What is the momentum of a girl jumping into a boat?

[vxr]

Homework Statement

A girl of mass $m_{1} = 50$ kg running with a velocity $v_{1} = 5$ m/s jumped into the boat of mass $m_{2} = 150$ kg. Determine a velocity $v_{2}$ at which a boat sailed away?

Relevant Equations

$p = mv$ and $m_{1}v_{1} = m_{2}v_{2}$ (I suppose)

Momentum of a girl:

$$p_{1} = m_{1}v_{1} = 50 * 5 = 250$$

Momentum of an idle boat:

$$p_{2} = m_{2}v_{2} = 150 * 0 = 0$$

So if the girl jumps into the boat, the two "systems" connect with each other. The momentum is passed onto the boat (?):

$$p_{1} = p_{2}$$

$$m_{1}v_{1} = m_{2}v_{2}$$

Substituting all known values:

$$250 = 150 v_{2}$$

$$v_{2} = \frac{5}{3} \quad \Big[ \frac{m}{s} \Big]$$

Does it look any good?

 

[Gaussian97]

Not really, you have to take into account that now the girl is in the boat, so in total, the mass is 50+150 kg.

The momentum at the beginning of the problem is $p_i=m_1v_1+m_2v_2=50\text{kg}\cdot 5\text{ms}^{-1}=250\text{kgms}^{-1}$. Then at the end both, the girl and the boat will move with the same velocity $v'_2$, so the momentum will be $p_f=(m_1+m_2)v'_2$. By conservation of momentum you have:
$$p_i=250\text{kgms}^{-1}=(m_1+m_2)v'_2=200\text{kg}\cdot v'_2\Longrightarrow v'_2=\frac{250}{200}\text{ms}^{-1}=\frac{5}{4}\text{ms}^{-1}=1.25\text{ms}^{-1}$$

 

[vxr]

Thank you. I have applied that logic to the following simple task:

A cart of mass $m_{c} = 160$ kg was running with a velocity (speed) $v_{c} = 2$ m/s. A boy of mass $m_{b} = 40$ kg catches up with him at velocity $v_{b} = 5$ m/s and drop-in. Determine the final velocity of cart with a boy.

Let:

- $p_{i}$ - initial momentum

- $p_{f}$ - final momentum, I assume equal to initial momentum?

- $v^{'}_{c}$ - velocity of a cart after the boy jumps in

Momentum at the beginning:

$p_{i} = m_{c}v_{c} + m_{b}v_{b} = 160 * 2 + 40 * 5 = 520 \frac{kgm}{s}$

After the boy drops-in:

$p_{i} = p_{f}$

$p_{f} = (m_{c} + m_{b})v^{'}_{c}$

$\frac{p_{f}}{m_{c} + m_{b}} = v^{'}_{c}$

$\frac{520}{160 + 40} = v^{'}_{c}$

$\frac{13}{5} = v^{'}_{c}$

Is this mistake-free?

 

[Gaussian97]

Yes, I think it's perfect!

 

[NP04]

Hi, I see those apostrophes after variables (ex. v') a lot on this website. Is anyone able to explain what it means? Thanks!

 

[Gaussian97]

Hi, I see those apostrophes after variables (ex. v') a lot on this website. Is anyone able to explain what it means? Thanks!

We simply use them to distinguish between two variables, so $v'_2$ is simply the velocity of the object 2. Since we have defined the velocity before and after we use de ' to distinguish. That's all.

 

[NP04]

Thanks I appreciate it.

 

[haruspex]

Hi, I see those apostrophes after variables (ex. v') a lot on this website. Is anyone able to explain what it means? Thanks!

It doesn’t always mean the same. In this thread it is being used just to create a name for a related variable. v is the initial velocity of one of the objects and v' is a later velocity of it.
In other contexts it might be used to indicate a derivative: if f(x)=x² then f'(x)=2x etc.