<p>It looks like you’ll have to solve for the final velocity first (only in the vertical direction), and then use it with a kinematic to solve for the time it takes for the object to hit the ground. Then plug it into r=d/t.</p>
<p>1998 #6) A wheel of mass M and radius R rolls on a level surface without slipping. If the angular velocity of the wheel is w, what is its linear momentum?
a) MwR
b) RM(w^2)
c) Mw(R^2)
d) M(w^2)(R^2)/2
e) 0</p>
<p>1998 #33) A wheel with rotational Inertia I is mounted on a fixed, frictionless axle. The angular speed w of the wheel is increase from 0 to w(f) in a time interval T. What is the average power input to the wheel during this time interval?
a) Iw(f)/2T
b) Iw(f)^2/2T
c) Iw(f)^2/2T^2
d) (I^2)w(f)/2T^2
e) (I^2)w(f)^2/2T^2</p>
<p>1998 6) Linear momentum is mv. We know that v=wR, so the linear momentum is M(wR), or answer choice a.</p>
<p>1998 33) So the average power is equal to (kinetic energy)/(change in time), and we know the change in time is T. Since kinetic energy=(1/2)(I)(w^2), the average power is equal to (1/2)(I)(w(f)^2)/T, or answer choice b.</p>
<ol>
<li>Two people are in a boat that is capable of a maximum speed of 5 kilometers per hour in still water, and wish to cross a river 1 kilometer wide to a point directly across from their starting point. If the speed of the water in the river is 5 kilometers per hour, how much time is required for the crossing?</li>
</ol>
<p>The answer is 10 hours. I thought it was like this:
You have your own boat speed and the river, so the total speed is 10 km/r.
d=st, t= 0.1 hours?</p>
<hr>
<ol>
<li>A projectile of mass M1 is fired horizontally from a spring gun that is initially at rest on a frictionless surface. The combined mass of the gun and projectile is M2. If the kinetic energy of the projectile after firing is K, the gun will recoil with a kinetic energy equal to:</li>
</ol>
<p>Not sure how to approach this problem besides that Kt = Kprojectile + Kgun.</p>
<hr>
<ol>
<li> Two artificial satellites, 1 and 2, orbit the Earth in circular orbits having radii R1 and R2, respec¬tively, as shown above. If R2 = 2R1, the accelerations a2 and a1 of the two satellites are related by which of the following?<br>
(A) a2 = 4a1 (B) a2 = 2a1 (C) a2 = a1 (D) a2 = a1/2 (E) a2 = a1/4</li>
</ol>
<p>The graph is not necessary.</p>
<hr>
<p>23 A bowling ball of mass M and radius R. whose moment of inertia about its center is (2/5)MR2, rolls without slipping along a level surface at speed v. The maximum vertical height to which it can roll if it ascends an incline is?</p>
<p>Actually it’s E for #6, also, think about it logically, if they want to reach the point directly across, they have to “nullify” the current of the river, the only way they can do this is by tilting their boat so it is in directly in opposition to the current, thus leaving no vertical component to their motion, leaving them in the same place. So they cannot reach the point directly across the river under these conditions.</p>
<p>And for 22: m<em>e is mass of the earth, m</em>s is mass of the satellite, I assume in this problem that the masses are equal, and a<em>c is centripetal acceleration, finally 1 and 2 indicate which satellites
Set gravitational force equal to centripetal force
For satellite 1: G(m</em>e)(m<em>s1)/(R1)^2=(m</em>s)(a<em>c1)
For 2: G(m</em>e)(m<em>s2)/(R2)^2=(m</em>s)(a<em>c2)
now isolate acceleration by dividing each side by m</em>s
which yields G(m<em>e)/(R1)^2=(a</em>c1) for satellite 1 and G(m<em>e)(R2)^2=(a</em>c2) for 2
substitute in 2R1=R2 in the second equation and divide equation 1 by equation 2, so you get
(1/(R1^2))/(1/(2R1^2))=(a<em>1)/(a</em>2)
rearranging gives 4=(a<em>1)/(a</em>2), thus 1/4a<em>1=a</em>2 so E</p>
<p>And for 23 we’re going to use conservation of energy, so U=K<em>rotational+K</em>linear, K rotational is 1/2Iw^2, Rw=v and I=2/5MR^2 so we can rewrite that as 1/2(2/5MR^2)(v/R)^2 which is 1/5Mv^2, substituting this into the original equation, we get mgh=1/2mv^2+1/5mv^2=> (cancelling m and combining like terms) gh=7/10v^2, so max height is h=7/10(v^2/g)</p>
<p>Does anyone know how close the 98/93/04/88 tests are to the modern test? I just got 33/35 on 98 mechanics and I’m really pumped :D, but I’m guessing it is more difficult now.</p>
<p>^^I’m not sure, but I don’t believe the tests can significantly change in difficulty over the years. Also, if you got a 33/35 on the MC, then all you would need is 8 more pts out of 45 pts on the FRQ for a 5.</p>
Yeah, E is right for that #6</p>![http://i.imgur.com/vqdYe.jpg[/url]](http://i.imgur.com/vqdYe.jpg%5B/url%5D){kind=link}
