# Find the magnitude and direction of the velocity

• inigooo
In summary, according to the conversation, the formula for V^2=u^2+2gh is used to find the velocity of an object after colliding with another object. The terms u, g, and h represent initial velocity, gravitational acceleration, and the height of the object, respectively. The formula for finding the magnitude and direction of the secondary ball's velocity after collision in terms of its velocity before contact 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, where m represents mass and v represents velocity. Additionally,

#### 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

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

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
inigooo said:
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

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

inigooo said:
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?

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

I dont

inigooo said:
I dont
Then look it up in your textbook.

malawi_glenn said:
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?

• Delta2
• Delta2
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 said:
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?

malawi_glenn said:
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

inigooo said:
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?

malawi_glenn said:
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

Can you use LaTeX?

What does a prime quantity refer to?

malawi_glenn said:
Can you use LaTeX?

What does a prime quantity refer to?
After collision

malawi_glenn said:
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
inigooo said:
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

inigooo said:
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

malawi_glenn said:
Almost. You are missing some important code to make it appear as latex
Is the answer a) unchanged, b) upward?

inigooo said:
Is the answer a) unchanged, b) upward?
Why do you think that is the correct answer?

malawi_glenn said:
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?

inigooo said:
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?

malawi_glenn said:
Cant it be 40% of initial speed?
40%?

inigooo said:
40%?
Or 75%

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

inigooo said:
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?

inigooo said:
Is this wrong?

malawi_glenn said:
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 said:
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…

inigooo said:
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?