Throw a small ball on a car with speed V that does not take into account drag

Since the ball is thrown when passing point A, it is obvious that to make it fall at point A, then the ball must do free fall motion to reach point A after being thrown, and to make the ball do free fall motion, then the horizontal velocity of the ball is 0, and before throwing, the ball has a velocity V that is the same as that of the train, assuming that the velocity x throws the ball from the car when passing through point A, according to the superposition of the velocity, the velocity after throwing is v+x, and the ball falls on point A, it can be seen from the front that v+x 0, so that we get: x -v, That is, the ball needs to be thrown at a velocity v, and the negative sign indicates that the throwing direction is opposite to the direction in which the train is running.

Hope this answer satisfies you.

This issue should be discussed in two cases.

The first one has been made clear, that is, the mass of the ball is much smaller than the mass of the car, and the throwing of the ball will not have a measurable effect on the speed of the car. In this way, it is relatively simple, as long as the velocity component of the ball in the horizontal direction is -v (the car is v relative to the ground).

Another situation is that the mass of the ball and the car is comparable, which date001 also said, assuming that the mass of the car is m, the ball is m, and the speed of the car after the ball is thrown is v1, then.

Momentum is conserved m+m)v=mv1+m0

Get v1=(m+m)v m

In this way, the ball only needs to be thrown at a speed relative to the speed of the cart v1.

Because the ball is thrown is a process, in this process, the speed of the car gradually increases, if the ball is thrown at the relative speed of -v, then the plan cannot catch up with the change.

Take the rook as m, the ball as m, and the velocity of the rook after the ball is thrown is v1, then.

Momentum is conserved. m+m）v=mv1+m0

Get v1=(m+m)v m

In this way, the ball only needs to be thrown at a speed relative to the speed of the cart v1.

However, before the ball is thrown, the speed of the rook is v, and the speed of the ball is also v, so it seems that it is only necessary to throw it in the opposite direction at the speed of v. However, according to the conservation of momentum, the speed of the car changes at the moment when the ball is thrown, so if the car is used as a frame of reference, the ...... should be thrown at the speed of v1

I've been thinking about this question for a long time, but I still don't understand it, and I hope that which master can come out to give you some advice, and how to understand it.

Since the ball is stationary with respect to the car, the velocity of the ball relative to point A is v

At point A, the throwing cut requires that the ball fall to point A, that is, the ball is stationary relative to A when thrown, because the velocity of A relative to the car is -v, so the speed of the ball in the direction of the car relative to the speed of the car should be -v, and the upstairs have forgotten that the velocity is loss, not scalar.

The relative velocity of the car is v The ball is the same speed as the car, that is, the relative velocity of the ball is v The speed of the ball is u and the mass is m The velocity of the ball is i and the mass is n (m+n)*v=m*u+n*i, and because it falls on a, then i =0 u=(m+n)*v m, so the relative speed of the ball is -m+n)*v m

Momentum Conservation m is the mass of the car and man m is the mass of the ball.

After throwing, the speed of the rook is v and the velocity of the ball is v; v is known.

m+m)v=mv car-mv ball ground should still be point a, so v ball ground = 0;

So (m+m)v=mv car land.

The amount of change in the momentum of the car to the ground p car to the ground = p to the ground of the sphere; Let the ball be thrown at the speed of v';Then there is p car land = mv car land - mv; p ball = 0-mv';

So m(v-car-v)=-mv'

v'=-m(v-v) m;

Of course, this is directly considered to be moving in the horizontal direction.

With the direction of motion of the car as the positive direction, the momentum of the ball is mv, and the ball needs to be stationary in the horizontal direction to let the ball fall at this point, and its momentum is 0, then the momentum change is 0-mv=-mv. Now we want to give this ball a momentum i, i=-mv. Just shoot it with V.

Traction speed + relative velocity = absolute velocity.

The trolley traction is v and later falls to a and the absolute speed is 0

So the relative velocity is -v

The velocity of the ball relative to the ground is 0, so throw the ball with a velocity of -v relative to the car.

1) A "car with speed V" must be a car moving on a horizontal plane, right? What if the car also has a sub-speed in the vertical direction (i.e. the car goes uphill and downhill)?

2) Does the ball have to be stationary at the moment of throwing? Can't the ball fall back to the throwing point if it is thrown vertically?

To sum up, in fact, as long as the velocity of the ball relative to the ground at the moment of throwing is vertically upwards (or 0).

However, the question does not state the direction of the car's V, so from the point of view of (1), this question cannot be done, in other words, it is an open question.

Backwards v is simple.

If the ball is not thrown, then the velocity of the ball is v, and now the ball is going to fall vertically, so there can be no velocity in the horizontal direction, so v backwards can cancel out the velocity in the horizontal direction.

This kind of topic has a formula and a detailed process, but it will be half the effort, and it will be confusing later.

Formulas are used as a last resort, starting from the concept and being clearly organized, formulas are only derived from concepts, and you still won't do this kind of problem after simply memorizing formulas.

Let the mass of the trolley be m and the mass of the ball be m, then v=mv m is obtained, according to the law of conservation of momentum

Halo, you people are really misleading, do you have to use momentum conservation for this question? Faint, the problem is very simple, just use a relative velocity of the car v, or the opposite speed of the ground stationary can make it stop, this is too simple, as for the angle, then you are thinking about the problem too difficult, if it is so difficult, we all have to prove that one plus one equals two, that is the business of scientists, it is useless for you to worry.

The ball is thrown to fall to point A, and its speed is related to the moment of throwing, which is related to the speed of travel.

Throws at velocity v in the opposite direction to the direction of movement of the car.

If you don't understand it, you can use the car as a reference frame, then point A moves to the back of the car with V, so you should also throw the ball at the same speed as point A.

Throw at ground-v velocity (the reference direction is the same as the reference direction of the vehicle's motion).

It is considered to be the ideal state, and the relative ground velocity v is thrown in the opposite direction.

No matter what speed you have, you can still move forward at a certain height and at a certain angle.

Size: V direction: Opposite to the direction in which the car is running.

As for proofs, they can be superimposed with vectors.

Is it that troublesome? As long as the speed exceeds v, it's fine, and the rest is a matter of direction.

Vertical to throw has nothing to do with v inertia can be done!!

It's speed v ah v=mv m it's that simple.

There are also people who ask such mentally retarded questions.

The ball falls from l=1 2gt 2 Time t= root chain is delayed (2l g).

In the t timecode, the trolley in the field is moved, s=vt+1, 2at, 2 = v, and the root number is (2l g)+al g

That is, this is the distance of the ball from its original position.

With respect to the trolley, the ball acceleration is the horizontal left component a and the vertical downward component g

The ball will move in a straight line with respect to the small slow car in the direction of the vector of a, g and finally fall to the position on the ground, and the starting point, the point on the ground directly below the starting point, forming a right-angled obstruction of the Lie triangle, and a g = x file l

x = al/g

1 liter = 1 cubic meter = 100km fuel consumption generated by the amount of hot luggage: q = 10. 100km in time:

t=100/90=10/9（h）=4000（s）。The work done by the engine is: w = pt = 20 103 4000 = 8 10 7 (j).

The efficiency of the engine is: =w q 100%=.

If we take the ground as the reference frame, then the big ball, the small ball, and the car are all moving in the same direction at the speed v of the car, so the ball and the car remain relatively stationary at this time.

And the velocity of the two balls is brought by the car. The big and small balls will maintain the original speed of the car and continue to move at the moment the car stops.

And the speed is the same. You can refer to the kinetic energy theorem to explain it.

They have the same velocity, different mass, just different kinetic energy raised. But they won't collide.

According to the original meaning of this problem, the friction coefficient between the plane and the ball on the car can be ignored, and the actual rolling friction is relatively small, and the friction can be approximated as zero. There is a mistake in your thinking, "but I think it takes longer for the ball with greater inertia to change to its original velocity than the small ball", this is an acceleration motion done under the action of an external force, and it has been explained in the question that the large and small balls are not affected by external forces, how can they be accelerated? The diameter of the large and small balls is not the same, at the same velocity, their rotational speed is not the same, the rotational speed of the large balls is smaller than that of the small balls, but the circumference of their rotation per unit time is the same, so the velocity will also be the same.

What is the original speed? The speed has always been the same, and they are all moving at the same speed, how can they collide?

Answer: At this point, the object will be moving in a circular motion at the next moment, therefore, the rope pull force increases to provide sufficient centripetal force, F direction = F pull - G ball = MV2 (L+R).

F-pull = MV2 (L+R) + G globules = MV2 (L+R) + mg

It's A, the same size and direction as a car.

At this time, the rope hanging the ball is oblique, and the tension of the rope is also oblique, and the tension of the rope is decomposed into two directions: vertical and horizontal. Where the vertical component and gravity are balanced, so there is no acceleration in the vertical direction, and the horizontal direction is used to provide the acceleration of the little hand Bichai ball with an a.

Therefore, if the acceleration a is larger, the slope of the rope closure will be greater, because the horizontal component of the rope tension is larger. So the pulling force of the rope will also be greater. If the acceleration is high or the ball is heavy, the rope may break.

It's hard to say, I didn't say it in the question, whether the ball is relatively stationary with the car at the beginning, then I will first think that it is relatively stationary.

First of all, analysis: Regardless of friction, then the trolley will do uniform acceleration linear motion.

There are two scenarios:

The ball is at the left end of the rook:

f=mal 3= a t1 (the time it takes for the ball to move to the left end of the cart is a bit unsightly, and the table below is not easy to play).

h= g t2 (the time when the ball starts to fall and the car moves).

v1=v+a (t1+t2)

Solve the equation: v1=?

The ball is at the right end of the car:

f=ma2l 3= a t1 (the time for the ball to move to the left end of the cart is a bit unsightly, and the following table is not easy to play).

h= g t2 (the time when the ball starts to fall and the car moves).

v1=v+a (t1+t2)

Solve the equation: v1=?

In addition, I said that if the car is already moving at a speed v, then suddenly put the ball at a stationary speed to the ground, it is also l 3, and the solution is different, I said that the situation of the ball at the left end is similar at the right end.

At this time, the ball is relative to the car, with v as the initial velocity, a=f m acceleration, and the acceleration to the left.

f=mal/3= v t1+ ½a t1²

h=½ g t2²

v1=v+a (t1+t2)

Solve the equation: v1=?

It's a car again, it's a ball again, it's not clear!