Samson


 * Week 3**

Part A: We know that A+B is 6i + 1j and A-B is -4i + 7j. I solved them separately for x components of A&B and y components of A&B and i got 1i+4j for A and 5i-3j for B. how did you do that ? -- note from the professor

Part B: we know the formula to find the magnitude which is |A| is (Ax^2+Ay^2)^(1/2) and in the way for B

Part C: to find the angle first find the dot product of A.B = AxBy+AyBx ( not correct -- has to be AxBx + AyBy ) and then use the other formula to get the angle A.B = |A||B| cos (theta) and then solve for theta.

Professor's Note : very Good effort !

Week 4

we know that initial velocity is 10.5 and the horizontal angle is 54 and the distance between the base of the tower and the place where the object landed is 15.3 so we can get the x component and y component of the velocity by Vox = Vocos(theta) and Voy = Vosin(theta) as we know the x and y components we can get the time by substituting the known values in x component equation deltax = Voxt + .5Ax(T^2) where A = 0 and now as we know the time we can use the y component equation deltay = Voyt + .5Ay(T^2) and solve for the distance delta y we get the delta y.

**Professor's Note :** Good effort !

Week 5 A) given delta x = 8m (given) delta y = 5.4m Vox = ? Voy = 0 Ax = 0 Ay = 9.8 we substitute known values in the equation (deltaY) = Voy * t + 0.5 * (9.81) * t^2 and solving for t we get "time" we know Velocity = distance / time we know time and distance we get Velocity because the fish is travelling in x direction initial and final velocities are the same B) we know initial velocity 7.6 from part a delta y = 2.7 using the equation (deltaY) = Voy * t + 0.5 * (9.81) * t^2 solving this we get time we get time we need distance along x direction so we know velocity and time using distance = velocity * time we get distance

the object has two forces Fn which acts upward and mg which acts down he gave us that when the cab is stationary the F = 65N because the cab is stationary f = mg we get a mass of 6.625 for part a: Fnet = ma We get Fn sin(90) + mg sin(270) = ma because acceleration = 0 Fn = mg Fn = 6.625 * 9.81 Fn = 65 for part b: Fn sin(90) + mg sin(270) = ma we get Fn - mg = m(a) Fn = 6.625*9.81 - 6.625*(-2.24) - - wrong ! ==> m (g+a) and a is negative as it is a deceleration ! we get Fn
 * Week 07**

Professor's Note : Sam, first part is correct, but in the second part, you are using the negative twice, and it kills the purpose of using negative in the first place !

Week 9

1) A cord is used to vertically lower an initially stationary block of mass M at a constant downward acceleration of g/4. When the block has fallen a distance d i) Find the work done by the cord's force on the block ii) Find the work done by the gravitational force on the block iii) Find the kinetic energy of the block iv) Find the speed of the block

part i Work done W = F*d*cos(theta) we know the distance is 'd' F = ma along the y direction T sin(90) + mg sin(270) = ma T - mg = ma  T =mg + ma  T = mg + m(g/4)

part ii W = F*d cos(theta) work done along the gravitational force therefore F = mg and theta = 0 W = mgd cos(0) W = mgd

part iv V^2 = Vo^2 +2a(s) V^2 = 0 +2(g/4)(d) V^2 = 2(g/4)(d) V = (2(g/4)(d))^(1/2)

part iii KE = (mv^2) / 2 KE = (m*(2(g/4)(d))) / 2

week 10

first we find the velocity of the ball just before it hits the block. we use the law of conservation of energy ½m1v12+ m1gh = ½m2v22+ m2gh Here initial velocity of the ball is zero the height of the ball just before it hits the block is zero now we solve for final velocity v2 we use the final velocity we got from above as the initial velocity of the when we solve for the final velocity of the ball and the block. We have two unknowns to find therefore we use to equations to find out the velocities Now we use law of conservation of energy during a elastic collision ½m1v12 + ½m2v22 = ½m1v1f2 + ½m2v22 Here the initial velocity of the block is zero. We make the final velocity of the ball as our subject and equate the rest of the formula to it Now we use the law of conservation of momentum M1v1 + m2 v2 = m1fv1f +m2fv2f Here we substitute the final velocity of the ball and solve for the final velocity of the block after collision And then we use the final velocity of the block and find the find the final velocity of the ball after collision