A.P. Physics B

6.5 #’s 2-4 6.5 #’s 2-4

By: Anna FindleyBy: Anna Findley

6.5 # 2

2. [CJ5 6.P.036.] A 47.0 g golf ball is driven from 2. [CJ5 6.P.036.] A 47.0 g golf ball is driven from the tee with an initial speed of the tee with an initial speed of 52.052.0 m/s and rises m/s and rises to a height of to a height of 24.624.6 m. m.

(a) Neglect air resistance and determine the kinetic (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.energy of the ball at its highest point.

(b) What is its speed when it is (b) What is its speed when it is 8.08.0 m below its m below its highest point?highest point?

Solution Part A

(a) Neglect air resistance and determine the (a) Neglect air resistance and determine the kinetic energy of the ball at its highest kinetic energy of the ball at its highest point.point.

KE + mgh=½ mvKE + mgh=½ mv22

KE+ (.047kg) (9.8) (24.6m) = ½ (.047kg) KE+ (.047kg) (9.8) (24.6m) = ½ (.047kg) (52.0m/s) (52.0m/s) 22

KE= 52.2 JKE= 52.2 J

Solution Part B

(b) What is its speed when it is (b) What is its speed when it is 8.08.0 m below m below its highest point?its highest point?

KE + mgh = ½mvKE + mgh = ½mv22 + mg (h-8.0) + mg (h-8.0) (52.2J) + (.047kg) (9.8) (24.6m) = (52.2J) + (.047kg) (9.8) (24.6m) =

½(.047kg) v ½(.047kg) v 2 2 + (.047kg) (9.8) (16.6m) + (.047kg) (9.8) (16.6m) V= 48.77m/sV= 48.77m/s

6.5 # 3

3. [CJ5 6.P.037.] A cyclist approaches the 3. [CJ5 6.P.037.] A cyclist approaches the bottom of a gradual hill at a speed of bottom of a gradual hill at a speed of 1111 m/s. The hill is m/s. The hill is 5.05.0 m high, and the cyclist m high, and the cyclist estimates that she is going fast enough to estimates that she is going fast enough to coast up and over it without peddling. coast up and over it without peddling. Ignoring air resistance and friction, find the Ignoring air resistance and friction, find the speed at which the cyclist crests the hill.speed at which the cyclist crests the hill.

Solution #3

KE + PE = KE + PEKE + PE = KE + PE ½ mv½ mv22+ mgh = ½ mv+ mgh = ½ mv22+ mg+ mg (½)(11(½)(1122) + 0 = (½)(v) + 0 = (½)(v22) + (9.8)(5)) + (9.8)(5) V = 4.8 m/sV = 4.8 m/s Remember: mass cancels out...Remember: mass cancels out...

6.5 # 4

*4. [CJ5 6.P.041.] A water slide is *4. [CJ5 6.P.041.] A water slide is constructed so that swimmers, starting from constructed so that swimmers, starting from rest at the top of the slide, leave the end of rest at the top of the slide, leave the end of the slide traveling horizontally. As the the slide traveling horizontally. As the drawing shows, one person hits the water drawing shows, one person hits the water 5.00 m from the end of the slide in a time of 5.00 m from the end of the slide in a time of 0.5000.500 s after leaving the slide. Ignoring s after leaving the slide. Ignoring friction and air resistance, find the height friction and air resistance, find the height HH in the drawing. in the drawing.

Solution # 4

Vx= d/tVx= d/t Vx= (5.00m) / (0.500s)Vx= (5.00m) / (0.500s) Vx= 10.0 m/sVx= 10.0 m/s

Vfy = Viy + atVfy = Viy + at Vfy = 0 + (9.8) (0.500s)Vfy = 0 + (9.8) (0.500s) Vfy = (4.9m/s)Vfy = (4.9m/s)

Solution # 4 Continued

Use the Pythagorean Theorem to find VfUse the Pythagorean Theorem to find Vf VxVx22 + Vfy + Vfy22 = vf = vf22

Vf = 11.14Vf = 11.14

VfVf22= 2ad= 2ad 11.1411.1422 = 2 (9.8) d = 2 (9.8) d D = 6.33D = 6.33