Impulse and momentum homework help

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SUMMARY

The discussion focuses on solving a physics problem involving impulse and momentum for two objects on a frictionless surface, where m1 > m2. The correct answers for the first scenario, where a constant force is applied to object 1, are (c) p1 > p2 and (e) K1 = K2. In the second scenario, after the same force is applied to object 2 for a time interval delta t, the correct answers are (b) p1 = p2 and (d) K1 < K2. The participants emphasize the importance of understanding the impulse-momentum law and the work-energy principle as alternative methods to analyze the problem.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Familiarity with the concepts of momentum (p = mv) and kinetic energy (K = mv² / 2)
  • Basic knowledge of impulse (I = integral of F * dt)
  • Ability to apply kinematic equations (v = at, v² = 2ad)
NEXT STEPS
  • Study the Impulse-Momentum Theorem in detail
  • Explore the relationship between force, acceleration, and time in physics problems
  • Practice solving problems using the work-energy principle
  • Learn how to apply kinematic equations in various scenarios
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Students studying physics, educators teaching mechanics, and anyone seeking to deepen their understanding of impulse and momentum concepts.

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Homework Statement



Two objects are at rest on a frictionless surface. m1 > m2
(1) When a constant force is applied to object 1, it accelerates through a distance d in a straight line. The force is removed from object 1 and is applied to object 2. At the moment when object 2 has accelerated through the same distance d, which statements are true?

(a) p1 < p2 (b) p1 = p2 (c) p1 > p2 (d) K1 < K2 (e) K1 = K2 (f) K1 > K2

(2) When a force is applied to object 1, it accelerates for a time interval delta t. the force is removed from object 1 and is applied to object 2. From the same list of choices, which statements are true after object 2 has accelerated for the same time interval delta t?

Homework Equations



delta p = I
I = integral of Ti to Tf (F * dt)
dp = F * dt
p = mv

The Attempt at a Solution

The answers for 1 is c and e ; for 2 is b and d

First, constant force means constant acceleration. F = ma, and F = d(mv) / dt = dp / dt
Since m1 > m2, this is why c is true for 1.

K = mv^2 / 2. If we have the same acceleration, then how do we determine which one has a greater velocity? The book cited W = Fd which is great, but I can be dumb and not remember this fact. So let's discuss the velocity for now.

For the second problem, the book use impluse and state their impluse are the same, so p1 = p2. Let's go back to our first problem.

How do we find the impluse for our first problem? Can't the impluse be the same even when the delta time is not the same?

For second problem:
So F is not constant, a is not constant either. again, how do we know the velocity?

I have a hard time applying things.

Thanks.
 
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Theer are a number of different ways to solve problems in physics, and after awhile, it does get confusing at first. But all roads lead to the same place. The Impulse-momentum law is really just F=ma. And work-energy methods are an alternative approach to Newton's laws and the kinematic motion equations. Let's lose the calculus here, and just concentrate on F=ma and v=at and v^2=2ad. That is one way to do it when in doubt. I find it helpful to put in values for m1 and m2 (like m1=2 units and m2=1 unit) and then crank our the momentum and KE's for each case, and see what happens.
 

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