PHYS 211 Exam 1 - Practice Test Solutions

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1 PHYS Exam - Practice Test Solutions A The ptern or structure of the graph is wh we should be looking. Based on the graph for accelerion, we can see th for the first 3 secs we have a constant negive accelerion (- m/s ). This should eque to a negive or downward slope for velocity during the first 3 secs. Only answers (a) and (b) show this. From t = 3 to t = 6, the accelerion becomes positive (+ m/s ), and so we expect a positive or upward slope, which again is present in both (a) and (b). However, since the magnitude of the accelerions is larger for the first three seconds we should expect to see a steeper slope from t = to t = 3 than from t = 3 to t = 6. This only happens in graph (a). B Remember people, the formula for accelerion is: Dv Average Accelerion, a = Dt However, we can t just plug in the velocities given since they re not all in the same dimension. The easiest way is to use the vector coordines for both velocities: V(initial) = (i + 4.j) m/s, V(final) = (3.i + j) ΔV = V(final) - V(initial) = (3.i i) + (j 4.j) = (3.i 4.j) m/s The next step is to write th vector in terms of its magnitude, using Pythagorean Theory to get ΔV = 5 m/s. Finally, we plug into the equion for accelerion: 5m / s a = =.5m s s Ó LionTutors 9

2 3E The area under the graph, the integral, from t= to t=3, will give us the displacement. The area equals 7m, so the final position t=3 is m, since the particle was originally x=5m. 4D Here the definition for average speed is total distance / total time. The informion given has everything we need except the distance travelled during the first portion of the trip, so we must calcule th first: Dx = vt = (8km/hr)(/hr) = 4km So we get average speed = [4 + + ] / [/ + /6 + ] = 6 5C There are two parts to this problem, the first requires figuring out the displacement and speed the cheetah reaches after the first.5s: x-axis V = m/s V =? a = 5 m/s t =.5 s Δx =? 3 out of 5 is enough and so we can solve for both using: Δx = v t + =6.875m v = v + =.5m /s Since we need to know how long it will take to reach 6m, we must now determine how much farther the cheetah needs to travel: = 43.5 m.5 m/s. The speed is constant, so the time required is: t = 43.5m /.5m/s =.9s Thus the total time taken to reach 6m equals =.5s +.9s = 3.4s Ó LionTutors 9

3 6B This is essentially a question dealing with -dimensional motion where both dimensions (x-axis and ) have their own constant accelerion. We are concerned with finding the y-coordine, which is like solving for Δy. The only problem is we are only given initial velocity and accelerion for the ( variables are not enough). Remember, when dealing with -D motion, you can always use the same time found in the other axis, in this case we can use the time for motion along the x-axis. All we need to do is find out wh th time is: x-axis V = m/s V = a = 6. m/s t =? Δx = 7 m 3 out of 5 is enough and so we can solve for time using: This gives us t = 3. s, which we can use for calculions involving the : V = 4. m/s V = a = 4. m/s t = 3. s Δy =? Again we use: Dx = vt + To get Δy = 3 m Dy = vt + Ó LionTutors 9 3

4 7C The velocity as the ball leaves the table is completely horizontal, so we should first look to the x-axis to calcule the value: Vx = Dx/t =.5/t Since we don t know the time, we must use the to get th informion: V = m/s V = a = -9.8 m/s t =? Δy = -.5m Again we use: Δy = v t + = t = Δy a =.55s Now we have time, we can plug it back into the x-axis equion to get velocity: Vx = Dx/t =.5/.55 =.76 m/s Ó LionTutors 9 4

5 8B The only way to travel due north is to counter the vector pointing east with a vector th has an x-component pointing west, as shown below: Specifically, we must ensure th x-component of the m/s vector is equal to 3.5 m/s, so th the only net vector is the y-component of the m/s which points north. So we get: sinq = 3.5 -æ 3.5 ö! q = sin ç = 7 è ø Which gets us 7 west of north. Ó LionTutors 9 5

6 9D A typical -D question where we re asked to determine Δy but don t have enough variables to do so. So, as usual, when dealing with projectile motion we can use the time from the other dimension, x-axis, to help us out. x-axis Dx Vx =, t so Dx t = = =.63s V cos 5 x Now we can go to the and solve for Δy: V = sin5 = 8.87 m/s V = a = -9.8 m/s t =.63 s Δy =? We then use: Dy = vt + = 3.65 m Ó LionTutors 9 6

7 C Remember th whenever you re asked to solve for final speed/velocity in any projectile motion problem, you must consider both the x-and y- components of the vector. So for this particular question, we can start with either the x- or. For the x-axis, the final velocity is the same as initial velocity since there s no accelerion along the horizontal axis. Therefore Vx = Vcosθ = 8cos55 = 6.6 m/s For the : V = 8sin55 =.9 m/s V =? a = -9.8 m/s t = 3 s Δy = And we can use the following to solve for Vy : Vy = V + = m/s Lastly, we solve for the resultant vector of the two velocity components using Pythagorean theory to get V = 7.3 m/s. Ó LionTutors 9 7

8 C Despite how morbid this question might be, the physics behind it is quite straightforward. The scenario suggests a -D motion type-problem where we re solving for time and ae given the following da: V = +8 m/s V = a = -9.8 m/s t =? Δy = - 4 m The only problem is th to solve for t we would have to set up a quadric eqn; which is time consuming, unless you plot the function on your graphing calculor and see where it crosses the x-axis. Alternively, we can first solve for V and then solving for time will be far easier. So we use: v = v + adx And get V = -9.8 (remember th it s negive because the man is falling down, not up) Now we can use any equion to solve for time: v = v + v - v t = a = =.s Ó LionTutors 9 8

9 B Since the two balls have different initial speeds, but the same constant accelerion due to gravity, the easiest way to solve this problem is to cree position functions for each of the balls and use them to solve for time. () Dy () Dy = 5t + = -t + The question asks for when the two balls will be 5m apart, so th means we want: Δy Δy = 5 Which can be re-written as: (5t + 5t = 5 t = s ) - (-t + ) = 5 Ó LionTutors 9 9

10 3B This is a particularly difficult question to deal with especially when you consider th you are not even given the height of the building. With only the initial velocities of the two stones, this question would be impossible to solve except th in this particular case the two velocities have equal magnitude. Why is th important?...we ll get to th in a minute. First, we must recognize a rule about projectile motion th stes th an object thrown up with any initial velocity, V, will be traveling a velocity of negive V when it falls back down to the height th it was initially thrown from. In other words, or to give an example relive to this question, the stone th is thrown upwards 5 m/s will reach its max height and then start falling back down towards earth with increasing negive velocity and when it gets to the position from which it was released (the start point of this question) it will be traveling -5 m/s. So why is this informion helpful?...well it tells us th the stone thrown upwards will eventually return to its starting point after some amount of time, and th th exact time it will the same velocity th the other stone had when it was initially released (-5 m/s).which means th it will follow the exact same ph as th stone and will take the exact same amount of time to reach the ground as th stone did when it was initially released. And so to find out how much time will pass between the two stones hitting the ground, we only have to calcule how much time the stone thrown upwards spends in the air before returning to its initial release point. For this scenario we have the following: V = 5 m/s V = -5 a = -9.8 m/s t =? Δy = Yes..we even have 4 variables as a reward for understanding physics. And we can use any equion to solve for time: v = v + v - v t = a = -5-5 = 3.6s Ó LionTutors 9

11 4D This question asks for the speed and represents -D motion. However, the equions for position in both axis are given, so we cannot rely on the typical equions for motion th we have used up to this point. The easiest solution is take the derivive of each position function to get velocity for each axis and then to calcule the specific velocities t = s: d x( t) = vx( t) = 4, dt v () = 4m / s x d y( t) = v y ( t) = 6t - 9, dt v () = 3m / s y Thus the resultant vector velocity (speed) is: = 5m / s 5B The is the best approach to analyzing the variables since it is due to the accelerion of gravity along the th the ball drops back down: So we get: V = vsinq V = a = -g t =? Δy = We then use the below equion which we can simplify to solve for time: Δy = v t + = (v sinθ)t + ( g)t gt = (v sinθ)t gt = v sinθ t = v sinθ /g Ó LionTutors 9

12 6C The easiest approach will be to cree position functions for each ball. First though, we must consider th the balls collide some point between and 5 m, called y. Th means th for ball A its displacement will be ΔyA = y = y And then for ball B its displacement will be ΔyB = y 5 Now th we ve clarified how the two displacements rele to the point of collision, y, we can go ahead and write our position functions: () Dy A () Dy B = y = V t + = y - 5 = t + If we then solve eqn () in terms of y, and set it equal to y in eqn () we get: y - 5= t + y = y = 5= V t = V t + Thus when we plug in t =.5 s we get: 5 V = = m.5 s Ó LionTutors 9

13 7B The approprie vector diagram should look as follows: To counter the easterly wind, the plane s velocity vector must have an x-component th is equal to m/s. So we get: 9sinq = q = sin - æ ö ç =.84 è 9 ø! Now th we have the angle we can calcule the velocity component th heads due north using V = 9cos.84 = 87.7 m/s Finally we use the equion for average velocity: Dx V =, t Dx t = = V = s = 66.5 min 8B (NOT ON TEST) Here we re looking for centripetal accelerion, which only requires velocity and radius. We get velocity by converting revolutions per second to meters per second: (pr ) (p.5) m v = = = p m s t s And so we get accelerion: v ac = r t (p ) =.5 = 8p m s Ó LionTutors 9 3