Aim: More conservation of Energy

Do Now:A 2 kg mass is dropped from a height of 10 m. What is the KE as it strikes the ground?ΔKE = ΔPEΔKE = mgΔhΔKE = (2 kg)(9.8 m/s2)(10 m)ΔKE = 196 J

HW #10Answer Key

1. A ball thrown vertically downward strikes a horizontal surface with a speed of 15 meters per second. It then bounces, and reaches a maximum height of 5 meters. Neglect air resistance on the ball.

a. What is the speed of the ball immediately after it rebounds from the surface?

b. What fraction of the ball's initial kinetic energy is apparently lost during the bounce?

ghv

ghv

mghmv

2

2/1

2/1

0

20

20

smv

msmv

/9.9

)5)(/8.9(2

0

20

2

22

2/1

2/12/1

i

fi

i

i

mv

mvmv

K

K

lostenergyK

K

56.0

)/15(2/1

)/9.9(2/1)/15(2/12

22

i

i

K

K

sm

smsm

K

K

Calculator

**5 minutes**

2. A 0.10-kilogram solid rubber ball is attached to the end of an 0.80-meter length of light thread. The ball is swung in a vertical circle, as shown in the diagram above. Point P, the lowest point of the circle, is 0.20 meter above the floor. The speed of the ball at the top of the circle is 6.0 meters per second, and the total energy of the ball is kept constant.

Calculator

**15 minutes**

a.Determine the total energy of the ball, using the floor as the zero point for gravitational potential energy.

ET = U + KET = mgh + ½ mv2

ET = (0.1 kg)(9.8 m/s2)(1.8 m) + ½ (0.1 kg)(6 m/s)2

ET = 3.6 J

b. Determine the speed of the ball at point P, the lowest point of the circle.

smv

msmkgJkg

v

mghEm

v

mghEmv

UEK

tpoinlowestatKUE

T

T

T

T

/2.8

)2.0)(/8.9)(1.0(6.31.0

2

2

2/1

)(

2

2

c. Determine the tension in the thread ati. the top of the circle

NT

smkgm

smkgT

mgr

mvT

r

mvmgT

r

mvF

maF

c

cc

5.3

)/8.9)(1.0(8.0

)/6)(1.0( 22

2

2

2

ii. the bottom of the circle.

NT

smkgm

smkgT

mgr

mvT

r

mvmgT

r

mvF

maF

c

cc

.5.9

)/8.9)(1.0(8.0

)/2.8)(1.0( 22

2

2

2

The ball only reaches the top of the circle once before the thread breaks when the ball is at the lowest point of the circle.

d. Determine the horizontal distance that the ball travels before hitting the floor.

y = vot + ½ at2

y = ½ gt2

t2 = 2y/gt2 = 2(0.2 m)/9.8 m/s2

t2 = 0.04 s2

t = 0.2 s

vx = x/tx = vxtx = (8.2 m/s)(0.2 s)x = 1.6 m

P

Calculator

**17 minutes**3.

ii. Calculate the value vmax of this maximum speed

smv

msmv

ghv

mghmv

UK

A

A

/42

)90)(/8.9(2

2

2/1

max

2max

max

2max

b. Calculate the speed vB of the car at point B.

smv

vkgmsmkgJ

mvmghE

JE

msmkgE

mghE

B

B

BBT

T

T

AT

/28

)700(2/1)50)(/8.9)(700(400,617

2/1

400,617

)90)(/8.9)(700(

22

2

2

Fg N

Fg = mg

Fg = (700 kg)(9.8 m/s2)

Fg = 6860 N

NN

Nm

smkgN

mgr

mvN

r

mvmgN

r

mvF

B

c

580,20

686020

)/28)(700( 2

2

2

2

Flatten out the track on either side of the loop so the bottom of the loop is on the ground, thus lowering point B. The work done by friction will reduce the total mechanical energy available at point B, so if the kinetic energy at point B is to remain the same, the potential energy at that point must be reduced.