How Do You Calculate the Mass of the Plane After Intercepting Luggage?

[Oxford365]

Homework Statement

A plane with an initial velocity of (225m/s ŷ + 325m/s Ẑ) intercepts a 20kg piece of luggage with initial velocity of (-75m/s ŷ - 15m/s Ẑ). After, plane is moving at speed 400 m/s. What is the mass of the plane?

Homework Equations

momentum y: Mplane(225m/s) + 20kg(-75 m/s) = Mplane(Vy) + 20kg(Vy)
momentum z: Mplane(325m/s) + 20kg(-15m/s) = Mplane(Vz) +20kg(Vz)
V = Vy^2 + Vz^2 , Tanθ= Vz/Vy

The Attempt at a Solution

I know the speed given after the interception is in magnitude form and it needs to be in component form and the speed of the plane and luggage are the same after.
For the plane I get V=395.3m/s @ 55.3°
For the luggage I get V=76.5m/s @ 11.3° (assuming this means the Luggage starts in Q1 and moves on the 11.3° angle into Q3?)
I tried doing conservation of momentum with plugging in 400 in the final velocities but I realized this does not make sense.

 

[Orodruin]

What is the momentum of the plane+luggage after the collision? How does this relate to the speed of the plane+luggage after the collision?

I know the speed given after the interception is in magnitude form and it needs to be in component form

Careful with nomenclature here, speed cannot be given on component form as it is a scalar! It is the magnitude of the velocity.

 

[Dr. Courtney]

Express in terms of the magnitudes and angles and solve.

 

[Oxford365]

Express in terms of the magnitudes and angles and solve.

This is where I seem to be lost. This is for a summer class and I haven't had much time/practice with this sort of problem.

 

[Oxford365]

What is the momentum of the plane+luggage after the collision? How does this relate to the speed of the plane+luggage after the collision?

Well I know the momentum after is Mplane(Vy) + 20kg(Vy) & Mplane(Vz) +20kg(Vz)
If I knew the angle of the 400m/s I could use sin and cosine to get the components. This is where I am lost

 

[Orodruin]

If I knew the angle of the 400m/s I could use sin and cosine to get the components.

The angle is irrelevant (although you can compute it from the angle of the initial momentum). You only need to find the magnitude of the momentum.

 

[Oxford365]

The angle is irrelevant (although you can compute it from the angle of the initial momentum). You only need to find the magnitude of the momentum.

So : Mplane(395.3m/s) + 20kg(76.5m/s) = Mplane(400m/s) + 20kg(400m/s) ?

 

[Orodruin]

So : Mplane(395.3m/s) + 20kg(76.5m/s) = Mplane(400m/s) + 20kg(400m/s) ?

No, you need to square the total momentum before the impact! A suggestion is: Do not calculate the direction of the momenta of the separate entities, it has no bearing on the problem - only the total momentum is of relevance.

 

[Oxford365]

No, you need to square the total momentum before the impact! A suggestion is: Do not calculate the direction of the momenta of the separate entities, it has no bearing on the problem - only the total momentum is of relevance.

We have never squared momentum in any case in class..and isn't calculating the momenta of separate objects crucial to finding the total momentum? I am not grasping your advice, could you dumb it down anymore?

 

[Orodruin]

We have never squared momentum in any case in class..and isn't calculating the momenta of separate objects crucial to finding the total momentum? I am not grasping your advice, could you dumb it down anymore?

How else do you usually find the magnitude of a vector other than adding the squares of the components, summing, and taking the square root of the result? You do not need to compute the magnitude of the individual momenta, you already have the momenta. What you need to do is to use vector addition to find the total momentum and you already have the vector components so the operation is trivial.

 

[Oxford365]

How else do you usually find the magnitude of a vector other than adding the squares of the components, summing, and taking the square root of the result? You do not need to compute the magnitude of the individual momenta, you already have the momenta. What you need to do is to use vector addition to find the total momentum and you already have the vector components so the operation is trivial.

Could you show me the first step or two? Maybe that will help.

 

[SammyS]

Homework Statement

A plane with an initial velocity of (225m/s ŷ + 325m/s Ẑ) intercepts a 20kg piece of luggage with initial velocity of (-75m/s ŷ - 15m/s Ẑ). After, plane is moving at speed 400 m/s. What is the mass of the plane?

Homework Equations

momentum y: Mplane(225m/s) + 20kg(-75 m/s) = Mplane(Vy) + 20kg(Vy)
momentum z: Mplane(325m/s) + 20kg(-15m/s) = Mplane(Vz) +20kg(Vz)
V = Vy^2 + Vz^2 , Tanθ= Vz/Vy

The Attempt at a Solution

I know the speed given after the interception is in magnitude form and it needs to be in component form and the speed of the plane and luggage are the same after.
For the plane I get V=395.3m/s @ 55.3°
For the luggage I get V=76.5m/s @ 11.3° (assuming this means the Luggage starts in Q1 and moves on the 11.3° angle into Q3?)
I tried doing conservation of momentum with plugging in 400 in the final velocities but I realized this does not make sense.

If the plane is actually intercepting the luggage, i.e. the plane captures the luggage so that it then becomes part of the plane's cargo, then is problem has no solution. The initial speed of the plane is less than 400 m/s. the initial velocity of the luggage is largely opposite that of the plane. The plane will slow down.

Maybe there is a language translation problem here.

 

[Oxford365]

If the plane is actually intercepting the luggage, i.e. the plane captures the luggage so that it then becomes part of the plane's cargo, then is problem has no solution. The initial speed of the plane is less than 400 m/s. the initial velocity of the luggage is largely opposite that of the plane. The plane will slow down.

Maybe there is a language translation problem here.

Lets say hypothetically it does work. What would the plan of attack be?

 

[haruspex]

Lets say hypothetically it does work. What would the plan of attack be?

Just work with the equations you posted in the OP. Forget the tan theta equation since that is the only one involving the angle and you don't care about that. This leaves you three equations with three unknowns.
You need to manipulate the equations so as to eliminate v_y and v_z. Do you understand how to eliminate variables in simultaneous equations?

 

[SammyS]

Lets say hypothetically it does work. What would the plan of attack be?

I see that you deleted most of what you posted when you did your Edit..

It might be helpful for others to know that you altered the values of your homework problem to get the parameters here. Unfortunately, I was not able to find the those values cached by my browser. At any rate, in the homework problem:

The initial speed of the plane is in excess of 400 m/s.
As I recall, the final speed was 400 m/s.
The mass of the luggage was something like 2000 kg.
Etc.​

For a problem such as this, you're not free to just pick these values at random.

 

[Oxford365]

I see that you deleted most of what you posted when you did your Edit..

It might be helpful for others to know that you altered the values of your homework problem to get the parameters here. Unfortunately, I was not able to find the those values cached by my browser. At any rate, in the homework problem:

The initial speed of the plane is in excess of 400 m/s.
As I recall, the final speed was 400 m/s.
The mass of the luggage was something like 2000 kg.
Etc.​

For a problem such as this, you're not free to just pick these values at random.

A spaceship with an initial velocity of (300 m/s y + 400m/s z) captures a 2000kg asteroid moving with an initial velocity of (-100m/s y - 50m/s z.) After, the ship is moving at 498m/s. What is the mass of the spaceship?

 

[Orodruin]

Well I know the momentum after is Mplane(Vy) + 20kg(Vy) & Mplane(Vz) +20kg(Vz)

Can you think of some alternative way of finding this momentum? Let us say, a conservation law of some sort?

 

[Oxford365]

Can you think of some alternative way of finding this momentum? Let us say, a conservation law of some sort?

I was thinking conservation of energy and using kinetic but kinetic energy is dissipated in this problem so I can't set kinetic initial equal to kinetic final

 

[Orodruin]

I was thinking conservation of energy and using kinetic but kinetic energy is dissipated in this problem so I can't set kinetic initial equal to kinetic final

So what other conservation laws do you know?

 

[Oxford365]

So what other conservation laws do you know?

I believe we have learned about momentum conservation, energy conservation, and angular momentum conservation.

 

[SammyS]

A spaceship with an initial velocity of (300 m/s y + 400m/s z) captures a 2000kg asteroid moving with an initial velocity of (-100m/s y - 50m/s z.) After, the ship is moving at 498m/s. What is the mass of the spaceship?

If you want to return to your airplane-luggage version, I think a little tweak will make it workable.

Changing the y component of the initial velocity of the plane from 225m/s to 235m/s gives a speed just over 401m/s. That should make the problem workable.

As far as finding a solution goes:
Your equations for components using momentum conservation should get you there. Square each equation then add them, (Maybe factor out V_x and V_y first.) You will then have V² on the right hand side (multiplied by some combination of masses). The only unknown will be M_plane. This gives a quadratic equation in M_plane. The result looks to be pretty messy.

Perhaps Orodruin is leading you to a cleaner method.

 

[Orodruin]

Perhaps Orodruin is leading you to a cleaner method.

No, this is pretty much it.

 

[Oxford365]

No, this is pretty much it.

So square my y-momentum equation and square my z-momentum equation and add them?

 

[SammyS]

So square my y-momentum equation and square my z-momentum equation and add them?

Yes.

You may want to factor out V_x and V_y from the right hand sides first.

Remember, V² = (V_x)² + (V_y)² .