what is the velocity of the plank relative to the ice surface

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Conservation of linear momentum/Galilean relativity problem

• Thread starter NoPhysicsGenius
• Commencement engagement Apr 23, 2008

[SOLVED] Conservation of linear momentum/Galilean relativity problem

Homework Statement

A 45-kg girl is standing on a plank that has a mass of 150 kg. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless supporting surface. The daughter begins to walk along the plank at a abiding velocity of one.5 m/southward relative to the plank.

(a) What is her velocity relative to the ice surface?

(b) What is the velocity of the plank relative to the ice surface?

Homework Equations

[tex]m_1v_1 = m_2v_2[/tex]
[tex]\overrightarrow{p_{1i}} + \overrightarrow{p_{2i}} = \overrightarrow{p_{1f}} + \overrightarrow{p_{2f}}[/tex]

[tex]\overrightarrow{r}' = \overrightarrow{r} - \overrightarrow{u}t[/tex]
[tex]\overrightarrow{v}' = \overrightarrow{v} - \overrightarrow{u}[/tex]

The Effort at a Solution

mass of girl:
[tex]m_g = 45 kg[/tex]

mass of plank:
[tex]m_p = 150 kg[/tex]

velocity of girl relative to plank:
[tex]v_{gp} = 1.5 thou/s[/tex]

velocity of plank relative to girl:
[tex]v_{pg} = ?[/tex]

conservation of momentum:
[tex]m_gv_{gp} = m_pv_{pg} \Rightarrow[/tex]
[tex]v_{pg} = \frac{m_gv_{gp}}{m_p} = \frac{(45 kg)(1.five m/s)}{150kg} = 0.450 m/southward[/tex]

And so I have the velocity of the plank relative to the daughter; but I have no idea how to find the velocity of the girl or the plank relative to the water ice. Can you help? Give thanks y'all.

Answers and Replies

I should likewise annotation that in the back of the book, the answer for (a) is 1.fifteen g/s and the respond for (b) is -0.346 m/south.

v_pg and v_gp are equal and opposite in sign. They Have to be, simply think almost it. They are the same relative velocity measured in two different frames. You lot can't combine them in a unmarried conservation of momentum equation. If vg and vp are the velocities relative to the ice, conservation of momentum tells y'all mg*vg+mp*vp=0. v_gp is but vg-vp.

v_pg and v_gp are equal and opposite in sign. They Accept to be, but think about it. They are the same relative velocity measured in two dissimilar frames. You can't combine them in a single conservation of momentum equation. If vg and vp are the velocities relative to the ice, conservation of momentum tells you mg*vg+mp*vp=0. v_gp is just vg-vp.

That makes sense. But at present I'one thousand really confused about how to proceed ...

But write an expression for conservation of momentum with all velocities with respect to the basis. The only unknown will be the velocity of the plank. Hint: Assume the plank goes left (negative) and the girl goes right (positive).

Just write an expression for conservation of momentum with all velocities with respect to the ground. The only unknown will be the velocity of the plank.

How tin the merely unknown be the velocity of the plank? What is the velocity of the girl relative to the basis?

You tin can express the girl's velocity with respect to the footing in terms of the plank'southward velocity.

Now I get information technology! ...

[tex]v_{gi} = v_{gp} - v_{pi}[/tex]
[tex]m_gv_{gi} = m_pv_{pi}[/tex]
[tex]\Rightarrow m_g(v_{gp} - v_{pi}) = m_pv_{pi}[/tex]
[tex]\Rightarrow m_gv_{gp} - m_gv_{pi} = m_pv_{pi}[/tex]
[tex]\Rightarrow m_gv_{gp} = m_gv_{pi} + m_pv_{pi}[/tex]
[tex]\Rightarrow v_{pi} = \frac{m_gv_{gp}}{m_g + m_p}[/tex]
[tex]\Rightarrow v_{pi} = \frac{(45 kg)(1.fifty g/s)}{45 kg + 150 kg} = 0.346 k/south[/tex]

So the velocity of the plank with respect to the ice is 0.346 m/s to the left. This is the answer for part (b).

To solve role (a) ...
[tex]v_{gi} = v_{gp} - v_{pi}[/tex]
[tex]\Rightarrow v_{gi} = 1.50 g/due south - 0.346 m/s = one.15 m/south (to the right)[/tex]

Thank you all for your fourth dimension and patience! Especially Doc Al!

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