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Comments (17)
Koh Xin Qi said
at 7:15 pm on Feb 21, 2013
I don't understand the solutions for WS 1.2 (questions 11&12b) and WS 1.4 (questions 4k&6)
johnlittlephysics said
at 9:31 pm on Feb 21, 2013
WS 1.2 Q11 Anwers:
The car moves northwards then southwards, along a straight road. We choose a direction to be positive, in this northwards.
Any vector in north direction is given a positive value. So initial velocity is positive, and final velocity (5 s later) is negative.
johnlittlephysics said
at 9:37 pm on Feb 21, 2013
WS 1.2 Q12(b) Answers:
From the table and part (a), the speed is observed to increase every second from 1 to 5 to 9 ... m/s.
Acceleration is the change in velocity every second.
Here the increase (change) in speed every second is 4 m/s.
Or acceleration is 4 m/s per second = 4 m per square second, a constant acceleration.
johnlittlephysics said
at 9:44 pm on Feb 21, 2013
WS 1.4 Q4(k):
Part (i): AB shows equally spaced dots, so throughout AB, including B, speed is constant = u = 9.3 m/s.
Part (j): CD also shows equally spaced dots, so through CD, including C, speed is constant = v = 52.5 m/s.
Hence, for BC: average a = (v - u)/t, where t is time interval between AB.
johnlittlephysics said
at 9:50 pm on Feb 21, 2013
WS 1.4 Q6:
Diagram shows 6 pictures of car placed together. Think of the pictures behaving like dots on the ticker tape, where spacing are of equal time intervals.
2 pictures taken by camera every 2 s, so time interval from a picture to the next is 1 s / 2 = 0.5 s.
From picture 1 to 6, there are 5 spacings, total time, t = 5 x 0.5 s.
From first to last lamp posts, total distance, d = 8 x 5.0 m. Hence, average v = d / t
Koh Xin Qi said
at 8:25 pm on Mar 5, 2013
Umm... Mr Ang, can you put the solution of WS 1.6 question 3e here with explanation? Because when I marked the worksheet, I couldn't find the solution to the question. Thank you
johnlittlephysics said
at 11:18 pm on Mar 5, 2013
The WS 1.6 answers are on iTunes U.
Q3e: - 10 m/s^2
The acceleration at 11s = acceleration from 10 to 12 s = gradient of the straight line QR = (v - u)/ t = (0 - 20)/ (12 - 10)
chinweiyee98@... said
at 7:35 pm on Apr 19, 2013
Mr ang, if there an object is not moving is it possible that there is friction acting on it just that the forces are balanced?
And precision is half the smallest reding on the scale right, so why is it so that the metre rule of which the smallest scale it can read is 0.1cm the precision is not 0.05 but 0.1?
johnlittlephysics said
at 12:19 am on Apr 20, 2013
1. If the object is resting on a horizontal surface with no other horizontal force (or no horizontal component of a force), there is no friction acting on it. Otherwise, friction is not balanced by any other force.
If an object is at rest on a slope , friction can act on it to keep it from slipping down the slope.
2. For most instruments, precision can be half the smallest division.
For metre rule, vernier calipers and micrometer screw, the precision is smallest division (0.1, 0.01 and 0.001 cm respectively). There is an uncertainty of half smallest division associated with the zero reading (with or without zero error). E.g. metre rule to measure a length between two markings 0.0 cm and 1.2 cm. Both the values 0.0 cm 1.2 cm have a possible error of 0.05 cm, so the maximum possible error is 0.05+0.05=0.1 cm for a measurement of a length (difference) between 0.0 and 1.2 cm.
- btw, Measurements is not main topic in BT1.
johnlittlephysics said
at 2:06 pm on Apr 20, 2013
302 Li June: Does deceleration always mean that an object is slowing down? What is the difference between deceleration and negative deceleration?
johnlittlephysics said
at 2:09 pm on Apr 20, 2013
1. Deceleration (or retardation) of an object means object is slowing down.
2. We normal don't say negative deceleration; usually we refer to positive or negative acceleration.
a. Acceleration can be positive or negative (sign) depending on the sign convention.
b. Whether an object is increasing or decreasing in speed depends on the actual direction of velocity of object compared to actual direction of acceleration (due to resultant force).
E.g. If upwards direction if positive, acceleration due to gravity a = -10 m/s^2 because weight of object (downwards) causes acceleration downwards (due to gravity).
- If object is thrown and moving upwards say 5 m/s (opposite in direction to actual downward acceleration due to gravity), it decelerates or slow down and we say it experiences a deceleration; note that a = -10 m/s^2 and u = 5 m/s (in opposite directions, so opposite in sign).
- If object is falling downwards say 7 m/s (same direction as actual downward acceleration), it accelerates or pick up speed and moves increasingly faster; note that a = -10 m/s^2 and v = -5 m/s (in same direction, so same in sign).
johnlittlephysics said
at 10:28 pm on Apr 20, 2013
302 Agalia: I don't really know how to interpret this graph. Could you explain it to me?
https://www.dropbox.com/s/a0vja37suilapng/Photo%2020-4-13%2010%2027%2011%20PM.jpg
johnlittlephysics said
at 10:36 pm on Apr 20, 2013
The ball is released, so it falls from t = 0 to 0.50 s when it's speed increases (velocity is negative because downwards is negative sign convention).
At t = 0.50 s, it hits the ground and changes direction abruptly as it bounces up. Hence it's velocity changes from negative (just before hitting ground) to positive (just after leaving the ground).
As it moves up its speed decreases to zero at its highest point at t = 0.90 s. Then it starts to fall again with increasing speed till t = 1.00 s.
johnlittlephysics said
at 11:07 am on Apr 21, 2013
302 Li June: I don't really understand the bonus question in physics quiz 4 on kinematics iii. May you please explain?
johnlittlephysics said
at 11:25 am on Apr 21, 2013
The ball is thrown downwards with an initial velocity say u.
See sketch at https://www.dropbox.com/s/p9v2i3hyps22vz3/Photo%2021-4-13%2011%2017%2030%20AM.jpg
Assuming no energy loss due to air resistance or collision with the ground, its speed just before hitting the ground v3 is equal to its speed just after leaving the ground v4.
When it bounce back to the same height (as when it was thrown), its speed will be the same as when it was thrown, so v1=u.
It will travel a little higher before reaching its maximum height when v=0.
Determine the speed v3, same as v4.
Consider the motion from ground to maximum height: initial speed = v4, final speed = 0, a = -10 m/s^2
Hence determine maximum height s is 12.8 m.
johnlittlephysics said
at 1:49 pm on Apr 21, 2013
302 Adeline: Could you explain q2 please?
See https://www.dropbox.com/s/xcjeu2mog9c98io/Photo%2021-4-13%201%2046%2048%20PM.jpg
johnlittlephysics said
at 2:03 pm on Apr 21, 2013
Box is initially moving upwards. Sign convention in question is upwards is positive.
u = 30 m/s, a = -4 m/s^2 (net force & acceleration is downwards), t = 3 s
Box would decelerate to v = u+at = 18 m/s just before rope snapped. Box would continue moving up.
Acceleration due to gravity (a = -10 m/s^2 due to weight) now acts to slow down the box, till it stops momentarily at its highest point, the it accelerates downwards.
Answer: D
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