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**In the diagram below, a cart is released from rest at a distance x1 from the first of two photogates. The cart then rolls down the ramp through both photogates. The distance between the photogates is x2. For this problem, you may assume that the cart is frictionless and the acceleration of the cart a=g sin THETA (in degrees).**

Question) If x1 = 20 cm, x2 = 50 cm, and THETA = 20 degrees, what is the average velocity of the cart between the photogates (over the distance x2)?

http://www.facebook.com/photo.php?pid=4024978&l=3ffbe03791&id=645560319 [Broken][/URL]

Question) If x1 = 20 cm, x2 = 50 cm, and THETA = 20 degrees, what is the average velocity of the cart between the photogates (over the distance x2)?

http://www.facebook.com/photo.php?pid=4024978&l=3ffbe03791&id=645560319 [Broken]

(Here is the link to the image:

**http://www.facebook.com/photo.php?pid=4024978&l=3ffbe03791&id=645560319 [Broken]**)

**This is what I did to solve this problem:**

v^2=v0^2 + 2ax

a=-9.8sin20degrees = 3.352

x1=20

x2=50

v0x=0

vx1^2 = 0 + 2(3.352)(20),

vx1^2= SQUARE ROOT of 134.072

= 11.5789 cm/s.

vx2^2 = 11.5789^2 + 2(3.352)(50),

vx2=SQUARE ROOT of 469.271

= 21.663 cm/s

v^2=v0^2 + 2ax

a=-9.8sin20degrees = 3.352

x1=20

x2=50

v0x=0

vx1^2 = 0 + 2(3.352)(20),

vx1^2= SQUARE ROOT of 134.072

= 11.5789 cm/s.

vx2^2 = 11.5789^2 + 2(3.352)(50),

vx2=SQUARE ROOT of 469.271

= 21.663 cm/s

Is this answer correct? I seriously doubt if I am using the correct equations. I can't talk to my professor until the day I have to turn my home work in, and I can't find any examples on google :) I will appreciate if somebody will confirm or correct this ASAP! :)

Thankyou,

Arshad_Physic

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