Question on Conservation of Momentum

[Guest Exodus]

can any1 explain to me conservation of momentum, how it works through a vector diagram. For example in the problem below.

 

Ball A of mass 5.0 kg moves at a speed of 4m/s. It collides with a second stationary ball B, also of mass 5.0kg. After the collision, ball A moves off in the direction 45 degrees to the left of its orginal direction. Ball B moves off in the direction 45 degrees to the right of ball A's original direction.

 

a. Draw a vector diagram to determine the momentum of ball A and of ball B after the collision.

b. What is the speed of each ball after the collision?

[swansont]

can any1 explain to me conservation of momentum' date=' how it works through a vector diagram. For example in the problem below.

 

Ball A of mass 5.0 kg moves at a speed of 4m/s. It collides with a second stationary ball B, also of mass 5.0kg. After the collision, ball A moves off in the direction 45 degrees to the left of its orginal direction. Ball B moves off in the direction 45 degrees to the right of ball A's original direction.

 

[b']a.[/b] Draw a vector diagram to determine the momentum of ball A and of ball B after the collision.

b. What is the speed of each ball after the collision?

 

If there is no net external force on the system, then momentum is conserved. Since it's a vector, it can be broken down into components, and those components must be conserved.

 

Ball A has 20 kg-m/s of momentum in the x direction, and zero in the y direction, before the collision. The system must always have this much momentum. Since you know the momentum of ball B after the collision, whatever is missing from he original must still be possessed by ball A.

[J.C.MacSwell]

can any1 explain to me conservation of momentum' date=' how it works through a vector diagram. For example in the problem below.

 

Ball A of mass 5.0 kg moves at a speed of 4m/s. It collides with a second stationary ball B, also of mass 5.0kg. After the collision, ball A moves off in the direction 45 degrees to the left of its orginal direction. Ball B moves off in the direction 45 degrees to the right of ball A's original direction.

 

[b']a.[/b] Draw a vector diagram to determine the momentum of ball A and of ball B after the collision.

b. What is the speed of each ball after the collision?

 

I think you have described a perfectly elastic collision.

[Johnny5]

can any1 explain to me conservation of momentum' date=' how it works through a vector diagram. For example in the problem below.

 

Ball A of mass 5.0 kg moves at a speed of 4m/s. It collides with a second stationary ball B, also of mass 5.0kg. After the collision, ball A moves off in the direction 45 degrees to the left of its orginal direction. Ball B moves off in the direction 45 degrees to the right of ball A's original direction.

 

[b']a.[/b] Draw a vector diagram to determine the momentum of ball A and of ball B after the collision.

b. What is the speed of each ball after the collision?

 

Law of conservation of momentum

 

The total momentum of a closed system is constant in time.

 

Mathematically:

 

[math] \vec P_i = \vec P_f [/math]

 

Initial system momentum equals final system momentum.

 

The first thing you have to do, is choose an inertial reference frame, in which to observe the collision.

 

Ball A, and ball B have an inertial mass of 5 kilograms each:

 

[math] M_A = 5 kg [/math]

[math] M_B = 5 kg [/math]

 

Now, you have ball A colliding with a stationary ball B.

 

This means that in the particular inertial reference frame you choose, the initial speed of ball b is zero, so that ball B is at rest in the frame, at least initially.

 

On the other hand, ball A has to initially have a speed of 4 meters per second.

 

So the frame I am going to carry out the analysis in, is going to have ball B initially located at the origin of the frame, and ball A is going to be moving upwards, along the y axis, with speed 4 meters per second.

 

So before collision, the total momentum of ball B in this frame is zero. This fact follows since the magnitude of its momentum is equal to its mass, times its speed, and initally its speed is 0, hence the magnitude of its momentum is initially zero, hence its vector momentum is initially zero, in this frame.

 

On the other hand, ball A has a nonzero speed, and a nonzero inertial mass, at least initially, so its initial momentum is nonzero, and will be represented by:

 

[math] \vec P_A = M_A \vec v_A = M_A v_A \hat j = 5*4 \hat j = 20 \hat j[/math]

 

And likewise the initial momentum of ball B in this inertial frame is:

 

[math] \vec P_B = M_B \vec v_B = 5*\vec 0 = \vec 0 [/math]

 

The intitial total system momentum is equal to the sum of the momentum of its parts:

 

[math] \vec P_i = \vec P_A + \vec P_B = 20 \hat j + \vec 0 = 20 \hat j [/math]

 

So as ball A creeps closer to ball B, the total system momentum does not change. But eventually there is a collision.

 

It cannot have been a head on collision, because after the collision ball A moves diagonally in the second quadrant of the XY plane, and ball B moves diagonally in the first quadrant of the XY plane, each ball at a 45 degree angle with the Y axis(which is the original line of motion of ball A).

 

The remainder of the problem is simple.

 

Simply express the momentum of the parts, (draw the vector diagram for this), and lastly set the sum of the momentum of the parts P_f.(after collision) equal to the original momentum P_i.

 

As a vector the momentum of ball A after collision is given by:

 

[math] \vec P^\prime_A = M_A \vec v^\prime_A [/math]

 

Now, I am assuming that the mass of ball A did not change during the collision, and that the mass of ball B didn't change during the collision.

 

As for the final velocity vector of ball A, you have it as 45 degrees from the Y axis. So break its final velocity vector into horizontal and vertical components.

 

[math] \vec v^\prime_A = -v_x_A \hat i + v_y_A \hat j [/math]

 

and do the same thing for the final velocity vector of ball B, you have it as 45 degrees from the Y axis, but in the first quadrant of the XY plane, so:

 

[math] \vec v^\prime_B = v_x_B \hat i + v_y_B \hat j [/math]

 

By the way, speed is a strictly positive quantity.

 

So now I am going to draw a picture of this, although unfortunately I cannot put it here.

 

The initial momentum in the i^ direction was zero, and this cannot have changed therefore:

 

[math] \vec 0 = -M_Av_x_A \hat i + M_Av_x_B \hat i[/math]

 

Dividing both sides by M_Ai^ we obtain:

 

[math] 0 = -v_x_A + v_x_B [/math]

 

From which it follows that:

 

[math] v_x_A = v_x_B [/math]

 

Now, the initial total system momentum (before collision) was 20 kilogram meters/second in the j^ direction, and this cannot have changed, therefore:

 

[math] 20 \hat j = M_Av_y_A \hat j + M_Bv_y_B \hat j [/math]

 

Dividing both sides by j^ we obtain:

 

[math] 20 = M_Av_y_A + M_Av_y_B [/math]

 

And the mass of each ball just happens to be the same... namely 5 kilograms, therefore:

 

[math] 20 = 5v_y_A + 5v_y_B [/math]

 

Dividing both sides by 5 kilograms, we obtain:

 

[math] 4 = v_y_A + v_y_B [/math]

 

Now, although you cannot see the vector triangles which I drew, I can explain them to you.

 

There is one vector triangle in the first quadrant, and its a right triangle, one leg is VxB, and the other leg is VyB, and both interior angles are 45 degrees.

 

Therefore we have:

 

[math] tan 45 = \frac{v_y_B}{v_x_B}=1 [/math]

 

And in the second quadrant there is a similiar right triangle, satisfying:

 

[math] tan 45 = \frac{v_y_A}{v_x_A}=1 [/math]

 

From which we may infer that:

 

[math] v_y_A = v_x_A [/math]

 

[math] v_y_B = v_x_B [/math]

 

And we already knew that:

 

[math] v_x_A = v_x_B [/math]

 

[math] 4 = v_y_A + v_y_B [/math]

 

So we now have enough information, to figure out the actual speeds of each ball.

 

The speed of ball A is given by:

 

[math] V_A =\sqrt{(v_x_A)^2+(v_y_A)^2} [/math]

 

And the speed of ball B is given by:

 

[math] V_B =\sqrt{(v_x_B)^2+(v_y_B)^2} [/math]

 

Now, [math] 4 = v_y_A + v_y_B [/math], and [math] v_y_A = v_x_A [/math] and [math] v_x_A = v_x_B [/math], therefore

 

Now, [math] 4 = v_y_A + v_y_B [/math], and [math] v_y_A = v_x_B [/math], therefore

 

Now, [math] 4 = v_x_B + v_y_B [/math]

 

Therefore:

 

[math] 4 - v_y_B = v_x_B [/math]

 

Therefore:

 

[math] V_B =\sqrt{(4 - v_y_B)^2+(v_y_B)^2} [/math]

 

And similarly we find that:

 

Since

 

[math] v_x_A = v_x_B [/math] and [math] v_x_B =v_y_B [/math] we have:

 

 

[math] v_x_A = v_y_B [/math]

 

And since:

 

[math] 4 = v_y_A + v_y_B [/math]

 

we have:

 

[math] 4 = v_y_A + v_x_A [/math]

 

From which it follows that:

 

[math] 4 - v_y_A = v_x_A [/math]

 

So that:

 

[math]

V_A =\sqrt{(4 - v_y_A)^2+(v_y_A)^2}

[/math]

 

Now, conservation of momentum isn't the only conservation law of physics used in analyzing collisions, there is also conservation of energy.

 

And the basic idea is the same, that the total energy of the system before collision, is equal to the total energy of the system after collision. Using conservation of energy, we will generate some more equations, in the same unknowns, allowing us to solve for the speed.

 

Conservation of system energy

 

The total energy of an object is a frame dependent quantity.

 

E = T + U

 

E is the total energy of the object, T is the kinetic energy of the object, and U is the potentially kinetic energy of the object.

 

Let us imagine that the collision is taking place in space, where there is no friction, and furthermore, that neither of the balls involved in the collision are spinning in the inertial frame before collision. Thus, initially, the potentially kinetic energy U of each ball is zero.

 

Thus, any energy which they have is purely translational kinetic energy T. I am going to use the classical formula for kinetic energy, from Newtonian mechanics, if one wishes to use relativistic mass, the calculations become needlessly complex. Here is the formula from Newtonian mechanics, for kinetic energy T:

 

[math] T = \frac{mv^2}{2} [/math]

 

m is the inertial mass of the object, and v is its speed in the inertial frame.

 

Now, ball b is initially at rest in the frame, so that means that its speed v, is equal to zero initially. After collision its speed is greater than zero in the frame.

 

As for ball A, initially, its speed is 4 meters per second in the frame, and after collision, its speed is something greater than zero, but less than 4 m/s.

 

So the initial total system energy is just the kinetic energy of ball A, there are no other contributions to the total system energy in the frame before collision. Thus, the initial total energy E of the system is:

 

[math] E_i = \frac{16M_A}{2} [/math]

 

(The number 16 is the initial speed squared 4*4=16)

 

And by the law of conservation of energy, this must equal the final system energy after collision E_f.

 

Now, the whole final system energy is equal to the sum of its parts. Let us presume that after collision neither ball is rotating (this is unrealistic, but it will simply further calculations).

 

The only way it would be literally true that after collision neither ball is rotating, would be for the collision to be head on. But if that was the case, then the balls would not depart at 45 degree angles.

 

At any rate, we already introduced symbols for the final speeds of both balls in the inertial frame, after collision.

 

The final speed of ball A in the frame was denoted by:

 

[math] V_A [/math]

 

The final speed of ball B in the frame was denoted by:

 

[math] V_B [/math]

 

Hence the kinetic energy of ball A in the frame after collision is:

 

[math] \frac{M_A(V_A)^2}{2} [/math]

 

And the final kinetic energy of ball B in the frame after collision is:

 

[math] \frac{M_B(V_B)^2}{2} [/math]

 

Since we presume that none of the initial energy was converted into rotational energy, the final total system energy is simply:

 

[math] E_f = \frac{M_A(V_A)^2}{2}+\frac{M_B(V_B)^2}{2} [/math]

 

And by the law of conservation of energy, the final system energy must equal the initial system energy, hence:

 

 

[math] E_i = \frac{16M_A}{2} = E_f = \frac{M_A(V_A)^2}{2}+\frac{M_B(V_B)^2}{2} [/math]

 

From which it follows that:

 

[math] \frac{16*5}{2} = \frac{5(V_A)^2}{2}+\frac{5(V_B)^2}{2} [/math]

 

Multiplying both sides of the previous equation by 2/5, leads to:

 

[math] 16 = (V_A)^2 +(V_B)^2 [/math]

 

And the previous equations we had were:

 

[math]

V_B =\sqrt{(4 - v_y_B)^2+(v_y_B)^2}

[/math]

 

[math]

 

V_A =\sqrt{(4 - v_y_A)^2+(v_y_A)^2}

 

[/math]

 

Squaring both sides of the previous two equations leads to:

 

[math]

(V_B)^2 = (4 - v_y_B)^2+(v_y_B)^2

[/math]

 

[math]

 

(V_A)^2 = (4 - v_y_A)^2+(v_y_A)^2

 

[/math]

 

From which we can now infer that:

 

[math] (V_A)^2 +(V_B)^2 = (4 - v_y_A)^2+(v_y_A)^2 + (4 - v_y_B)^2+(v_y_B)^2 = 16 [/math]

 

Recall that:

 

[math]

4 = v_y_A + v_y_B

[/math]

 

Hence:

 

[math]

v_y_A = 4 - v_y_B

[/math]

 

and

 

[math]

v_y_B = 4 - v_y_A

[/math]

 

Therefore:

 

[math] (4 - v_y_A)^2+(v_y_A)^2 + (v_y_A)^2+(4 - v_y_A)^2 = 16 [/math]

 

From which it follows that:

 

 

[math] 2[(4 - v_y_A)^2+(v_y_A)^2] = 16 [/math]

 

From which it follows that:

 

[math] (4 - v_y_A)^2+(v_y_A)^2 = 8 [/math]

 

From which it follows that:

 

[math] 16+(v_y_A)^2-8v_y_A +(v_y_A)^2 = 8 [/math]

 

From which it follows that:

 

[math] 2(v_y_A)^2- 8v_y_A + 8 = 0 [/math]

 

 

Let X denote the unknown quantity we are looking for, hence

 

[math] X \equiv v_y_A [/math]

 

So we can rewrite the equation a little nicer as:

 

[math] 2X^2- 8X + 8 = 0 [/math]

 

And we can divide both sides by 2 to obtain

 

[math] X^2- 4X + 4 = 0 [/math]

 

Which can be easily factored.

 

Any rational roots of the previous equation will be of the form p/q where p is one of the factors of the constant term(which is 4), and q is one of the factors of the leading coefficient(which is now 1, because we divided by 2).

 

The positive factors of 4 are found from:

 

1*4

2*2

 

and the negative factors are found from:

 

-1*-4

-2*-2

 

We now have to find which pair sums to the coefficient of the linear term (-4x).

 

The answer is (-2)+(-2) = -4

 

Thus, we have the following true statement:

 

[math] X^2- 4X + 4 = (X-2)(X-2)=0 [/math]

 

So we have one repeated root, which counts as two roots in a theorem supposedly proven by Carl Friedrich Gauss 4 different times, the first of which was his dissertation for his doctorate (1799).

 

Replacing X by [math] v_y_A [/math] we have:

 

[math] (v_y_A-2)^2=0 [/math]

 

From which it follows that:

 

[math] v_y_A-2 =0 [/math]

 

therefore:

 

[math] v_y_A=2 [/math]

 

So since:

 

The following code was used to generate the image you clicked on:

 

[math]

 

 

(V_A)^2 = (4 - v_y_A)^2+(v_y_A)^2

 

 

[/math]

 

It follows that:

 

[math] (V_A)^2 = (4 - 2)^2+2^2 = 4+4 = 8

[/math]

 

From which it follows that:

 

[math] V_A = \sqrt{8} = \sqrt{(2)^2 \cdot 2} =\sqrt{(2)^2} \cdot \sqrt{2}=2\sqrt{2}

[/math]

 

Radical two is approximately equal to 1.4, and two times 1.4 is equal to 2.8.

 

And similarly, you will find that the speed of ball 2 in the frame, after collision is 2.8 meters per second. You should check that for yourself.

 

Regards