How Do I Solve These Physics Problems on Center of Mass and Momentum?

In summary: A stiff-legged landing means that you don't bend your knees at all. This means that the force of the ground on you is pointed upward. So, the force of the ground on you in this situation is equal to the weight of the man multiplied by the cosine of the angle of impact. That gives you a force of 9.81 Newtons. 3) The impulse applied to the golf ball by the floor is equal to the change in momentum of the golf ball (9.81 Newtons) times the duration of the impulse (0.064 seconds).
  • #1
lonelygirl
7
0
I have been missing for class 3days and i was get lost because my daughter gets sick in the past weeks :cry: :( until todate. The problems i need everyone helps below are just a sample exam. I have contact my professor about my family problems and he was nice, gave me the sample exam which it's similar to the exam this Monday. This chapter is the hardest chapter I ever seen :( Please walk me through this

1) A dump truck is being filled with sand. The sand falls straight downward from rest from a height of 2.00 m above the truck bed, and the mass of sand that hits the truck per second is 55 kg/s. The truck is parked on the platform of a weight scale. By how much does the scale reading exceed the weight of the truck and sand?

2) When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bed your knees upon landing to reduce the fore of the impact. A 75 kg man just before contact with the ground has a speed of 6.4m/s
a. In a stiff-legged landing he comes to a halt in 2 ms. Find the average net force that acts on him during this time.
b. When he bends his knees, he comes to a halt in 0.1 s. Find the average net force.
c. During the landing, the froce of the ground on the man points upward, while the force due to gravity points downward. The average net froce acting on the man included both of these roces. Taking into account the directions of teh frocesw, find the force of teh ground on the man in parts a & b

3) A golf ball strikes a hard, smooth floor at an angle of 30 degree and as the drawing shows, rebounds at the same angle. The mass of the ball is 0.047kg, and its speed is 45m/s just beforece and after stirking the floor. What's the magnitude of the impulse applied to the golf ball by the floor?

All helps are appreciated.
 
Physics news on Phys.org
  • #2
Let's take 3) first:
How is impulse related to change in momentum?
 
  • #3
arildno said:
Let's take 3) first:
How is impulse related to change in momentum?

Here is how I've done to #3

J = 2(0.047kg)(45m/s)cos30 = 3.66m/s
 
  • #4
Want me to do all of this for you, and email it to you? When do you need it by?
 
  • #5
It looks like you've got the basic idea right, however:
1) Your unit for impulse should be kg*m/s (you've written m/s in the last expression)
2) The ball struck at 30 degrees angle TO THE FLOOR.
Therefore, you must use sin(30)=1/2, rather than cos(30) (do you see why?).
 
  • #6
OptimusPrime said:
Want me to do all of this for you, and email it to you? When do you need it by?

No, she doesn't want you to do it all for her! She wants you and us to give her some hints and help so she can learn to do it herself.
 
  • #7
Awwwwww my daughter got fever again :( I'm going to die, i don't know if i can make it on Monday

I'm so frustrating now

If i still can't figure it out by tonight I'll just ask for the answer then because i don't have time on the weekend :(

Here's my email: lonelygal_2@yahoo.com

Thank you guys
 
  • #8
I was only trying to be helpful. Forget it. :bugeye:
 
  • #9
:cry: help me out please.....
 
  • #10
1) You cannot know what the force looks like for each grain of sand that hits the truck. Fortunately, you don't have to know. You just need to find the averge force over a period of time while the sand is falling. For some amount of mass of sand [tex] \Delta M [/tex]. you can find the velocity with which it hits the truck, and hence its momentum. All that momentum gets lost in the collision. The change in momentum is equal to the implulse, which is the average force applied times the duration of that force, [tex] \Delta t [/tex]. The ratio [tex]\frac{ \Delta M}{ \Delta t} [/tex] is a given of the problem.

That should give you a good start.
 

Related to How Do I Solve These Physics Problems on Center of Mass and Momentum?

1. What is the center of mass?

The center of mass is the point in an object or system where the mass is evenly distributed in all directions. It is also known as the center of gravity, as it is the point where the object's weight is concentrated.

2. How is the center of mass calculated?

The center of mass is calculated by finding the weighted average of the positions of all the particles in the object or system. This is done by multiplying each particle's mass by its distance from a chosen reference point and then dividing the sum of these products by the total mass of the object or system.

3. What is linear momentum?

Linear momentum is a measure of an object's motion. It is defined as the product of an object's mass and its velocity. In other words, it is the quantity of motion an object has in a particular direction.

4. How is linear momentum related to center of mass?

The linear momentum of a system is equal to the product of the total mass of the system and the velocity of its center of mass. This means that if the center of mass is moving, the entire system will have a linear momentum, even if some parts are stationary.

5. How can the center of mass and linear momentum be used in real life?

The concepts of center of mass and linear momentum are used in various fields, including physics, engineering, and sports. For example, in sports, understanding the center of mass and its relation to linear momentum can help athletes improve their performance and balance. In engineering, these concepts are used in designing structures and machines to ensure stability and efficiency. In physics, they are essential in understanding the motion of objects and predicting their behavior.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
3K
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Back
Top