Momentum of falling objects on impact; p = mv not working?

[exi]

Got the first part; the second part of the question asks what angle of inclination the bale will strike, given between -180° and 180°. Drawn out, this seems to resemble a right triangle with a 65 m height, 200.4188 m length (initial X velocity = 55 m/s, time of 3.6422 s), and the inverse tangent of that is 17.970° - yet neither that nor ~162° seem to work. Why not?

Homework Statement

A plane, flying due east at 55 m/s, drops a bale of hay from an altitude of 65m.

Acceleration due to gravity: 9.81 m/s².

If the bale of hay weighs 173 N, what's the momentum of the bale the moment it strikes the ground? Answer in kg∙m/s.

The Attempt at a Solution

Since momentum is just the product of mass and velocity, what I tried to do is consider y-axis movement only; that is,

$$V_0 = 0 \frac{m}{s}$$

$$a = 9.8 \frac{m}{s^2}$$

$$\Delta y = 65 m , \mbox so:$$

$$v^2 = v_0^2 + 2a\Delta x$$

Solving that, velocity comes out to 35.711 m/s. Since the bale's weight is 173 N, then 173 / 9.8 = 17.653 kg. Multiplying that weight by the above velocity would seem to me that it should produce the correct answer, but it does not; why?

 

[ice109]

because momentum is a vector and you've only considered the y component. you have to consider both and then find the resultant vector

 

[exi]

because momentum is a vector and you've only considered the y component. you have to consider both and then find the resultant vector

My mistake; got it now.

Thanks for the reminder.

 

[exi]

Edited original post to try and figure out what I'm screwing up now.

 

[Astronuc]

Did you get the answer?

momentum and velocity are vectors.

In Cartesian coordinates, $$v\,=\,\sqrt{v_x^2\,+\,v_y^2\,+\,v_z^2}$$, where v is the speed or magnitude of the velocity, and then momentum is just the product of the mass (scalar) and velocity (vector).

In the OP, just use v_x and v_y.

 

[exi]

Did you get the answer?

momentum and velocity are vectors.

In Cartesian coordinates, $$v\,=\,\sqrt{v_x^2\,+\,v_y^2\,+\,v_z^2}$$, where v is the speed or magnitude of the velocity, and then momentum is just the product of the mass (scalar) and velocity (vector).

In the OP, just use v_x and v_y.

Yes, I did - it's the angle of impact that's got me now.

 

[ice109]

$$tan(\theta)=\frac{opp}{adj}$$

 

[Dick]

$$tan(\theta)=\frac{opp}{adj}$$

True. But apply this to the final velocity vector itself. The path of the bale (approximate hypotenuse) of the triangle you are looking at is actually curved.

 

[Astronuc]

One has v_x and v_y

So, what does one do with v_y / v_x with components measured at impact.

 

[exi]

One has v_x and v_y

So, what does one do with v_y / v_x with components measured at impact.

Not sure what you're getting at. Using a straight line for the approximate trajectory of the bale would seem to suffice, and since I had what I thought to be both vertical and horizontal displacement, it would seem to me that taking the tangent of those figures would've done it - but it doesn't.

 

[Doc Al]

Treating the trajectory as a straight line would a lousy and unnecessary approximation. Forget about displacement. Focus on the velocity at impact. (Reread Astronuc's hints.)

 

[exi]

If only I were graded on my ability to needlessly complicate things.

$$65.5761cos\theta = 55$$

$$\theta = cos^{-1}\frac{55}{65.5761} = 32.994^{\circ}$$

Much appreciated.