Hovercraft B/C

[Zioly]

So close (Click to Show Hidden Text)

The units of impulse are N*s.

Sorry about that, but wouldn't it be N/s? Also, what unit is momentum in? I had a question on a test the other day and was uncertain... as you can tell, impulse and momentum are concepts I'm not as familiar with.

No, it's N*s. As you said, the unit of momentum is kg*m/s. Since Impulse is the change in momentum, it will have the same unit. A Newton is kg*m/s^2. Therefore, the unit for momentum is the Newton times a unit of time.

Oooohh, I didn't realize that a Newton is how much a kilo can be accelerated! Thanks for the tip!

[HandsFreeCookieDunk]

Sorry about that, but wouldn't it be N/s? Also, what unit is momentum in? I had a question on a test the other day and was uncertain... as you can tell, impulse and momentum are concepts I'm not as familiar with.

No, it's N*s. As you said, the unit of momentum is kg*m/s. Since Impulse is the change in momentum, it will have the same unit. A Newton is kg*m/s^2. Therefore, the unit for momentum is the Newton times a unit of time.

Oooohh, I didn't realize that a Newton is how much a kilo can be accelerated! Thanks for the tip!

Yup, right from F=ma

[Zioly]

Sorry, I thought that since I got it wrong, someone else could have a go at the right answer and put up the next question.

You drop a ball from a 26 meter tall building on earth. Find the velocity when the ball hit the ground. Assume no air resistance and gravity as 9.81. Show all work.

[RestingDoll]

Sorry, I thought that since I got it wrong, someone else could have a go at the right answer and put up the next question.

You drop a ball from a 26 meter tall building on earth. Find the velocity when the ball hit the ground. Assume no air resistance and gravity as 9.81. Show all work.

v^2=vi^2+2a(delta_y)
v^2=0-2g(26)
v^2=510.12
v=22.5858 m --> v~23 m

[RestingDoll]

Sorry, I thought that since I got it wrong, someone else could have a go at the right answer and put up the next question.

You drop a ball from a 26 meter tall building on earth. Find the velocity when the ball hit the ground. Assume no air resistance and gravity as 9.81. Show all work.

Wait lol how do I hide it. Edit: nvm, figured it out

[Zioly]

Sorry, I thought that since I got it wrong, someone else could have a go at the right answer and put up the next question.

You drop a ball from a 26 meter tall building on earth. Find the velocity when the ball hit the ground. Assume no air resistance and gravity as 9.81. Show all work.

v^2=vi^2+2a(delta_y)
v^2=0-2g(26)
v^2=510.12
v=22.5858 m --> v~23 m

All correct, except for that one little misclick. (You typed "-" instead of "+" in line 2. Your go!)

[RestingDoll]

A small 60 g point mass starts sliding down a frictionless surface at 30 cm in height, then sticks to the end of a uniform rod of mass 120 g and length 40 cm. The rod rotates about a point "O" lengthwise and its moment of inertia about that point is given by 1/3 mr^2. After the point mass sticks to the rod, it pivots to an angle theta, before momentarily stopping. Find theta. (Hint: Don't forget that the potential energy stored after it pivots is based on the center of mass of the entire system.)

[Tom_MS]

A small 60 g point mass starts sliding down a frictionless surface at 30 cm in height, then sticks to the end of a uniform rod of mass 120 g and length 40 cm. The rod rotates about a point "O" lengthwise and its moment of inertia about that point is given by 1/3 mr^2. After the point mass sticks to the rod, it pivots to an angle theta, before momentarily stopping. Find theta. (Hint: Don't forget that the potential energy stored after it pivots is based on the center of mass of the entire system.)

I'm getting (Click to Show Hidden Text)

75.5 degrees. If I'm correct, the 1/3mr^2 only matters to determine where the center of mass is.

[RestingDoll]

A small 60 g point mass starts sliding down a frictionless surface at 30 cm in height, then sticks to the end of a uniform rod of mass 120 g and length 40 cm. The rod rotates about a point "O" lengthwise and its moment of inertia about that point is given by 1/3 mr^2. After the point mass sticks to the rod, it pivots to an angle theta, before momentarily stopping. Find theta. (Hint: Don't forget that the potential energy stored after it pivots is based on the center of mass of the entire system.)

I'm getting (Click to Show Hidden Text)

75.5 degrees. If I'm correct, the 1/3mr^2 only matters to determine where the center of mass is.

Interesting... I think you may be forgetting that the center of mass of the system is different than the center of mass of the rod.