- #1
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!