Question

DATA AND RESULTS

Mass of Ball: .059 kg

Trial	Initial Height (yi)	Final return height (yr)	Time to ground	Time to return
1	4 m	2.2 m	.882 s	.644 s
2	3.5 m	2 m	.830	.615 s
3	3 m	1.7 m	.772 s	.561 s

1. The potential energy of the object at its highest point _2.32_Trial #1

2. The kinetic energy of the object just before impact _2.32 _

3. The velocity of the object just before impact, using kinetic energy _8.87 _

4. The “kinetic energy” of the object just after impact _33.80_ (Hint: neglect air resistance and think about the height it rebounds to)

5. The “rebound” velocity of the object _10.62_

6. The loss in energy _0 + 2.32 = 33.80 + 0 + loss_

Trial #2

1. The potential energy of the object at its highest point _2.03_

2. The kinetic energy of the object just before impact _2.03_

3. The velocity of the object just before impact, using kinetic energy _8.30 _

4. The “kinetic energy” of the object just after impact _-32.97_

5. The “rebound” velocity of the object __

6. The loss in energy __

Trial #3

1. The potential energy of the object at its highest point _1.74_

2. The kinetic energy of the object just before impact _1.74_

3. The velocity of the object just before impact, using kinetic energy _7.68 _

4. The “kinetic energy” of the object just after impact _-1.28_

5. The “rebound” velocity of the object _6.59_

6. The loss in energy __

Can someone please help me on this to double check I am doing this correctly? Thank you!

Answer 1

Trial #1

1.We know the potential energy of the object at its highest point =mgh

given mass=0.059kg;maximum height=4m

=>mgh=0.059*9.8*4=2.3128

2.We know the kinetic energy of the object just before impact is equals to the potential energy at its heighst point due to conservation of energy

SO,kinetic energy of the object just before impact=2.3128

3. The velocity of the object just before impact, using kinetic energy=(2gh)^0.5

  =(2*9.81*4)^0.5 =8.85

4. The “kinetic energy” of the object just after impact =The potential energy of the object when it reaches its first maximum height after impact

So, here first maximum height after impact=Final return height h1=2.2m

The potential energy of the object when it reaches its first maximum height after impact=mgh1=(0.059*9.81*2.2)=1.27

hence the “kinetic energy” of the object just after impact=1.27

5. The “rebound” velocity of the object =(2gh1)^0.5

=(2*9.81*2.2)^0.5

^=6.5699

6. The loss in energy =KE before impact -KE after impact=2.31-1.27=1.04

Trial #2

1.We know the potential energy of the object at its highest point =mgh

given mass=0.059kg;maximum height=3.5m

=>mgh=0.059*9.8*3.5=2.02

2.We know the kinetic energy of the object just before impact is equals to the potential energy at its heighst point due to conservation of energy

SO,kinetic energy of the object just before impact=2.02

3. The velocity of the object just before impact, using kinetic energy=(2gh)^0.5

  =(2*9.81*3.5)^0.5 =8.28

4. The “kinetic energy” of the object just after impact =The potential energy of the object when it reaches its first maximum height after impact

So, here first maximum height after impact=Final return height h1=2m

The potential energy of the object when it reaches its first maximum height after impact=mgh1=(0.059*9.81*2)=1.15

hence the “kinetic energy” of the object just after impact=1.15

5. The “rebound” velocity of the object =(2gh1)^0.5

=(2*9.81*2)^0.5

^=6.26

6. The loss in energy =KE before impact -KE after impact=2.02-1.15=0.87

Trial #3

1.We know the potential energy of the object at its highest point =mgh

given mass=0.059kg;maximum height=3m

=>mgh=0.059*9.8*3=1.73

2.We know the kinetic energy of the object just before impact is equals to the potential energy at its heighst point due to conservation of energy

SO,kinetic energy of the object just before impact=1.73

3. The velocity of the object just before impact, using kinetic energy=(2gh)^0.5

  =(2*9.81*3)^0.5 =7.67

4. The “kinetic energy” of the object just after impact =The potential energy of the object when it reaches its first maximum height after impact

So, here first maximum height after impact=Final return height h1=1.7m

The potential energy of the object when it reaches its first maximum height after impact=mgh1=(0.059*9.81*1.7)=0.98

hence the “kinetic energy” of the object just after impact=0.98

5. The “rebound” velocity of the object =(2gh1)^0.5

=(2*9.81*1.7)^0.5

^=5.77

6. The loss in energy =KE before impact -KE after impact=1.73-0.98=0.75