Ballisotcardiogram and mass of blood

[quicknote]

When the heart beats it expels a mass m of blood into the aorta with speed v, and the body and platform move in the opposite direction with speed v. The bllod velocity can be determine independently by observing the Doppler shift of ultrasound. Assume that v is 50.0 cm/s in one trial. The mass of the subject plus the platform is 54.0kg. The platform moves 6x10^-5 m in 0.160s after one heart beat.
Calculate the mass m of blood that leaves the heart.


initially I used the equation m1v1=m2v2 and calculated the mass of blood to be 0.041kg.

But I'm not sure if you can use conservation of momentum in this question. A friend of mine used (ma)_body=(ma)_blood and calcualted the mass of the blood to be 0.08kg...but I'm not sure if this makes sense.

 

[Andrew Mason]

When the heart beats it expels a mass m of blood into the aorta with speed v, and the body and platform move in the opposite direction with speed v. The bllod velocity can be determine independently by observing the Doppler shift of ultrasound. Assume that v is 50.0 cm/s in one trial. The mass of the subject plus the platform is 54.0kg. The platform moves 6x10^-5 m in 0.160s after one heart beat.
Calculate the mass m of blood that leaves the heart.
initially I used the equation m1v1=m2v2 and calculated the mass of blood to be 0.041kg.
But I'm not sure if you can use conservation of momentum in this question. A friend of mine used (ma)_body=(ma)_blood and calcualted the mass of the blood to be 0.08kg...but I'm not sure if this makes sense.

I am not very clear on the setup here but it seems that the heart is expelling blood into the aorta which is still connected to the body. So let's assume that the expelled blood continues to move with its initial speed for a long time (ie does not give it back to the body). Then it becomes a conservation of momentum problem.

$$M_{body}v_{body} = - m_{blood}v_{blood}$$

The first step is to determine the final speed of the body/platform. If it moves $6\times 10^{-5} m.$ in .160 s. you can find its average speed: v = d/t. But it starts at 0 and it ends up with $v_f$ to obtain that average. So you really can't use $v_f = d/t$. We can't use .5 d/t either because we don't know the duration that the expulsive force and we don't know that it is constant.

After we find v of the body/platform, it is just a matter of plugging it into the above equation to find the mass of the blood.

AM

 

[quicknote]

Thanks for getting back to me.
I ended up using v = d/t.
I know you said that we couldn't really use it but there isn't really any other way to solve it based on the information that was given.
I was also doing a search on ballistocardiograms and this site has the formula:
$$F_{body} = -F_{blood}$$
$$M_{body+table}a_{body+table} = - m_{blood}a_{blood}$$
http://www.upscale.utoronto.ca/GeneralInterest/Harrison/BCG/BCG.html
But I think this is useless info since we don't have acceleration either. And I don't think we can calculate the acceleration...
Thanks again!

 

[Andrew Mason]

Thanks for getting back to me.
I ended up using v = d/t.
I know you said that we couldn't really use it but there isn't really any other way to solve it based on the information that was given.
I was also doing a search on ballistocardiograms and this site has the formula:
$$F_{body} = -F_{blood}$$
$$M_{body+table}a_{body+table} = - m_{blood}a_{blood}$$
http://www.upscale.utoronto.ca/GeneralInterest/Harrison/BCG/BCG.html
But I think this is useless info since we don't have acceleration either. And I don't think we can calculate the acceleration...
Thanks again!

Since the forces are always equal and opposite within any inertial system, and since they both have the same duration, $F_1\Delta t = - F_2\Delta t = m_1v_1 = -m_2_v_2$

But to solve using force, one has to know the duration of the force and the way the force varies with time. It is much easier to use momentum in which you need only know the final speed and the mass.

AM