Find the magnitude and direction of the velocity

[inigooo]

Homework Statement

A ball is hanging from a cord of negligible weight. A secondary ball with the same weight as the first ball is measured 3m above the hanging ball, and just touching the cord, is released from rest and gains its velocity before striking the hanging ball. If the effect of friction is negligible, and the impact is perfectly elastic, what is
a) the magnitude of the velocity of the secondary ball in terms of its velocity before contact;
b) the direction of the velocity of the secondary ball in terms of its velocity before contact

Relevant Equations

My answers are a) 7.67 m/s and b) downward, however my teacher says it’s wrong

V^2= u^2 +2gh

 

[malawi_glenn]

Show your work, not just your answers

 

[inigooo]

v^2= u^2 +2gh
v^2= 0^2+2(9.81)(3)
v=sqrt(2x9.81x3)
v=sqrt(58.86)
v=7.67m/s, downward
here is my solution

 

[malawi_glenn]

Given one signigicant digit, don't answer with three.

Are you asked about the magnitude and direction of the velocity directly after the collision with the primary ball?

I do not think you are supposed to answer with numbers

a) the magnitude of the velocity of the secondary ball in terms of its velocity before contact;
b) the direction of the velocity of the secondary ball in terms of its velocity before contact

 

[inigooo]

no, it is ’in terms of its velocity before contact’

 

[malawi_glenn]

no, it is ’in terms of its velocity before contact’

Such question means that you should find a ratio or similar. Let's say the velocity just before the impact was $\vec v_1$ and the velocity just after the impact was $\vec v_2$. You should just answer what $|\vec v_2| / |\vec v_1|$ is.

Similarly for the b)-part. Is the direction of $\vec v_2$ same, opposite, 90 degree, etc, compared to $\vec v_1$?

The problem says that the collision is elastic. Do you know what that means and how to use that information here?

 

[inigooo]

The problem says that the collision is elastic. Do you know what that means and how to use that information here?

 

[inigooo]

I dont

 

[malawi_glenn]

I dont

Then look it up in your textbook.

 

[inigooo]

Then look it up in your textbook.

Pls. How can i get the magnitude and direction of the velocity of the secondary ball in terms of its velocity before contact?

 

[malawi_glenn]

Pls. How can i get the magnitude and direction of the velocity of the secondary ball in terms of its velocity before contact?

State what an elastic collison is.
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

 

[inigooo]

V^2=u^2 +2gh
v= sqrt(2x9.81x3)
v=7.67m/s
a) 7.67m/s
cos theta = 1
or theta = 90 degree
b) along 90 degrees with vertical line

however, this computation is for after collision. how can i compute for the magnitude and direction of the velocity of the secondary ball in terms of its velocity before contact?

 

[inigooo]

State what an elastic collison is.
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.68678

Where bodies separates completely after collision. Linear momentum is conserved and total kinetic energy IS conserved

 

[malawi_glenn]

Where bodies separates completely after collision. Linear momentum is conserved and total kinetic energy IS conserved

Yes. How can you use these two conditions on this problem?

 

[inigooo]

Yes. How can you use these two conditions on this problem?

The total kinetic energy before and after collision will remain the same, and the total momentum along a particular direction before and after collision will remain the same

 

[malawi_glenn]

The total kinetic energy before and after collision will remain the same, and the total momentum along a particular direction before and after collision will remain the same

Write down the equations for the linear momentum before and after the collision, same with kinetic energy. Then solve that system of equations.
Note: what will happen to the string supporting the primary ball?

 

[inigooo]

Write down the equations for the linear momentum before and after the collision, same with kinetic energy. Then solve that system of equations.
Note: what will happen to the string supporting the primary ball?

1/2m1v1^2+1/2m2v2^2=1/2m1v1’^2+1/2m2v2’^2
m1vx1+1/2m2vx2=m1v’x1 + m2v’y2
m1vy1+1/2m2vy2=m1v’y1 + m2v’y2
v2^2=u^2+2gh
v2=sqrt(6x9.81)
0+1/2mv2^2=1/2mv’1^2+1/2mv’2^2
v2^2=v’1^2 + v’2^2
0+0=mv’1-mv’2sintheta
sintheta=v’1/v’2
mv2=mv’2costheta
costheta= v2/v’2
v2^2-v’2^2+v2^2=v’2^2
v’2=v2
v’2=sqrt(58.86)
costheta=1 or theta = 90Degrees

 

[malawi_glenn]

Can you use LaTeX?

What does a prime quantity refer to?

 

[inigooo]

Can you use LaTeX?

What does a prime quantity refer to?

After collision

 

[malawi_glenn]

After collision

https://www.physicsforums.com/help/latexhelp/

 

[inigooo]

https://www.physicsforums.com/help/latexhelp/

1/2 m_1v_1^2 + 1/2 m_2v_2^2 = 1/2 m_1v_1’^2 + 1/2 m_2v_2’^2
m_1v_x1 + 1/2 m_2v_x2 = m_1v’_x1 + m_2v’_y2
m_1v_y1 + 1/2 m_2v_y2 = m_1v’_y1 + m_2v’_y2
v_2^2 = u^2 +2gh
v_2=sqrt(6x9.81)
0+1/2 mv_2^2 = 1/2 mv’_1^2 + 1/2 mv’_2^2
v_2^2 = v’_1^2 + v’_2^2
0+0= mv’_1 - mv’_2 sintheta
sintheta = v’_1 / v’_2
mv_2=mv’_2 costheta
costheta= v_2 / v’_2
v_2^2 - v’_2^2 + v_2^2 = v’_2^2
v’_2 = v_2
v’_2 = sqrt(58.86)
costheta=1 or theta = 90Degrees

 

[inigooo]

I don't know what the final answer is

1/2 m_1v_1^2 + 1/2 m_2v_2^2 = 1/2 m_1v_1’^2 + 1/2 m_2v_2’^2
m_1v_x1 + 1/2 m_2v_x2 = m_1v’_x1 + m_2v’_y2
m_1v_y1 + 1/2 m_2v_y2 = m_1v’_y1 + m_2v’_y2
v_2^2 = u^2 +2gh
v_2=sqrt(6x9.81)
0+1/2 mv_2^2 = 1/2 mv’_1^2 + 1/2 mv’_2^2
v_2^2 = v’_1^2 + v’_2^2
0+0= mv’_1 - mv’_2 sintheta
sintheta = v’_1 / v’_2
mv_2=mv’_2 costheta
costheta= v_2 / v’_2
v_2^2 - v’_2^2 + v_2^2 = v’_2^2
v’_2 = v_2
v’_2 = sqrt(58.86)
costheta=1 or theta = 90Degrees

i don't know what the final answer is

 

[malawi_glenn]

1/2 m_1v_1^2 + 1/2 m_2v_2^2 = 1/2 m_1v_1’^2 + 1/2 m_2v_2’^2
m_1v_x1 + 1/2 m_2v_x2 = m_1v’_x1 + m_2v’_y2
m_1v_y1 + 1/2 m_2v_y2 = m_1v’_y1 + m_2v’_y2
v_2^2 = u^2 +2gh
v_2=sqrt(6x9.81)
0+1/2 mv_2^2 = 1/2 mv’_1^2 + 1/2 mv’_2^2
v_2^2 = v’_1^2 + v’_2^2
0+0= mv’_1 - mv’_2 sintheta
sintheta = v’_1 / v’_2
mv_2=mv’_2 costheta
costheta= v_2 / v’_2
v_2^2 - v’_2^2 + v_2^2 = v’_2^2
v’_2 = v_2
v’_2 = sqrt(58.86)
costheta=1 or theta = 90Degrees

Almost. You are missing some important code to make it appear as latex

 

[inigooo]

Almost. You are missing some important code to make it appear as latex

Is the answer a) unchanged, b) upward?

 

[malawi_glenn]

Is the answer a) unchanged, b) upward?

Why do you think that is the correct answer?

 

[inigooo]

Why do you think that is the correct answer?

Since it says that ‘in terms of its velocity before contact’? Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1?

 

[malawi_glenn]

Since it says that ‘in terms of its velocity before contact’? Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1?

Cant it be 40% of initial speed?

 

[inigooo]

Cant it be 40% of initial speed?

40%?

 

[malawi_glenn]

40%?

Or 75%

 

[inigooo]

Or 75
why would the the magnitude of its velocity is 75% or 40% of initial speed?

 

[inigooo]

Since it says that ‘in terms of its velocity before contact’? Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1?

Is this wrong?

 

[malawi_glenn]

Is this wrong?

Why do you think your answer is correct?

 

[inigooo]

Why do you think your answer is correct?

Since it says that ‘in terms of its velocity before contact’. Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1.

 

[inigooo]

Since it says that ‘in terms of its velocity before contact’. Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1.

Pls. Help me point out what is wrong, if it is wrong…

 

[malawi_glenn]

Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1.

Why is it so?

If it is 0m/s how can ut have a direction?