Forces and Motion - Chapter Review Questions (SPM Fm 4 Chapter 2)

Updated on 21/7/2013...more updates to follow thereafter...

A)  Linear Motion, Ticker Tape, Distance, Displacement, Speed, Velocity and Acceleration

1.      2012 P1 Q6 at pg 332: Diagram shows a man walks from O: 2m towards west and then (2 + 7)m towards east. What is the displacement of the man? Answer: C, 7 m to the east (because displacement is the shortest straight line distance two reference points which are, in this case, the starting point O and the final point B. hence taking towards west or A as negative and towards east or B as positive, we have: -2m + (2 + 7)m = 7m i.e. displacement 7m to the right)

2.      2012 P1 Q11 at pg 334: Which ticker tape shows movement with uniform velocity and then deceleration? Answer: Tape in B (because the distance between initial intervals is the same meaning uniform velocity and then the distance get smaller and even smaller for later intervals, implying decreasing speed over time i.e. deceleration)

B)  Motion Graphs

1.      2012 P1 Q9 at pg 333: Diagram shows the velocity-time graph of a bus with passengers on board. The driver applies brakes after 10 minutes.  The graph shows an upward climbing straight line from the origin to velocity 6 m/s at time 10 s, then followed by a downward inclined straight line ending at velocity 0 at time 12 s. Q: What is the velocity of the passengers immediately after the brake is applied? Answer: C (because inertia will cause the passengers to move at the velocity just before the brake is applied i.e. at 6 m/s hence C)

2.      2012 P1 Q10 at pg 333: Diagram shows a mango falling – Q: Which acceleration-time graph represents the motion of the mango? Answer: B (for short distance fall, air resistance (drag) doesn’t lower the mango’s acceleration, g, much – so, the seemingly horizontal straight line in B represents its seemingly constant acceleration.)

C)  Inertia

1.      2012 P1 Q5 at pg 332: Shown 4 events / phenomena: A, an archer shooting an arrow; B a boy floating on a float; C, a ping pong ball floating under a tap with water flowing downwards around the ball; and D, shaking tomato sauce out of a tomato sauce bottle. Q: Which phenomenon shows the effect of inertia? Answer: D (because inertia will cause the sauce to move out at the speed just before the bottle’s movement is abruptly stopped)

D)  Momentum – Elastic and Inelastic Collisions; Explosions

1.      2012 P1 Q7 at pg 332: Question shows a trolley with mass 1.5 kg and velocity 2 ms^-2 collides with a stationary trolley of mass 1.0 kg. After the collision, both trolleys move together. Q: What is the velocity of both trolleys after the collsion? The collision is inelastic (trolleys stick together after collision) BUT the nature of the collision – whether elastic or inelastic- is immaterial because Law of Conservation of Momentum applies in either cases. The velocity after collision can found by equating momentum before collision to momentum after collision. Thus, m₁v₁ = (m₁ + m₂)v and v = 3/2.5 = 1.2 m/s – hence C)

2.      2012 P2 Secion A Q5 at pg 350~352 (8 marks): Collision and Conservation of Momentum - Diagram shows collision of bowling ball and bowling pin – Table shows momentum before and after collisions for ball and pin – Q:

·        What is the meaning of momentum?

·        Based on diagram and table, determine total momentum of ball and pin before and after collision (simple adding of values given in table)

·        Compared added values (same before and after collision)

·        Based on aforesaid comparison,  state a conclusion about total momentum (conserved or remained the same before and after collision)

·        Name the physics principle involved in foregoing conclusion (Law of conservation of momentum)

·        State 1 condition needed for the foregoing physics principle to apply (no other external forces act on  the bowling ball and pin)

·        Given: after collision, total kinetic energy of the ball and pin decreases – state the type of collision involved (inelastic collision)

3.      2011 P1 Q5 at pg 280:

Diagram shows 3 identical coins at rest on a horizontal surface. Q: What happens when 1 coin collides into the 2 coins which abut each other? (Ans B – situation of approximate elastic collision – much like in the case of Newton’s cradle – momentum and kinetic energy conserved)

E)   Effects of a Force – Balanced and Unbalanced Force

·        Newton’s 1st Law of Motion:

o   An object, if at rest, tends to remain at rest and if in motion, tends to move at the same speed in a straight line unless there a net force acting on it. Therefore,

§  A net force is needed to change velocity i.e. magnitude and/or direction

§  A net force changes momentum and causes impulse

§  A net force changes an object’s state of motion i.e. its state of equilibrium – static or dynamic

o   Any object with mass has inertia (covered in a separate segment) i.e. the natural resistance to change its current state of motion

·        Newton 2nd Law of Motion

o   An object subjected to a net force always experiences an acceleration (or rate of change in velocity) which is:

§  directly proportional to the net force

§  inversely proportional to its mass

§  summarised as: F = ma or a = F/m = (mv – mu)/t divided by m

1.      2012 P1 Q8 at pg 333: Diagram shows a woman pushing a trolley with force F. Q: Which of the 4 option shows the total downward force? Answer: A (because the downward force comprises the downward component of the push F (F_y) the weight of the trolley (mg) – thus A: F_y + mg)  

2.      2011 P1 Q6 at pg 281: Diagram shows a boy pulling a block on a rough surface with force F. Q: Diagram in which options shows action of forces acting on the block. (Answer D – My comments: The boy’s upward diagonal pulling force F tends to do 2 things: 1) the horizontal component F_h triggers “Friction” - if F_h = Friction, the block remains at it prevailing state of motion (not given) – either at rest or motion with uniform velocity; and, 2) the vertical component F_v tends to lift the block up against the downward gravitational force on the block – with the boy pulling, the “Normal reaction’ is lower than that without the boy pulling the block. F_v  + Normal reaction = Weight of the block. When F_v = 0 (boy not pulling), Normal reaction = Weight of the block)

F)   Impulse and Impulsive Force

1.      2012 P2 Section C Q11 a) and b) at pg 363~366 (5 marks):

·        Diagram shows golfer continues swing after golf ball is hit in what is called ‘follow through’

a)      What is the meaning impulse? (1 mark)

Answer: Impulse is the change in momentum (= mv – mu)

b)      Explain how follow through can increase impulse? (4 marks)

Answer: Follow through can increase impulse because:

§  Impulse = Ft = mv - mu.

§  With follow through, the time, t, that the swing force F acts on the golf ball increases;

§  which means Ft increases or the change in momentum (mv – mu) increases. ;

§  This results higher impulse - that is higher final velocity (v)  for the golf ball since initial momentum mu of the golf ball is zero and mass of the ball m is constant.

2.      2011 P1 Q4 at pg 280:

Diagram shows a long-jumper bends his legs upon landing. Q: Bending his legs to reduce what? A. Impulse B. Impulsive force C. Velocity D. Time of impact? Answer: B (Because bending his legs increases the time of impact thereby reducing the impulsive force on his legs)

G)  Force and Motion Affect Safety Features in Vehicles

Questions coming up soon…check out the blog

H)  Gravity

I)     Forces in Equilibrium

1.      2012 P1 Q 4 at pg 332: Diagram shows 4 forces acting on a stationary object – Q: Which statement about the forces is correct? Answer: C (because when an object is stationary, the 4 forces are in equilibrium, meaning the vertical opposite forces F₁ = F₂ (not a given option) and the horizontal opposite forces F₃ = F₄ (option C). – hence C)

J)    Work, Energy, Power and Efficiency

1.      2012 P2 Sec C Q11 d) at pg 365~366 (5 marks): Diagram shows a 60-kg man sliding down a water slide from top point A to bottom point B. Q:

a.       energy transformation – potential to kinetic;

b.      calculation of gravitational potential energy;

c.       calculation of speed ignoring friction.

(Note: To find the speed, I suggest you use the Law of conservation of energy: GPE (mgh) = KE (1/2mv²); If you use the equation of linear motion (v² = u² + 2as), you appear to suggest that all the GPE is converted into vertically downward component speed v? If so, where does the energy that causes horizontal displacement come from - if not from the GPE? The v will obviously not be in the same direction as the direction of a i.e. g, acceleration due to gravity)

K) Importance of Maximising Efficiency of Devices

 Q & A coming up soon...

L)   Elasticity

....Lots of Q & A coming up soon...

Linear Motion

 Kinematics - Linear Motion

1.1  Linear motion and Non-Linear Motion:

·        Linear motion is motion in a straight line – For examples: A passenger is carried by an escalator; or an athlete running a 100 m race; whereas

·        Non-linear motion is motion not in a straight line - A top spinning; or the earth orbiting the Sun.

1.2  Kinematics and Dynamics:

·        Kinematics is the study of the motion of an object without considering the force acting on it. Therefore, the equations of motion do not have force F as a variable in them.

·        Dynamics is the study of motion and the forces acting on the object.

1.3  The physical quantities involved in linear motion:

·        Distance and Displacement

·        Speed and Velocity

·        Acceleration and Deceleration

·        Time

The Physical Quantities Explained:

Suppose an object at point O moves east 100 metres (m) to point A in 14 seconds (s); then it immediately moves west from point A to point B 40 m away in 8 s:

·        Distance:

o       The total distance moved by the object = (100 + 40) m = 140 m;

o       Distance is the total length of the path travelled by an object in motion from 1 position to another position;

o       Distance is a scalar quantity – it does not take into account the direction of motion.

o       In linear motion: Distance = Magnitude of Displacement (since direction of motion remains constant)

·        Displacement:

o       The total displacement of the object from point O = (100 – 40) m = 60 m east of point O;

o       Displacement is a measure of how far and the direction in which an object has been displaced from a reference point (e.g. original position) due to the motion

o       Displacement measures the straight-line distance and the direction between the initial position and the final position of an object due to the motion.

o       Displacement is a vector quantity because it has both the magnitude and the direction.

·        Speed:

o       The object’s Average Speed from O to B = 140 m / 22 s = 6.36 m s^-1:

§         Its Av. Speed from O to A = 100 m / 14 s = 7.14 m/s

§         Its Av. Speed from A to B = 40 m / 8 s = 5 m/s

o       Thus, Average Speed, v, = Total Distance Travelled, s (m) / Time Taken, t (s).

o       Constant Speed: If an object moves equal distances in equal time intervals, then it is moving with constant speed – otherwise, it is moving with non-uniform speed.

o       Speed is a measure of how fast an object moves – the rate of distance travelled in the motion.

o       Speed is a scalar quantity – it measures only the magnitude of distance moved over time with no regard to direction of motion.

o       In linear motion: Speed = Magnitude of Velocity (since the direction of motion remains constant)

·        Velocity:

o       Its average velocity from O to B = (100 – 40) m / 22 s or 2.73 m/s  due east of O:

§         Its Av. Velocity from O to A = 100 m / 14 s = 7.14 m/s due east of O;

§         Its Av. Velocity from A to B = - 40 m / 8 s = - 5 m/s (west of A)

o       Thus, Average Velocity, v = Total Displacement, s (m) / Time Taken, t (s).

o       Constant Velocity v means equal displacements in equal time intervals.

o       Constant Velocity: If an object moves with equal displacement in equal time intervals, then it is moving with constant velocity – otherwise, it is moving with non-uniform velocity.

o       Velocity is a measure of how fast an object is displaced (the rate of displacement of the object) from a reference position (its initial position).

o       Velocity, unlike speed, is a vector quantity because it measures both the magnitude and direction of displacement due to motion.

o       A change in velocity means either a change in speed or a change in direction (or both) of motion.

·        Speed and Velocity in Linear Motion:

o       In a linear motion, the direction of motion remains unchanged: The speed of linear motion equals to the magnitude of velocity since the distance travelled per unit time in a linear motion is the same as the displacement per unit time along the straight line. Thus, in linear motion:

v = v = Total distance travelled, s (m) / Time taken, t (s)

o       In linear motion, a change in velocity can only means a change in the speed with no change in direction of motion and this change of speed can only be due to acceleration (increase in speed over time) or deceleration (decrease in speed over time).

·        Acceleration and Deceleration:

o       Acceleration is defined as the rate of change of velocity (or rate of change of speed, in linear motion):

Acceleration, a = Change in Velocity / Time Taken

                         = (Final Velocity, v – Initial Velocity, u) / Time, t

                      a = (v – u) / t …1st Equation of Motion

o       Acceleration is positive if initial velocity, u, increases with time, t, to final velocity, v: that is, if v > u, a is +ve. 

o       Conversely, acceleration is negative if velocity decreases to final velocity, v, from initial velocity, u: that is, if v < u, a is –ve, in which case, it is known as deceleration or retardation.

o       Constant (or uniform) acceleration:

·        If an object moves with equal change in velocity (or speed in linear motion) in equal time intervals, then it is moving with constant acceleration – otherwise, it is moving with non-uniform acceleration.

·        An object moves with constant acceleration, a, for time, t, will change the initial velocity, u, to final velocity, v, as follows:

v = u + at …2nd Equation of Motion; Therefore,

t = (v – u) / a …3rd. Equation of Motion 

o       Zero acceleration: If velocity is constant or uniform (or, speed is constant or uniform in the case of linear motion), the acceleration is zero since change in velocity (or, speed) is zero.

o       Acceleration or deceleration is a vector quantity with SI unit metre per second per second, m s^-2.

o       Equations of Motion are derived on the basis of constant or zero acceleration – please see later.

·        Time (Scalar Quantity in second (s)):

·        Time interval is an important quantity in the study of motion – it is measured by a stopwatch or by the use of “Ticker-Timer”.

·        There are 2 types of stopwatches depending of the accuracy needed:

Accuracy Needed              Type of Stopwatch Used

0.1 s ~ 0.2 s                       Analogue (mechanically-operated)

0.01 s                                 Digital (electronically-operated)

·        Apart from stopwatch, another device that we use to measure time interval of linear motion is the “ticker-timer” – please see the ensuing.

·        We know that an object that moves with constant acceleration, a, for time, t, will change its initial velocity, u, to final velocity, v:

v = u + at … 2nd Equation of Motion (as in foregoing);

         Therefore, time, t = (v – u) / a …3rd Equation of Motion

·        Ticker-Timer

Ticker-timer is a device that can be used to determine a number of quantities relating to linear motion of an object (including time interval), namely:

a)      Time Interval of the Motion

b)      Displacement of the Object

c)      Velocity of the Object

d)      Acceleration of the Object

e)      Type of Motion of the Object

·        The above quantities can be determined because ticker-timer has a metal strip with a pin that vibrates up and down at 50 Hz (which is the frequency of the 12 V or 6 V ac power supply). Each time the pin moves down at the interval of 1/50 seconds (or 0.02 s), it makes a dot on the pre-carbonated ticker tape which passes beneath it as the tape is pulled by the moving object to which the tape is attached. Thus,

a)      The time interval which elapses between successive dots is 1/50 second or 0.02 s (T = 1/f) – therefore, the time interval of the motion between any 2 points / dots can be determined by: Multiplying the number of dots after the 1^st point / dot until the other point / dot by 1/50 s; Therefore:

1 dot-space (1-tick space) = 1 x 0.02 s = 0.02 s

2 dot-space (2-tick of time) = 2 x 0.02 s = 0.04 s

5 dot-space (5-tick of time) = 5 x 0.02 s = 0.1 s

10 dot-space (10-tick space) = 10 x 0.02 s = 0.2 s

b)      Displacement of the object between any 2 points can also be determined by measuring the distance between the 2 points on the ticker tape

c)      Velocity:

·        The velocity of the objects between 2 points / dots is the displacement between the 2 points / dots over the number of tick-time over the 2 points / dots;

·        The average velocity between any 2 points is the total displacement between the 2 points of the tape over the total time intervals elapsed (i.e. the number of tick-time) between the 2 points.

·        Constant (or, Uniform) Velocity: When the ticker tape shows equal displacement over equal time intervals, the object is moving linearly with constant or uniform velocity.

d)      Acceleration or Deceleration:

·        The acceleration (or deceleration) of the objects between 2 successive intervals of motion on the ticker-tape is the change in velocities between the 2 intervals over the time between the mid-points of the 2 intervals. The velocities can be determined as in the foregoing.

·        The average acceleration (or deceleration) of a moving object between any 2 intervals of motion (note: intervals not points) can be determined by comparing the corresponding velocities for the 2 intervals over the time between the mid-points of the 2 intervals - increasing velocity means acceleration and decreasing velocity means deceleration.

·        Constant (or uniform) Acceleration: When the ticker tape shows equal increase (or decrease) in velocity over equal time interval, the object is moving linearly with constant acceleration (or constant deceleration, which ever is appliacable)

e)      From the foregoing, it is therefore clear: That the type of motion of an object – whether it is moving linearly at constant velocity (zero acceleration), at irregular acceleration or deceleration or at constant acceleration– can be seen and determined from the pattern of the dots on the ticker tape.

·        Equations of Motion

We have already learnt from the foregoing 3 basic equations of linear motion where the object moves with constant (or uniform) acceleration, a, with initial velocity, u, final velocity, v, for time, t:

1)      a = (v – u) / t …1^st  Equation of Motion

2)      v = u + at …2^nd Equation of Motion; and

3)      t = (v – u) / a …3^rd. Equation of Motion 3

The 4^th Equation of Motion is about displacement or distance travelled, s, of the object under the same state of motion:

4)      s = Average Velocity x Time Taken

      s = ½ (u + v)t … 4^th Equation of Motion

     

The 5^th and 6^th Equations of Motion are obtained from the 4^th Equation [s = ½ (u + v)t] by substituting the 2^nd Equation (v = u + at) and the 3^rd equation [t = (v - u)/a] respectively into the 4^th Equation. Thus,

5)   s = ½ (u + v)t…from 4^th Equation

s = ½ [u + (u + at)]t …(substitute 2^nd equation into the 4^th)

s = ½ [2u + at]t

s = ut + ½ at^2 …5^th Equation of Motion

(or, s = 1/2 (g) (t^2), free fall under gravity: 2005 P1 Q4. pg 3)

6)      s = ½ (u + v)t…from 4^th Equation

      s = ½ (u + v)(v – u)/a …(substitute 3^nd equation into the 4^th)

      s = ½ (v^2 – u^2)/a …(Form 3 algebraic expansion)

2as = v^2 – u^2

 v^2 = u^2 + 2as … 6^th Equation of Motion

Science is best learnt by understanding rather than by memorizing formulae (rote learnig). I believe, by understanding alone, you should be able to easily derive the first 4 equations of motion. In solving a kinematics problem, just ask yourself this: What is the variable that the question wants me to find the value?

·        If acceleration, a: then, use 1^st equation;

·        If final velocity, v: use 2^nd equation;

·        If time, t: use 3^rd equation;

·        If displacement, s: use 4^th equation.

At times, you need to find the value of another variable before you can solve your problem using the above simplified method – ample examples in my handouts.