How to do physics problems 5, how to do physics problems

The most important thing to do physics problems is to build a physical model (spatial imagination) and be proficient in knowledge points.

Also, there are more physics question types that are usually seen. Master the steps for solving a type of question. For example, to do a mechanics problem, the first step is to conduct a force analysis; In the second step, the motion analysis is carried out; The third step is to list the equations.

Find out all the known conditions and bring formulas. It is essential to understand each of the movements mentioned in the question.

First, the examination of the topic. It is not a simple reading question, but it should be purposeful, such as whether it is smooth, whether the charge is fixed, unit conversion problems, and so on.

2. Construct a sports scenario.

Physics is closely related to real life, and constructing a dynamic scenario can help us sort out the whole movement process and reduce some mistakes that we take for granted.

Here's a very simple example:

Imagine the wheels rolling in your head, build a physical scenario, and you can't go wrong.

3. Judge whether you can write.

Determine Test Centers - Invoke Knowledge Network - Connect the knowledge network to the test center.

Fourth, don't just think about it.

Try to list all the equations that can be listed, draw all the force analysis diagrams, try the whole method isolation method, etc., and sometimes you will find that the idea is suddenly opened.

5. Test-taking. There are also a lot of tips when it comes time to take the exam:

1 Limit method.

2 Hypothetical method.

3 Algebray.

4 Exclusion.

Starting Point - Learn to define concepts.

Maps - Build a knowledge framework.

How to break the level - Physical scenario construction method.

How to beat the level - physical thinking mode and model.

How to beat the level - notes and error book.

Finish - Physics problem with five steps.

If you can analyze all the physical processes in a physics problem, then you can basically get it right.

First of all, ask yourself, do you analyze each physical process clearly when doing physics problems? Do you even see what physical processes there are in this question? So:

1. Sports scenario analysis method.

As we mentioned above, constructing a physical scenario, starting from the natural phenomenon of reverence, and then applying the concept.

The first step is to identify how many movements there are, and the second step is to analyze each movement in detail.

So how to analyze, what to analyze, and how to analyze clearly?

The simplest and most straightforward way is to use everything that can be used by the knowledge network.

For example, a mechanics problem can be operated like this:

Linear or curvilinear motion.

Whether it is a circular motion or a flat throwing motion.

Can you use a uniform speed movement formula?

There is no velocity correlation included.

Find a good force analysis of the research object.

Can it be analyzed with energy relationships?

Can it be analyzed with momentum relations?

Work and power analysis is required.

Is there a known model that can be applied?

As long as every physical process has been analyzed in this way, and the analysis is completely clear and thorough, then no matter what the question asks or requires, it can be found 100% vertically. Of course, after doing a lot of questions, you can quickly see what kind of knowledge you need to use.

This problem is currently partially used to calculate friction, which is based on the formula 1 2at 2=sa=(f-f) m

Calculate a=10m s2 and f=10n.

In order for the object to reach the edge of the wall, the velocity is greater than or equal to 0, which can be divided into two processes, one is when f acts (time t1), and the other is when f does not act (t2), and the sum of the work done with f and f is 0

i.e. f·l=f·s

l=20/3，a=(f-f)/

m，1/2at1^2=l

a'=f/m=5m/s^2,1/2a't2 2=s-l calculates t1=(4 3) 1 2

t2=（16/3）^1/2

In order to minimize the thrust action time but still reach the wall, the velocity of the object to the wall should be exactly zero, and the object should be accelerated by the thrust for the first time: a1=2s t2=10

m s2 is obtained from f-f=ma1, f=f-ma1=10n, and the velocity of the object when the external force f is removed is v

then v=a1t1=a2t2, where a2=f m=5m s2 so there is t22t1

s=s1+s2=

a1t12+

a2t22 i.e. 20=

10t12+

5(2t2)2

Solution. t1=Answer:

Got: a2s t

m s is composed of f-f=ma

Get f=f-ma

Let n let the velocity of the object when the external force f is removed be v then v a

tat where a

f/m=5m/s

Hence t2ts=ssa

t at is 20 10t

5 (2t solved.) t

1 15s I don't know if it can be displayed.

Both of these are Ruqing problems with the relative movement of laughing oak Minguan, and the solution is as follows:

1) 9km h=two opposite directions: the speed of riding relative to the speed of the truck v=, the elapsed time t=, then there is s=vt=5m;

2) The two directions are the same: the speed of the bicycle is relative to the speed of the truck v=8 4=2m s, so the speed of the bicycle v'=2+10=12m/s

Energy is conserved, and the formula is not easy to write (pegging u2 r) t=c m (t2-t1).

That is, the change in the time of r times u = c times m times temperature.

With shouting, the base entry value is Zheng Jinke.

According to the density of copper is grams, per cubic centimeter, then the volume of copper contained in it is calculated to be equal to the mass of the ball, divided by the density of copper. After calculation, the volume of copper that can be obtained is two cubic centimeters.

The total volume of Yu Tong's total sphere is five cubic centimeters, so the volume of the hollow part is that five cubic centimeters - 2 cubic centimeters are equal to three cubic centimeters.

The hollow part is this, three cubic centimeters, if you fill in water, according to the density of the water is cubic kilograms, per cubic meter, you can get, the mass of water is three grams. So the total mass, kilogram + 3 grams, is equal to grams.

Solution: From =m v, v is obtained = m =......=2cm and v-ball = 5cm v-solid, so it is hollow.

v empty = v ball - v real = ......=3cm³

m water = water v air = ......=3g

m = m ball + m water = ......=

Step by step to score points.

Cm.

5-2 3 square centimeters.

The density of water is 1

So the hollow part is 3 square centimeters.

When the water is full, the mass is equal to.

Solution: The volume of 178g of copper v=m p copper = 178g So the ball is hollow, and the volume of the hollow part v = 30cm3-20cm3 = 10cm3 After being filled with aluminum, the mass of aluminum m aluminium = m aluminum p empty =

Total mass m total = m copper + m aluminum = 178g + 27g = 205g

The mass of the copper ball divided by the volume is equal to 2, so the hollow is 5-2=3, and only the hollow part can be filled with water, so it is only 3g, and finally added to the original mass.

Solution: The actual volume of copper: v copper = m copper =

Therefore, the volume of the hollow part v empty = 5cm3-2cm3 = 3cm3 after being filled with water, the mass of water m water = m water empty = 1g cm3x3cm3 = 3g total mass of copper balls filled with water: m total = m copper + m water =

A: 3 cubic centimeters; Gram. Analysis:

Solution: The volume of a solid copper ball: v solid = m copper =

Then the hollow volume is v empty = 5cm3-2cm3 = 3cm3 after being filled with water, the mass of water m water = water v empty = 1g cm3x3cm3 = 3g total mass after filling with water: m total = m copper + m water =.

Known: m = gram, p = gram cubic centimeter, v real = 5 cubic centimeter compared v real and v should be.

Solution: should have = m p = cubic centimeter, v empty = v real - v should have = 5-2 = 3 cubic centimeters, fill the middle with water, inject the quality of water.

m1 = v empty p1 = 3x1 = 3 grams.

The total mass of the copper ball after water injection m2 = m + m1 = grams.

Solution: The volume of 178g of copper v=m p copper = 178g So the ball is hollow, and the volume of the hollow part v = 30cm3-20cm3 = 10cm3 After being filled with aluminum, the mass of aluminum m aluminium = m aluminum p empty =

Total mass m total = m copper + m aluminum = 178g + 27g = 205g

That is, from the point of application as a perpendicular line to the extension line of the force.