UTF-8 U+6211 U+662F wrote:Oh whoops, haha.
Find the final speed of a 100 kg roller coaster car that has dropped 50 m (from rest), gone back up 30 m, dropped 60 m, gone back up 30 m, and then at that level been pushed for 5 seconds with a force of 50 Newtons.
2.5 m/s
No, not quite. Remember that the car has dropped and therefore gained speed.
Re: Hovercraft B/C
Posted: March 14th, 2018, 3:16 pm
by MattChina
UTF-8 U+6211 U+662F wrote:
MattChina wrote:
UTF-8 U+6211 U+662F wrote:Oh whoops, haha.
Find the final speed of a 100 kg roller coaster car that has dropped 50 m (from rest), gone back up 30 m, dropped 60 m, gone back up 30 m, and then at that level been pushed for 5 seconds with a force of 50 Newtons.
2.5 m/s
No, not quite. Remember that the car has dropped and therefore gained speed.
33.8m/s?
Re: Hovercraft B/C
Posted: March 14th, 2018, 3:18 pm
by UTF-8 U+6211 U+662F
MattChina wrote:
UTF-8 U+6211 U+662F wrote:
MattChina wrote:
2.5 m/s
No, not quite. Remember that the car has dropped and therefore gained speed.
33.8m/s?
Yep (at least that's what I remember). Your turn!
Re: Hovercraft B/C
Posted: March 14th, 2018, 3:39 pm
by MattChina
A hovercraft with a mass of 100 kg going at 4m/s has its thrust turned off abruptly. It is traveling across the ice with a friction coefficient of 0.38. How far will it travel across the ice?
Re: Hovercraft B/C
Posted: March 14th, 2018, 6:53 pm
by UTF-8 U+6211 U+662F
MattChina wrote:A hovercraft with a mass of 100 kg going at 4m/s has its thrust turned off abruptly. It is traveling across the ice with a friction coefficient of 0.38. How far will it travel across the ice?
a = 9.8 m/s^2 * 0.38 = 3.724 m/s^2
0 = (4 m/s)^2 - 2 * 3.724 m/s^2 * d
d = 2.15 m
Re: Hovercraft B/C
Posted: March 14th, 2018, 7:26 pm
by MattChina
UTF-8 U+6211 U+662F wrote:
MattChina wrote:A hovercraft with a mass of 100 kg going at 4m/s has its thrust turned off abruptly. It is traveling across the ice with a friction coefficient of 0.38. How far will it travel across the ice?
a = 9.8 m/s^2 * 0.38 = 3.724 m/s^2
0 = (4 m/s)^2 - 2 * 3.724 m/s^2 * d
d = 2.15 m
Correct. Your Turn.
Re: Hovercraft B/C
Posted: March 14th, 2018, 7:32 pm
by UTF-8 U+6211 U+662F
An elastic collision happens over a period of five seconds, between a 1 kg ball and a 5g pencil. What is the average force exerted on the pencil?
Re: Hovercraft B/C
Posted: March 14th, 2018, 9:08 pm
by Justin72835
UTF-8 U+6211 U+662F wrote:An elastic collision happens over a period of five seconds, between a 1 kg ball and a 5g pencil. What is the average force exerted on the pencil?
Do you need to know anything about initial and final velocities to solve this question?
Re: Hovercraft B/C
Posted: March 15th, 2018, 1:45 pm
by UTF-8 U+6211 U+662F
Justin72835 wrote:
UTF-8 U+6211 U+662F wrote:An elastic collision happens over a period of five seconds, between a 1 kg ball and a 5g pencil. What is the average force exerted on the pencil?
Do you need to know anything about initial and final velocities to solve this question?
Whoops: They are moving toward each other, the ball with a speed of 5 m/s and the pencil with a speed of 0.4 m/s.
Re: Hovercraft B/C
Posted: March 15th, 2018, 2:18 pm
by Justin72835
UTF-8 U+6211 U+662F wrote:
Justin72835 wrote:
UTF-8 U+6211 U+662F wrote:An elastic collision happens over a period of five seconds, between a 1 kg ball and a 5g pencil. What is the average force exerted on the pencil?
Do you need to know anything about initial and final velocities to solve this question?
Whoops: They are moving toward each other, the ball with a speed of 5 m/s and the pencil with a speed of 0.4 m/s.
Alright, first find the final velocities of both of the objects.
Plugging in the given values and solving for the final velocities gives 4.946 m/s for the ball and 10.346 m/s for the pencil. Now you can find the average force on the pencil:
This gives an average force of 0.0107 N. You can check this answer by doing the same calculation on the ball (Newton's Third Law).