Sophomore physics problems, solutions and processes

Step 1: Calculate the velocity of the ball before it collides with the iron using the law of conservation of energy.

mgh = 1 2mv2 v = root number under (2gh) = 4m s Step 2: Calculate the velocity of the iron after the ball collides with the iron using the law of momentum.

m1v1=-m1v2+m2v3 v1 is the velocity of the ball before the collision, v2 is the velocity of the autumn after the collision, and v3 is the velocity of the iron block.

Calculated v3=

Step 3: Calculate the distance of the iron block to the ground Using the flat throwing motion to solve.

1.Calculate the landing time first t=under the root number (2h g)=2.Calculate the horizontal displacement s=v3*t=

Give points Give points.

Release conservation of mechanical energy:

m gh = m v v v = (2gh) = 4m s positive impact momentum conservation:

m v = m v + m v v = m v -v ) m = flat throwing motion after touch:

y=h=½gt² ∴t=√﹙2h/g﹚=

x=v₂t=

When the ball is released, the conservation of mechanical energy is satisfied: ep=ek;

i.e.: 1 2m1v2=m1gl;

The solution v is: 4m s;

When it is in contact with an iron block, it satisfies the conservation of momentum:

That is: m1v=m2v2-m1v1

The solution is v2 = the time it takes for the iron to fall to the ground.

1/2gt2=

The solution is t=so the horizontal distance is:

The velocity of the ball falling to the lowest point: mgl=, v=sqrt(2gl)=4m s, the ball bounces back, and the velocity of the iron after being hit by the momentum theorem: m1*4=m2v-m1*2

Can you attach an image of the idea of your question that is not very clear, is it a ball on the side of the table? The ball tied to the rope just touched the ball on the side of the table after it was released? If yes, the problem involves knowledge of the kinetic energy theorem, the law of conservation of momentum and flat throwing motion.

Use the kinetic energy theorem to analyze the small ball, mgh=1 2mv2, v=4m s, and then use the law of conservation of momentum to analyze the iron block and the ball system: find the velocity of the iron block, and finally use the flat throw, vertical displacement to calculate the time, t=, so the iron block landing point is away from the table.

OAB is an equilateral triangle, so AB is 60cm apart and 120 angles to the vertical line. b is otherwise stationary, and the direction of the resultant force of the Coulomb force and gravity should be opposite to the direction of the pulling force, so the magnitude of the Coulomb force is equal to the gravity of the b ball.

f=mg From the direction of the pulling force, the magnitude of the F-pull is equal to the Coulomb force and the tensile direction component of the gravity, and it is easy to obtain F-pull=2mgCos60

For the force analysis of the B ball, the B ball is affected by three forces, the gravitational pull force and the Coulomb force, and the vector triangle of these three forces is similar to the triangle OAB, so mg = Coulomb force = the pull force of the ob rope, and the amount of charge and the tensile force of the ob line are obtained.

Momentum is conserved. m1v1+m2v2=(m1+m2)v

600*15-400*10=1000*v

v=5cm s is the direction of the 600g slider.

1) Calculate the time the volleyball is in the air by 1 2gt 2=h to know t=

So the horizontal flyout s=vt= i.e. a horizontal constant velocity and a vertical direction uniform acceleration.

2) The title should mean that people are running at a constant speed, and what is the approximate speed. As you can see, the volleyball stays in the air for a t= time, so a person two meters away should run at a speed of v=s t= to catch it.

AB rod slides horizontally to the right at a constant speed, e=blv

i=e/(r+r)

u=e-ir

f = the magnitude of the induced current in the bilab rod i= , the direction from b to a (ampere's rule), and the velocity of the ab rod sliding horizontally to the right at a uniform speed v=

When the AB rod moves at a uniform speed, the horizontal external force f= and the direction is to the left (left-hand rule).

You should go to the junior high school classmates for advice, definitely.

Picture: No? I did a lot of questions, and I think it should be like this.

The current is from b to a, i=u r=

f=bil=

f direction to the left.

f=b squared l square v r+r

v=10m/s

(1) Let the total mass of the aircraft be m, according to the law of the ox 2 to obtain a =, so the fall height within 30s h1 = 1 2at 2 = 4320m;

2) The speed of the aircraft after 30s is known by (1) v=at=;

The aircraft must not be lower than 200m, which means that the speed must be reduced to 0 before the height of 200m; When the critical value is taken to a height of 200m, the speed is just 0;

Let the acceleration of the subsequent uniform deceleration be a'then there are 2a'h=v^2;where h=6000-h1-200=1480m; Solution: a'=28m/s2;

Let Hawking have a mass of m again, because it is the same acceleration as an airplane, and he is supported by his own gravity g and the chair's support force n (and his pressure on the seat are action and reaction forces), according to Ox's law: n-mg=ma';Get n=m(g+a')=;So the conclusion is that the pressure is at least twice his gravity.

（1）f=

g-f=ma Fall acceleration a=

h=at^2/2=

2) It is easy to know that the velocity is zero at 200m, 2as=-v0 2 acceleration a=(

n+f-g=ma n=

n g = the pressure on the seat and the elastic force of the seat on the person are the relationship between the action force and the reaction force n' = n, so it is a multiple of the gravitational force.

b When Coulomb force is not present, Fa pull=2g

FB pull=g, when Coulomb force is present, FA pull'=2g

fb pull' = g + (f coulomb force).

Fa Pull = Fa Pull ' = 2g The force is equal.

In summary, choose B

If you choose B, if you look at the two balls as a whole, Fa is constant, and if you look at the two balls as a whole, there are forces that repel each other, so Fb