# Impulse and momentum homework help

• jwxie

## 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.

Theer are a number of differnt 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.