A fast response earns you another question! :-)
1. A geocacher on the hunt for a First To Find moves along a straight line of bushes, such that their velocity v m/s, given in terms of time t seconds, is v=(1/2)t^2 – 3t +4. A= the total distance traveled between t=0 and t=4, in metres.
2. Two particles, G and H, of masses 2m and m, are on a smooth horizontal table. H is at rest and G collides with it when it is moving with speed u. After the collision, the total kinetic energy of the two particles is (3/4) m*u^2 (where * means multiply). The coefficient of restitution between the two particles is 1/B. Find B.
3. A hollow right circular cone, with base of radius r and height h, is fixed with its axis vertical and vertex downwards. A particle moves with constant speed v in a horizontal circle of radius (1/3)*r on the smooth inner surface of the cone. V=sqrt(h*g/C), where sqrt means square root. Find C.
4. A particle of mass 4kg is moving along the horizontal x-axis under the action of a single force which acts in the positive x-direction. At time t seconds, the force has magnitude (1+3t^(1/2))N. When t=0, the particle has speed 2m/s in the positive x-direction. D=the work done by the force in the interval from t=0 to t=4, written to one significant figure. Find D.
5. A light elastic string has natural length 10m and modulus of elasticity 130N. The ends of the string are attached to fixed points P and Q, which are at the same horizontal level and 12.6m apart. An object is attached to the midpoint of the string and hangs in equilibrium at a point 1.6m below PQ. Find E, the weight of the object, to 2 significant figures.
These are mainly physics problems at this point -- but two of them only require basic calculus, which I can do right away. I am sorry for the delay, due to air-traveling, so I will answer these two and follow-up when I have time to look in a physics reference for the other three.
#1 just asks you to find the distance traveled. Since velocity is the rate of change of position, the position is the integral of velocity. Note that since you want "total distance" you look at speed (absolute value of velocity).
If you have f(t) = (1/2)t^2 - 3 t + 4, then F(t), the integral, is (1/6)t^3 - (3/2)t^2 + 4t + C. The +C is irrelevant for our purposes because we are doing a definite integral. Notice, however, that from t=0 to t=2, f(t)>0, but for t=2 to t=4, f(t)<0.
Change in distance from t=0 to t=2 is F(2)-F(0) = 10/3.
Change in distance from t=2 to t=4 is F(2)-F(4) = 2/3.
Note: It's reversed (not F(4)-F(2)) because you want it to be positive (as noted above).
The answer is the sum, 12/3 = 4.
#4 Work is defined as force * distance, so to get the total work done, you need to integrate F(x)dx. However, you don't have F in terms of x (the x coordinate), you have it in terms of the time t. In that case, you really just use a change-of-variables to obtain F(t)v(t)dt.
You also aren't given v(t), but you have the mass and the force, which means you have the acceleration:
a(t) = (1+3√(t))/4
From this, you obtain v(t) = (t+√(t^3))/4 + C, and C=2 based on v(0)=2.
Finally, you can compute:
F(t)v(t) = (1+3√(t)) [(t+√(t^3))/4 + 2 ]
and integrate this (I'll spare the gory details) to get 90.
: Here are the rest of the problems.
#2 You need two equations: One for kinetic energy, one for momentum.
Let's start with some notation, let's say CR is the coefficient you want, vG is the speed of G after the collision and vH for H. Then by definition, you have:
CR = ( vH - vG ) / ( u - 0 )
And if you could find vG and vH you'd get it. Momentum is conserved, so:
2 m u = 2 m vG + m vH
And kinetic energy can be computed (and set equal to the given quantity):
(3/4) m u^2 = m vG^2 + (1/2) m vH^2
You can solve this system of two equations to get two solutions:
vG = (1/2)u, vH = u
vG = (5/6)u, vH = (1/3)u
The second one is not possible. If G collides with H and they continue, they'll be headed in the same direction. But H is in front of G, so it has to have the larger momentum.
So CR = ( vH - vG ) / ( u - 0 ) = ( u - u/2 ) / u = 1/2.
You actually want B = 1/CR, so B=2.
#3 This example is solved many places, e.g. here
. It's just using the definition of centripital force. You will get C=3 following a formula of this sort.
#5 You can draw a diagram (shaped a bit like the letter M) and notice that the wire will pull the mass in two different directions (up and to the left, up and to the right). The pull it gives will be k(L-5), where L is the length of the wire from the top of the pole to wherever the mass is. The natural length of the wire is 10, so half of it is "trying" to be length 5. The "k" is the "spring constant" of this wire, which can be computed using the given information as 130/10 = 13.
We know L = √(1.6^2+6.3^2)=6.5
However, the two pulls from the elastic cancel each other out left-to-right and only wind up pulling upwards with force 2k(L-5)cos(θ), where θ is the angle at which the wire is bent downwards. No need to worry, we have the sides of that triangle (12.6/2 = 6.3 and 1.6, which gives cos(θ)).
From there you just plug-and-chug to get:
w = 2k(L-5) cos(θ) = 2(13)(6.5-5)(6.3/6.5) = 37.8.