## Thursday, 2 May 2013

### 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 (Fy) the weight of the trolley (mg) – thus A: Fy + 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 Fh triggers “Friction” - if Fh = 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 Fv 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. Fv  + Normal reaction = Weight of the block. When Fv = 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...