Physics momentum and acceleration

In summary: Yes, it seems reasonable that the target would reach a velocity of 5.37 m/s in this short amount of time.
  • #1
troy132
7
0

Homework Statement


You throw a 0.42-kg target upward at 15 m/s. When it is at a heigh of 10 m above the launch position and moving downward, it is struck by a 0.338-kg arrow going 27 m/s upward. Assume the interaction is instantaneous.
a)What is the velocity of the target and arrow immediately after the collision?
b)What is the speed of the combination right before it strikes the ground?

Homework Equations


Not a homework problem. This is practice questions for an upcoming midterm and I am not sure how to go about this. An answer with an explanation would be great but if all you can provide is an answer I would be grateful.

The Attempt at a Solution


I am not sure where to begin. Conservation of momentum maybe? but even with that I am not sure how to bring gravity into it.
 
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  • #2
troy132 said:

Homework Statement


You throw a 0.42-kg target upward at 15 m/s. When it is at a heigh of 10 m above the launch position and moving downward, it is struck by a 0.338-kg arrow going 27 m/s upward. Assume the interaction is instantaneous.
a)What is the velocity of the target and arrow immediately after the collision?
b)What is the speed of the combination right before it strikes the ground?

Homework Equations


Not a homework problem. This is practice questions for an upcoming midterm and I am not sure how to go about this. An answer with an explanation would be great but if all you can provide is an answer I would be grateful.

The Attempt at a Solution


I am not sure where to begin. Conservation of momentum maybe? but even with that I am not sure how to bring gravity into it.
Even for practice questions, the rules of PF state that the poster is responsible for providing his own solutions. Once you do that, if you get stuck or want to know if you've done things correctly, other members can then offer comments and suggestions.

So, you've thrown the target upward at a certain initial velocity. It's now falling back to earth. What's the velocity of the target when it's 10 meters off the ground?
 
  • #3
For part A I need to find the velocity immediately after the instantaneous collision
 
  • #4
troy132 said:
For part A I need to find the velocity immediately after the instantaneous collision
Yes, but first you need to determine the velocity of the target, as it is falling back to earth, when it reaches 10 m above the ground.

You can't calculate the combined momentum of the target and the arrow if you don't know the velocity of the target. :sorry: :wink:
 
  • #5
Ok how in the world do I do that? What I think I need to do is find the max height it reaches, from there I am not too sure.
 
  • #6
troy132 said:
Ok how in the world do I do that? What I think I need to do is find the max height it reaches, from there I am not too sure.
Have you studied anything about objects being thrown up in the sky? Projectile motion? Gravity?
 
  • #7
Yes I have. Gravity is constant 9.81m/s downwards, the part I am struggling with is finding velocity on its way down. Am I on the right track by determining its maximum height?
 
  • #8
troy132 said:
Yes I have. Gravity is constant 9.81m/s downwards, the part I am struggling with is finding velocity on its way down. Am I on the right track by determining its maximum height?
That's a start.
 
  • #9
Ok i have calculated the max height to be 11.47 meters. Now i find the velocity its traveling when it falls the distance of 1.47 m?
 
  • #10
troy132 said:
Ok i have calculated the max height to be 11.47 meters. Now i find the velocity its traveling when it falls the distance of 1.47 m?
Yep. You want the velocity of the target when it is 10 meters off the ground.
 
  • #11
Ok I've found velocity to be 5.366m/s by vf^2=vi^2 + 2ad and then found time using v=d/t and solving as 0.27 seconds. I am not too sure what to do from this point. Does time help with anything or is it an unnecessary value
 
  • #12
Update: Just worked around a bit and i think i got the answer. Do you get 9.1m/s in your calculations?
 
  • #13
troy132 said:
Ok I've found velocity to be 5.366m/s by vf^2=vi^2 + 2ad and then found time using v=d/t and solving as 0.27 seconds. I am not too sure what to do from this point. Does time help with anything or is it an unnecessary value
I'm not following what you did here. You can't use v = d/t here because velocity is not constant.

What's the velocity of the target when it reaches the highest point above the ground?

The target falls a distance of 1.47 m from this point. Does it seem reasonable that it would reach a velocity of 5.37 m/s in this short distance?
 

FAQ: Physics momentum and acceleration

1. What is momentum in physics?

Momentum in physics is defined as the product of an object's mass and velocity. It is a measure of the object's motion and is a vector quantity, meaning it has both magnitude and direction. In simpler terms, momentum is the amount of force an object has when it is moving.

2. How is momentum calculated?

Momentum (p) can be calculated by multiplying an object's mass (m) by its velocity (v). The formula for momentum is: p = m x v. The SI unit for momentum is kilogram-meters per second (kg·m/s).

3. What is the relationship between impulse and momentum?

Impulse and momentum are closely related concepts in physics. Impulse is defined as the change in an object's momentum over time, and can be calculated by multiplying the force applied to an object by the time it is applied. This means that a larger force applied over a shorter period of time can result in the same change in momentum as a smaller force applied over a longer period of time.

4. How does acceleration affect an object's momentum?

Acceleration is the rate at which an object's velocity changes. In terms of momentum, acceleration can either increase or decrease an object's momentum depending on its direction. When an object experiences acceleration in the same direction as its velocity, its momentum increases. Conversely, when an object experiences acceleration in the opposite direction of its velocity, its momentum decreases.

5. Can momentum be conserved in a closed system?

Yes, momentum is conserved in a closed system, meaning that the total momentum of all objects in the system remains constant. This is known as the law of conservation of momentum. This law applies to all types of collisions, including elastic and inelastic collisions.

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