- #1

twofish

- 25

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Hi all,

I'm having a tough time with a below question and would like some hints on where I've gone wrong if my answer is incorrect.

I think the reason it's difficult is because it intuitively doesn't make sense, ...heading 90 degrees south after initially travelling due east, not to mention that it changes the hypotenuse of my vector addition triangle from the Pa to P`b.

Here is the question;

"Ball A, rolling west at 3.0 m/s, has a mass of 1.0 kg. Ball B has a mass of 2.0 kg and is stationary. After colliding with ball B, ball A moves south at 2.0 m/s. Calculate the momentum and velocity of ball B after the collision.

This is my answer and methodology so far..

pa + pb = p`a + p`b (vector sum) with pb = 0

pa = mava = (1.0 kg) * (3.0 m/s) = 3.0 kg*m/s (this is the end total that must be met to satisfy conservation of momentum law)

p`a = ma`+va` = (1.0 kg) * (2.0 m/s) = 2.0kg*m/s

pa_x + pb_x = p`a_x + p`b_x

pa_y + pb_y = p`a_y + p`b_y

now I know the following.

pa_x = 3.0, p`a_x = -3.0 (since ball a is now going due south and has no eastward momentum)

pb_x = 0, p`b_x = ?

pa_y = 0, p`a_y = 2.0 (since ball a is now going due south)

pb_y = 0, p`b_y = -2.0 (since vector sum of vertical components must be 0)

Now I do this, I've changed signs to represent direction of travel for -x, +x, -y and +y ..they should all work out in the end?

pa + pb = p`a + p`b OR (pa_x + pa_y) + (pb_x + pb_y) = (p`a_x + p`a_y) + (p`b_x + p`b_y)

(-3.0 + 0) + (0 + 0) = (-3 + (-2)) + ( X + 2)

X = 0, or p`b_x = 0 ..so I now know that Ball B has no horizontal momentum, in theory.

Great ..do I care? I have no idea where to go from here with this methodology...

Alternatively I don't have to do any of the above save for;

p`a = ma`+av` = (1.0 kg) * (2.0 m/s) = 2.0kg*m/s.

Now I have two sides of a triangle, with pa = 3.0kg*ms and p'a = 2.0kg*m/s, p'b (Hypotenuse) uknown.

By Pythagoras theorem I get.. p'b = [tex]\sqrt{13}[/tex] or (3.6kg*m/s) which is the momentum of ball B.

Divide by 2.0 kg and I get 1.8m/s as the velocity.

Is this correct?

Thanks....

I'm having a tough time with a below question and would like some hints on where I've gone wrong if my answer is incorrect.

I think the reason it's difficult is because it intuitively doesn't make sense, ...heading 90 degrees south after initially travelling due east, not to mention that it changes the hypotenuse of my vector addition triangle from the Pa to P`b.

Here is the question;

"Ball A, rolling west at 3.0 m/s, has a mass of 1.0 kg. Ball B has a mass of 2.0 kg and is stationary. After colliding with ball B, ball A moves south at 2.0 m/s. Calculate the momentum and velocity of ball B after the collision.

This is my answer and methodology so far..

pa + pb = p`a + p`b (vector sum) with pb = 0

pa = mava = (1.0 kg) * (3.0 m/s) = 3.0 kg*m/s (this is the end total that must be met to satisfy conservation of momentum law)

p`a = ma`+va` = (1.0 kg) * (2.0 m/s) = 2.0kg*m/s

pa_x + pb_x = p`a_x + p`b_x

pa_y + pb_y = p`a_y + p`b_y

now I know the following.

pa_x = 3.0, p`a_x = -3.0 (since ball a is now going due south and has no eastward momentum)

pb_x = 0, p`b_x = ?

pa_y = 0, p`a_y = 2.0 (since ball a is now going due south)

pb_y = 0, p`b_y = -2.0 (since vector sum of vertical components must be 0)

Now I do this, I've changed signs to represent direction of travel for -x, +x, -y and +y ..they should all work out in the end?

pa + pb = p`a + p`b OR (pa_x + pa_y) + (pb_x + pb_y) = (p`a_x + p`a_y) + (p`b_x + p`b_y)

(-3.0 + 0) + (0 + 0) = (-3 + (-2)) + ( X + 2)

X = 0, or p`b_x = 0 ..so I now know that Ball B has no horizontal momentum, in theory.

Great ..do I care? I have no idea where to go from here with this methodology...

Alternatively I don't have to do any of the above save for;

p`a = ma`+av` = (1.0 kg) * (2.0 m/s) = 2.0kg*m/s.

Now I have two sides of a triangle, with pa = 3.0kg*ms and p'a = 2.0kg*m/s, p'b (Hypotenuse) uknown.

By Pythagoras theorem I get.. p'b = [tex]\sqrt{13}[/tex] or (3.6kg*m/s) which is the momentum of ball B.

Divide by 2.0 kg and I get 1.8m/s as the velocity.

Is this correct?

Thanks....

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