greater than its own weight? And finally, if the ball is gaining speed upwards after normal force becomes maximum, should there still not be a resultant upward force on the ball to accelerate it upwards?

I would really appreciate if you could answer these questions with detail as these are difficult concepts to grasp. ThankYou.

The reason why the ball compresses (though this is not really important here) is due to the upward force exerted by the table on the ball.

DeleteActually, such details are not important. The important thing to remember are the last lines on the explanations above.

"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."

As long as the ball is not in contact with the table, the table cannot exert a force on it. Then, as they come in contact, the speed of the ball changes, causing its momentum to change and thus the force on the ball changes. The change is as shown by the graph.