Calculate change in KE and momentum (impulse)

[jfnn]

Homework Statement

I have attached the problem.

A 14,000 kg tractor traveling north at 21 km/h turns west and travels 26 km/h. Calculate the change in the tractor's

a. kinetic energy
b. linear momentum (magnitude AND direction)

Homework Equations

[/B]
delta KE = (1/2)mv^2 - (1/2)mvi^2

change in linear momentum (delta p) = m(vf-vi)

The Attempt at a Solution

I defined east as positive and north as positive.

I first converted the velocities to m/s to get: 5.83 m/s North and -7.22 m/s West.

a) I used: delta KE = (1/2)mv^2 - (1/2)mvi^2

Since kinetic energy is a scalar quantity, I do not have to worry about direction. So... I will just plug in the values all given to find the change in KE.

(1/2)(14000)(-7.22)^2 - (1/2)(14000)(5.83)^2

After computing I get 1.27*10^5 J --> Can anyone verify?

b) This part is where I get lost... The change in linear momentum is the same thing as impulse. Impulse is m(vf-vi) --> I know I cannot just subtract the velocities that are given because they are not in same direction. They have direction and magnitude. I must use vectors to find the vf-vi.

In vector notation, vf-vi is that same things as saying vf+(-vi)

Turns out that the initial velocity only has a y component, which makes its velocity (after computing with the vectors) = 5.83 m/s

The final velocity only has an x component and a 0 y component, which makes its velocity ( after computing with a y and x component) = -7.22 m/s

After getting this far, if I did it right, I assume that now I must do another vector problem with the vector final + (- vector initial)

I draw the final vector in the negative direction (west) and lable it with -7.22, then I flip the direction of the north vector (v initial) making it south to subtract from the v final. I get a vector pointing west, plus a vector pointing south. I draw an arrow from the tail of first to tip of last.

I then have a right angle, do pythagorean theroem to find

86.1173 = resultant^2 ... I take the square root to get 9.27

Is this value vf-vi?

If so, I multiply this by the mass of the truck to get the change in momentum?

So... (9.27) * 14000 kg = 1.30*10 ^5 kg*m/s

To get the direction, I do the arctan of what values? Is it the velocity of final/intial??

Thanks for the help!

 

[BvU]

In vector notation, vf-vi is that same things as saying vf+(-vi)

Yes. If you make a skeetch you will also see which ##\arctan## you need for the angle

 

[jfnn]

Yes. If you make a skeetch you will also see which ##\arctan## you need for the angle

Okay so I would use the values though for the velocity? Because momentum is in the same direction as the velocity right?

So it will be the arc tan of (initial velocity / the final velocity) (I drew out the triangle and tan is the opposite over adjacent)

The opposite is the velocity initial and the adjacent is the velocity final.

So theta = arctan(5.833/-7.22) ?

Does this seem correct?

 

[BvU]

Okay so I would use the values though for the velocity? Because momentum is in the same direction as the velocity right?

Yes

Note:

5.83 m/s North and -7.22 m/s West.

but you meant -7.22 m/s East (but you processed correctly).

So theta = arctan(5.833/-7.22) ?

Does this seem correct?

No. Can you show the drawing ? In which quadrant does the vector ##\Delta\vec v## end up ?
And: in which quadrants do arctan results end up ?

 

[jfnn]

Yes

Note:

but you meant -7.22 m/s East (but you processed correctly).No. Can you show the drawing ? In which quadrant does the vector ##\Delta\vec v## end up ?
And: in which quadrants do arctan results end up ?

Isn't it drawn like this?

The resultant is in quadrant three?

 

[jfnn]

Isn't it drawn like this?

The resultant is in quadrant three?

Oh wait.. the downwards line has to be negative as well?
So it is arc tan (-5.8333/-7.22)

Is that better now?

 

[jfnn]

Oh wait.. the downwards line has to be negative as well?
So it is arc tan (-5.8333/-7.22)

Is that better now?

Thus it is 38.9 degrees below negative x axis.

 

[haruspex]

Thus it is 38.9 degrees below negative x axis.

Yes.