Lower Secondary Science

[Dr.033430Daniel]

red rose:

Q1. Tom uses a stopwatch to measure the time between 2 lamp-posts on a car journey. As the car speeds up, 2 consecutive times are 1.2s and 1.0s. Later, he finds out that the lamp-posts are 30m apart.
(a) Calculate the acceleration of the car and its speed at the first lamp-post.
(b) Assuming the same acceleration as found in (a), find the time the car took to travel the 30m before the first lamp-post.

For Question 1, part a)
His speed in the middle of the first time interval is 30m/1.2s = 25 m/s.
His speed in the middle of the second time interval is 30m/1s = 30 m/s
To find the acceleration, take change in speed over change in time. The change in speed is 5m/s. The change in time is 0.6s+0.5s = 1.1s. (the amount of time between the middle of the two time intervals). The acceleration is then 5/1.1 = 4.5 m/s^2.

To get the velocity at the moment he passes the first lamp post, take 25m/s to be a final velocity 0.6s later. Use: v= u+at. Solve for u, the initial velocity at the time he passes the first lamp post.

25m/s = u + 4.5(0.6s). u = 22.3m/s

Now with this velocity for part b you can solve using one of the kinematics equations. It will become the final velocity for part b. Assume constant acceleration.

This problem is unusual for secondary level in that it involves realizing that the speed you calculate is in the middle of a time interval. It takes a while to \"see\" this. I used to teach a laboratory class at University in the USA where we did these kind of calculations for falling objects.

[red rose]

Dr.Daniel:

red rose:

Q1. Tom uses a stopwatch to measure the time between 2 lamp-posts on a car journey. As the car speeds up, 2 consecutive times are 1.2s and 1.0s. Later, he finds out that the lamp-posts are 30m apart.
(a) Calculate the acceleration of the car and its speed at the first lamp-post.
(b) Assuming the same acceleration as found in (a), find the time the car took to travel the 30m before the first lamp-post.

For Question 1, part a)
His speed in the middle of the first time interval is 30m/1.2s = 25 m/s.
His speed in the middle of the second time interval is 30m/1s = 30 m/s
To find the acceleration, take change in speed over change in time. The change in speed is 5m/s. The change in time is 0.6s+0.5s = 1.1s. (the amount of time between the middle of the two time intervals). The acceleration is then 5/1.1 = 4.5 m/s^2.

To get the velocity at the moment he passes the first lamp post, take 25m/s to be a final velocity 0.6s later. Use: v= u+at. Solve for u, the initial velocity at the time he passes the first lamp post.

25m/s = u + 4.5(0.6s). u = 22.3m/s

Now with this velocity for part b you can solve using one of the kinematics equations. It will become the final velocity for part b. Assume constant acceleration.

This problem is unusual for secondary level in that it involves realizing that the speed you calculate is in the middle of a time interval. It takes a while to \"see\" this. I used to teach a laboratory class at University in the USA where we did these kind of calculations for falling objects.

Your help is greatly appreciated, Dr Daniel.

If it's not too much to ask, can you help with Q2 too? Thanks and have a good day!

[verydad]

red rose:

Hi! Wondering if anyone can help me out with these 2 upper sec Physics problems.


Q1. <Answered previously>

Q2. A juggler throws a ball up in the air with initial speed of 5.00m/s from a height of 1.20m. At the instant the ball reaches its maximum height, the juggler throws up another ball with the same speed and from the same height. At what height will the balls pass each other? (g=10.0m/s2)

Thank you in advance!

Q2 is a three-step question, which I know some students in my classes have difficulty breaking down into steps to solve them. Being able to break down the question into simpler steps is one of the important critical thinking skills in science questions. The steps are:
A. Find the distance and time taken from where the ball is thrown to the maximum height reached;
B. Using simultaneous equations, solve for the height above the 'throwing point' where the balls meet;
C. Add the 1.2m to the height reached to get the total height above ground.

Step A:
1. Ignore the 1.20 m for simplicity. Take upwards as positive. Then, find the maximum height reached as follows.
Using v = u + at, find the time taken where v = 0
0 = 5 – 10t(max)
t(max) = 0.50 s
Since t = 0.50 s, using s(max) = ½(u+v)t
s(max) = ½(5)(0.50) = 1.25 m.

Step B:
After getting the maximum height reached, two equations can be written out.
First, for the ball falling: s - smax = ½(-10)t2
s = 1.25 – 5t2 – (1)
Second, for the ball going up: s= ut + ½(at2)
s = 5t + ½(10)(t2) – (2)

Equating (1) = (2)
1.225 – 5t2 = 5t + ½(10)(t2)
Solve t = 0.25 s
Sub t = 0.25 s, s = 5t + ½(10)(t2) = 0.9375 m = 0.94 m (2 s.f.)

Step
Answer: 0.94 m + 1.20 m = 2.14 m


Hope that helps.

[red rose]

Would appreciate help in the following Physics problems:

1) A 5kg block of wood is pushed with a force of 60N along a horizontal flat surface at a constant speed of 10m/s. State the frictional force acting on the wood when it is pushed with a force of 40N.
2) A 6kg bowling ball collides with a single 1.5kg bowling pin. Determine the peak acceleration of the pin if the peak acceleration of the ball during the collision was 2m/s (assume no resistive force acting on both ball and pin).

Thank you in advance.

[mum_sugoku]

red rose:

Would appreciate help in the following Physics problems:

1) A 5kg block of wood is pushed with a force of 60N along a horizontal flat surface at a constant speed of 10m/s. State the frictional force acting on the wood when it is pushed with a force of 40N.
2) A 6kg bowling ball collides with a single 1.5kg bowling pin. Determine the peak acceleration of the pin if the peak acceleration of the ball during the collision was 2m/s (assume no resistive force acting on both ball and pin).

Thank you in advance.

1) If contact surface remains the same, frictional force would remain unchanged, ie 60N.

2) Assuming force acting on pin = force acting on ball, then,
m(ball)a(ball)=m(pin)a(pin),
6x2= 1.5 a(pin)
hence a(pin) = 8 m/s^2

[njcstudent]

Hi, I am an IP student in Sec 1. Can anyone please recommend any good reference books and assessment books?

[Dr.033430Daniel]

mum_sugoku:

red rose:

Would appreciate help in the following Physics problems:

1) A 5kg block of wood is pushed with a force of 60N along a horizontal flat surface at a constant speed of 10m/s. State the frictional force acting on the wood when it is pushed with a force of 40N.
2) A 6kg bowling ball collides with a single 1.5kg bowling pin. Determine the peak acceleration of the pin if the peak acceleration of the ball during the collision was 2m/s (assume no resistive force acting on both ball and pin).

Thank you in advance.

1) If contact surface remains the same, frictional force would remain unchanged, ie 60N.

2) Assuming force acting on pin = force acting on ball, then,
m(ball)a(ball)=m(pin)a(pin),
6x2= 1.5 a(pin)
hence a(pin) = 8 m/s^2

For question #1, if the block starts out moving at constant speed and the pushing force is reduced from 60N to 40N, then the force of friction will be 60N and the block will decelerate and come to rest. This is probably the way to interpret the question based on the way it reads, although it is not 100% clear.
If, on the other hand, the block is initially at rest and is pushed with a force of 40N on the same surface, the force of friction will also be 40N and the block won’t move.
I mention this because last year I saw a test where they were getting at this second point.

[Dr.033430Daniel]

njcstudent:

Hi, I am an IP student in Sec 1. Can anyone please recommend any good reference books and assessment books?

I had an njc student in Sec1 last year and they were not following a standard syllabus and were doing very hard material early on. One of their modules was called \"basic atmosphere\".
Best advice I can give is to build up your basic physics and chemistry understanding. The textbook Interactive Science for Inquiring Minds is not bad. Try to understand measurement, density, speed and then go through elements and compounds and atoms and molecules.
You will probably need a higher level book for O-Level, such as Physics Insights and Chemistry Insights as a reference for harder topics.
On assessments, The red book Lower Secondary Science by GS Cheng is good.

[mum_sugoku]

:oops: :oops: :oops:

Dr.Daniel:

mum_sugoku:

[quote=\"red rose\"]Would appreciate help in the following Physics problems:

1) A 5kg block of wood is pushed with a force of 60N along a horizontal flat surface at a constant speed of 10m/s. State the frictional force acting on the wood when it is pushed with a force of 40N.
2) A 6kg bowling ball collides with a single 1.5kg bowling pin. Determine the peak acceleration of the pin if the peak acceleration of the ball during the collision was 2m/s (assume no resistive force acting on both ball and pin).

Thank you in advance.

1) If contact surface remains the same, frictional force would remain unchanged, ie 60N.

2) Assuming force acting on pin = force acting on ball, then,
m(ball)a(ball)=m(pin)a(pin),
6x2= 1.5 a(pin)
hence a(pin) = 8 m/s^2

For question #1, if the block starts out moving at constant speed and the pushing force is reduced from 60N to 40N, then the force of friction will be 60N and the block will decelerate and come to rest. This is probably the way to interpret the question based on the way it reads, although it is not 100% clear.
If, on the other hand, the block is initially at rest and is pushed with a force of 40N on the same surface, the force of friction will also be 40N and the block won’t move.
I mention this because last year I saw a test where they were getting at this second point.[/quote]Agree --> action=reaction. (Thanks for pointing that out, frankly I've never thought of the second scenario at all :oops: )

[mum_sugoku]

Dr.Daniel:

njcstudent:

Hi, I am an IP student in Sec 1. Can anyone please recommend any good reference books and assessment books?

I had an njc student in Sec1 last year and they were not following a standard syllabus and were doing very hard material early on. One of their modules was called \"basic atmosphere\".
Best advice I can give is to build up your basic physics and chemistry understanding. The textbook Interactive Science for Inquiring Minds is not bad. Try to understand measurement, density, speed and then go through elements and compounds and atoms and molecules.
You will probably need a higher level book for O-Level, such as Physics Insights and Chemistry Insights as a reference for harder topics.
On assessments, The red book Lower Secondary Science by GS Cheng is good.

My DS is also currently a year 1 student in an IP school. Agree that their science is rather tough. Take my DS for instance, while he used to score good grades in science in his pri school, he has managed to get only 'C's and 'D's for his science assessment so far ..