Question

60 with the horizontal. The 0 cm mark is at the bottom. The lo er photogate is at the 23.1 cm mark and the upper photogate is at the 66.2 cm mark a In procedure 2: suppose the inclined plane is tipped so it makes an angle θ the track, Assume (potential energy is zero). The bottom of the track is at U = 0 The total mass of the cart is 449 g The track is totally frictionless NOTE: Be careful of units! a) Find the potential energy of the cart At the lower photogate: U1 = At the upper photogate: U2- b) Suppose the cart is released at rest at some point above the upper photogate. If the cart passes through the upper photogate at speed 0.581 m/s, at what speed will the cart pass through lower photogate? m/s

Answer 1

For photogate 1
Height = 23.1 * sin 6 = 2.41 cm

For photogate 2
Height = 66.2 * sin 6 = 6.92 cm

a) PE at gate 1 = mgh = 0.449 * 9.81 * 2.41*10^-2 = 0.106 J
PE at gate 2 = 0.449 * 9.81 * 6.92*10^-2 = 0.30 J

b) Recall that the loss of PE = gain in KE

Gain in KE between the two gates is 0.30-0.106 = 0.194 J

KE = (1/2) m v^2

v = √ (2KE / m ) = √( 2 * 0.194 / 0.449) = 0.93 m/s
This would be the velocity if the cart fell vertically downward
Since it is rolling down a plane, this velocity is reduced be a factor of the sine of the angle of inclination

So v through gate 1 = 0.93 sin 6= 0.0972 m/s