<p>Hi. I'd appreciate if someone could help me with these AP Physics C problems involving momentum. I've gotten some help with it, but I just can't seem to understand it and I have a test tommorow :-(:</p>
<p>1) A 25 kg child is in a 10 kg sled that travels 1 m/s east on ice. The child throws a 3 kg snowball at 40 m/s at an angle of 50 degrees north of east. Find the velocity of the sled after the snowball leaves the childs hand.</p>
<p>2) Cube B is at rest on the edge of a frictionless horizontal table that is 5 m above the ground. It has a mass of 5 kg. Cube A, with a mass of 20 kg travels towards it at 20 m/s. The two cubes collide and the force that cube A exerts on B during the collision is given by F(t)=3x10^6t-2x10^8t^2 i (N). The collision lasts 10 milliseconds.</p>
<p>a) Find the velocity of cube B immediately after the collision
b) Find the velocity of cube A immediately after the collision
c) Find the coefficient of restitution for the collision
d) Find the speed with which each cube hits the ground</p>
<p>I really would appreciate any help I can get. Thanks!</p>
<p>"sled travels 2.84 m/s at 67 degrees south of west"</p>
<p>Wait... wouldn't that imply that the sled is moving downwards intothe ice? I'm having trouble with that part, could you explain more on why it would move in a downward direction as well as to the west?</p>
<p>Tamirms:
If you're looking at movement on a 2-dimensional surface, why in the world would moving south imply that you had to go into the ice?</p>
<p>Think of it as movement on a flat map, or moving on the surface of the earth. It's like you were in Chicago, and began driving towards western Texas.</p>
<p>Yea, based on the law of conservation of momentum, because originally there was no momentum in the y-direction (the sled was traveling due east) in order for there to be conservation the momentum of the sled sould have to be equal and opposite to the momentum of the snow ball. Since the snowball is traveling north, the sled must travel south.</p>
<p>Well, given that the OP mentioned that
"the child throws a 3 kg snowball at 40 m/s at an angle of 50 degrees north of east."
I assume that he's throwing the snowball in both the x and y direction, and that north signifies an upward direction, as in towards the sky. However, the way in which you use south when you say:
"...why in the world would moving south imply that you had to go into the ice? Think of it as movement on a flat map, or moving on the surface of the earth. It's like you were in Chicago, and began driving towards western Texas."<br>
I think you take the north-south axis to be what I consider the z axis.</p>
<p>40m/s(sin50)3kg
^
^ >>40m/s(cos50)3kg
CART -->1m/s * 35kg There is eastward motion and northward motion
Now that the snowball moves where the upward and right arrows go, the sled needs to move south in some magnitude to counter the 40m/s(sin50)3kg (snowball y movement) and then either slow down or reverse direction, to counter snowball x movement</p>
<p>the cart then does this velocity wise. To get momentum, <em>35kg
cos(67)</em>2.87m/s <<Cart
/
sin(67)*2.87m/s
So it moves downward and backwards.</p>
<p>J (Impluse)=integral(F(t)) from 0 to .010
Since J=Favg*deltaT, you can solve for velocity that way using an equation; i have to eat dinner now; I'll post it in a bit :-P. The answer I got, though, is like 16.7 m/s.</p>
<p>a) Find the velocity of cube B immediately after the collision
b) Find the velocity of cube A immediately after the collision</p>
<p>You were right for part a.
for part a and b take the integral of F from 0 to .010, that'll be the change in momentum cube A so then you can find velocity for cube A.
To find velocity for cube B set the change in momentum of cube B equal to the negative value of the change in momentum of A (ie -1 * integral of F from 0 to .010).</p>
<p>c) Find the coefficient of restitution for the collision
don't know how to do that, haven't covered it in Ap physics B</p>
<p>d) Find the speed with which each cube hits the ground</p>
<p>So you found the horizontal component velocity in parts a and b. To find the vertical components set PE = KE, since all the potential energy will be used up in KE right before the cube hit's the ground.
mgh = .5mv^2
sqrt(2gh) = v
then just plug in the given height of 5 m and 9.81.
now that you have the vertical component use pythagoras:
v^2 = vy^2 + vx^2
v = sqrt(vy^2 + vx^2)</p>
<p>I looked up coefficient of restitution online, turned out to be pretty simple.
so for part c:
c) Find the coefficient of restitution for the collision</p>
<p>coefficient =(velocity of part a - velocity of part b)/(0-20)</p>