PHYSICS 191/193 REVIEW QUESTIONS TEST#1

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1 PHYSICS 9/93 REVIEW QUESTIONS TEST#. The brakes on your car can slow you at at rate of 5. m/s. If you are going 37 km/h and suddenly see a radar trap, what is the minimum time in which you can get your car under the 9 km/h speed limit? (The answer reveals the futility of braking to keep your high speed from being detected with a radar or laser gun.). In the figure below, a red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis. At time t =, the red car is at x red = and the green car is at x green = m. If the red car has a constant velocity of km/h, the cars pass each other at x = 44.5 m; if the red car has a constant velocity of 4 km/h, they pass each other at x = 76.6 m. What are: (a) the initial velocity and (b) the acceleration of the green car? 3. A hot-air balloon is ascending at the rate of m/s and is 8 m above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground? 4. In the sum A+ B = C, A has a magnitude of. m and is angled 4. o counterclockwise from the + x direction, and C has a magnitude of 5. m and is angled. o counterclockwise from the x direction. What are: (a) the magnitude; and (b) the angle (relative to +x) of B? 5. Two vectors are given by A = 3.î + 5. ĵ and B =.î + 4. ĵ. Find: (a) A B ; (b) A B A+ B i B; and, (d) the component of A along the direction of B. i ; (c) ( )

2 6. An ion s position vector is initially r = 5.î 6. ĵ +. ˆk, and sec later it is r = -.î + 8. ĵ. ˆk, all in meters. In unit-vector notation, what is its v during the seconds? 7. In the figure below, a stone is projected at a cliff of height h with an initial speed of 4. m/s directed at angle θ = 6. o above the horizontal. The stone strikes at point A, 5.5 sec after launching. Find: (a) the height h on the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground. 8. A rifle that shoots bullets at 46 m/s is to be aimed at a target 45.7 m away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

3 PHYSICS 9/93 TEST # Review Questions. We choose the positive direction to be that of the initial velocity of the car (implying that a < since it is slowing down). We assume the acceleration is constant and use the five fundamental equations. (a) Substituting v = 37 km/h = 38. m/s, v = 9 km/h = 5 m/s, and a = 5. m/s into v f = v + at, we obtain 5 m/s 38 m/s t = = 5. s. 5. m/s (b) We take the car to be at x = when the brakes are applied (at time t = ). Thus, the coordinate of the car as a function of time is given by (Equation [4]) in SI units. bg b g x = 38 t + 5. t. Let the origin of the x coordinate system correspond to the initial position at (t = ) of the RED car (x R =). The initial position of the GREEN car is x Go = m. For the first case, let v of the red car be the km/h =5.56 m/s velocity (corresponding to a passing point of x = 44.5 m) and let v be the 4 km/h =. m/s velocity (corresponding to a passing point of x = 76.6 m) of the red car. We have two equations that apply to the GREEN car when it reaches points x and x (based on Equation [4]): x = x Go + v Go t + a t where t = x v = 8. sec x = x Go + v Go t + a t where t = x v = 6.9 sec Here a is the acceleration of the GREEN car. We simultaneously solve these equations and obtain the following results: (a) v Go = -3.9 m/s (or roughly 5 km/h along the x direction). (b) a = -. m/s (acceleration is in the x direction).

4 3. We neglect air resistance, which justifies setting a = g = 9.8 m/s (taking up as the +y direction) for the duration of the motion. We are placing the coordinate origin on the ground. We note that the initial velocity of the package is the same as the velocity of the balloon, v = + m/s and that its initial coordinate is y = +8 m. (a) We solve y = y + vt gt for time, with y =, using the quadratic formula (choosing the positive root to yield a positive value for t). (b) Using Equation () leads to Its final speed is 4 m/s. b gb g v + v + gy t = = g 98. v f = v gt = (9.8)(5.4) = 4 m/s. = 54. s 4. The angle between C and the +x axis is 8 +. =. (a) The x component of B is given by C x A x = 5. cos. cos 4 = 3.3 m, and the y component of B is given by C y A y = 5. sin. sin 4 =.8 m. Consequently, its magnitude is ( 3.3) + (.8) = 6.6 m. (b) The two possibilities presented by a simple calculation for the angle between B and the +x axis are tan [(.8)/( 3.3)] = 8.9, and = 9. We choose the latter possibility as the correct one since it indicates that B is in the third quadrant (indicated by the signs of its components). We note, too, that the answer can be equivalently stated as We apply the formulae for vector and scalar products. (a) A B = ( A ) kˆ xby AyBx since all other terms vanish, due to the fact that neither A nor B have any z components. Consequently, we obtain [(3.)(4.) (5.)(.)]kˆ=.kˆ. (b) A B = Ax Bx + AyBy yields (3.)(.) + (5.)(4.) = 6. (c) A+ B= (3. +.) ˆi + ( ) ˆj ( A + B) B = (5.) (.) + (9.) (4.) = 46.

5 (d) Several approaches are available. In this solution, we will construct a ˆB unit-vector and dot it (take the scalar product of it) with A. In this case, we make the desired unitvector by We therefore obtain ˆ ˆ ˆ B. i + 4. j B = =. B (.) + (4.) AB ˆ (3.)(.) + (5.)(4.) = A B = = 5.8. (.) + (4.) 6. Using the definition of average velocity as the change in the position vector divided by the change in time, we have v (.i ˆ+ 8.j ˆ.k) ˆ (5.i ˆ 6.j ˆ +.k) ˆ = = (.7i ˆ+.4j ˆ.4k) ˆ m/s. 7. (a) Using the standard coordinate system (origin at initial point), we solve for y = h: h= v sinθt gt which yields h = 5.8 m for v = 4. m/s, θ = 6. and t = 5.5 s. (b) The horizontal motion is steady, so v x = v x = v cos θ, but the vertical component of velocity varies according to Equation () for the y-component. Thus the speed at impact is ( θ) ( θ ) v= v cos + v sin gt = 7.4 m/s. (c) We use Equation (5) with v y = and y = H. Thus ( v θ ) sin H = = 67.5 m. g

6 8. The coordinate origin is at the end of the rifle (the initial point for the bullet as it begins projectile motion), and we let θ be the firing angle. If the target is a distance d away, then its coordinates are x = d, y =. The projectile motion equations lead to d = v t cos θ and = vt sinθ gt. Eliminating t leads to v sinθ cosθ gd =. Using sinθ cosθ = sinbθ g, we obtain v gd (9.8)(45.7) sin ( θ ) = gd sin( θ ) = = v (46) 3 which yields sin( θ) =. and consequently θ =.66. If the gun is aimed at a point a distance above the target, then tan θ = d so that = d tanθ = 45.7 tan(.66 ) =.484 m = 4.84 cm.

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