In procedure 2: suppose the inclined plane is tipped at angle = 2.2 o . The 0 cm
ID: 1442978 • Letter: I
Question
In procedure 2: suppose the inclined plane is tipped at angle = 2.2 o. The 0 cm mark is at the bottom. (NOTE: In the lab, the 0 cm mark will be at the top.)
Photogate 1 is at the 31.5 cm mark, and photogate 2 is at the 60.5 cm mark along the track, Assume
- The bottom of the track is at U = 0 J (potential energy is zero).
- The total mass of the cart is 517 g.
- The track is totally frictionless
NOTE: Be careful of units!
Suppose the cart is released at rest at some point above photogate 2 (the higher point on the track). If the cart passes through photogate 2 at speed 0.66 m/s, at what speed will the cart pass through photogate 1?
v1 = ____________ m/s
Explanation / Answer
acceleration of the cart, a = g*sin(2.2)
= 9.8*sin(2.2)
= 0.376 m/s^2
v2 = 0.66 m/s
distance between photo gates, d = 60.5 - 31.5
= 29 cm
= 0.29 m
v1 = ?
apply, vf^2 - vi^2 = 2*a*d
v1^2 - v2^2 = 2*a*d
v1 = sqrt(v2^2 + 2*a*d)
= sqrt(0.66^2 + 2*0.376*0.29)
= 0.81 m/s <<<<<<<<<<<<----------------Answer