Motion of Objects: Calculating Displacements, Speeds, and Accelerations - Prof. John Laird, Assignments of Physics

Download Motion of Objects: Calculating Displacements, Speeds, and Accelerations - Prof. John Laird and more Assignments Physics in PDF only on Docsity! 4-1 Chapter 4 Questions All three projectiles start with the same horizontal and vertical speeds. The horizontal speed of(4) each stays constant (ignoring air resistance; even if there is air resistance, all three experience the same effects before they land). Therefore any difference in total speed results from the vertical speed. All three reach zero vertical speed (and minimum total speed) at the top. From that point on, the further they fall, the faster the vertical speed (and the greater the total speed). Projectile (a) falls the most from the top and (c) the least, so the order from greatest to least final speed is (a), (b), and (c) . (a) For windows 1, 2, and 3, the tangerine is crossing vertically, so the vertical speed determines the(6) crossing time. The vertical speed decreases as the tangerine rises, so the crossing time increases. Therefore the order from longest to shortest time is 3, 2, 1 . (b) The horizontal speed stays constant (ignoring air resistance, or slows somewhat if we include air resistance), so the total speed varies as the vertical speed. So the order from greatest to least speed is 1, 2, 3 . (c) For windows 4, 5, and 6, the tangerine is crossing horizontally, so the horizontal speed determines the crossing time. Ignoring air resistance, the horizontal speed stays constant, so the crossing time is the same for all three. (If air resistance is included, the horizontal speed gradually decreases, so the order would be 6, 5, 4.) (d) The vertical speed increases as the tangerine falls, so the total speed increases. Therefore the order from greatest to least speed is 6, 5, 4 . (a) The airplane’s velocity when the bundle is released is horizontal, so the bundle’s initial velocity(9) is horizontal. Thus the vertical component is zero (b) and the horizontal component is the speeds of the plane, 350 km/h . (c) Ignoring air resistance, the horizontal speed does not change as the bundle falls, so its horizontal speed just before it hits the ground is still 350 km/h . (d) The time of fall depends only on the vertical motion, not on the horizontal motion. So if the horizontal speed is increased, the time of fall stays the same . Problems The speed stays constant at v = 60 km/h. There are three legs of the trip: leg 1 has time t1 = 40.0(7) min E, leg 2 has time t2 = 20.0 min at 50.0 ◦ E of N, and leg 3 has time t3 = 50.0 min W. We’re asked to find the average velocity for the whole trip. The average velocity is the total displacement over the total time: ~vavg = ~r/t. The total time ∆t = t1 + t2 + t3 = 40.0 + 20.0 + 50.0 min = 110.0 min. We must calculate the displacement for each leg, using ~r = ~vt. Thus r1 = v1t1 = (60 km/h)(40.0 min)(1 h/60 min) = 40 km E. For leg 2, r2 = v2t2 = (60 km/h)(20.0 min)(1 h/60 min) = 20 km, 50◦ E of N. For leg 3, the displacement r3 = v3t3 = (60 km/h)(50.0 min)(1 h/60 min) = 50 km W. 4-2 EW NS r1 +40.000 0.000 r2 +20 sin 50 ◦ = +15.321 +20 cos 50◦ = +12.856 r3 −50.000 0.000 r +5.321 +12.856 (a) The total displacement r = √ r2x + r 2 y = √ (5.321 km)2 + (12.856 km)2 = 13.91 km. The mag- nitude of the average velocity vavg = r/t = [(13.91 km)/(110.0 min)](60 min/1 h) = 7.59 km/h . (b) The direction of the average velocity is the same as that of the displacement: θ = tan−1(rx/ry) = tan−1 (5.321 km)/(12.856 km) = 22.5◦ east of north . The bicyclist has initial position ~r1 = 40.0 m E and velocity ~v1 = 10.0 m/s S, and final position ~r2(14) = 40.0 m N and velocity ~v2 = 10.0 m/s E. The elapsed time t = 30.0 s. (a) The displacement from the initial position to the final position is 40 m W and 40 m N. (We can express the positions using the unit vectors, ~r1 = (40 m)̂i and ~r2 = (40 m)̂j, so the displacement ∆~r = ~r2 − ~r1 = 40̂j− 40̂i = −40̂i + 40̂j m.) From the Pythagorean theorem, the total displacement ∆r = √ ∆r2x + ∆r 2 y = √ (−40 m)2 + (40 m)2 = 56.6 m . (b) The direction θ = tan−1(|∆ry/∆rx|) = tan−1 (40 m)/(40 m) = 45.0◦ north of west . (c) (d) The average velocity ~vavg = ∆~r/t, so the magnitude vavg = ∆r/t = (56.57 m)/(30.0 s) = 1.89 m/s and the direction is the same as that of the displacement, 45.0◦ north of west . (e) Using the unit vectors, the velocities are ~v1 = −10̂j and ~v2 = 10̂i in m/s, so the change in velocity ∆~v = ~v2 − ~v1 = 10̂i− (−10̂j) = 10̂i + 10̂j m/s, i.e. 10 m/s east and 10.0 m/s north. From the Pythagorean theorem, the change in velocity ∆v = √ ∆v2x + ∆v 2 y =√ (10.0 m/s)2 + (10.0 m/s)2 = 14.14 m/s. The average acceleration ~aavg = ∆~v/t, so the magnitude aavg = ∆v/t = (14.14 m/s)/(30.0 s) = 0.471 m/s 2 . (f) The direction is the same as that of the change in velocity: θ = tan−1(∆vy/∆vx) = tan−1 (10 m/s)/(10 m/s) = 45.0◦ north of east . The initial vertical position y0 = 45.0 m and the final position is the ground, y = 0, so the dis-(21) placement ∆y = −45 m. The initial velocity v0 = 250 m/s and is horizontal, so the initial x and y velocities v0x = 250 m/s and v0y = 0. Ignoring air resistance, the accelerations are ax = 0 and ay = −9.8 m/s2. (a) To find the time the projectile is in the air, we only need to consider the vertical motion. From eq. 2-15, ∆y = v0yt+ 1 2 ayt 2, so with v0y = 0, the time t = √ 2∆y ay = √√√√ 2(−45 m) −9.8 m/s2 = 3.03 s . (b) Using eq. 2-15 in the x direction, ∆x = v0xt+ 1 2 axt 2 = (250 m/s)(3.03 s) + 0 = 758 m . (c) From eq. 2-11, the vertical velocity just before it hits the ground vy = v0y + ayt = 0 + (−9.8 m/s2)(3.03 s) = −29.7 m/s. The magnitude (i.e. the speed) is 29.7 m/s . 4-5 Questions The three paths reach the same height, so the vertical motions are the same. Only the horizontal(7) motions are different. (a) The time of flight depends only on the vertical motion, so all three are the same . (b) The vertical motions are the same, so the initial vertical speeds must all three be the same . (c) The times of flight are the same, so the difference in horizontal displacement is due to different horizontal speeds (which remain constant, ignoring air resistance). Thus the order from greatest to least horizontal speed is 3, 2, 1 . (d) The total initial velocity is the combination of the horizontal and vertical components. Because the initial vertical speeds are the same, the differences arise from the horizontal speeds, so the order is again 3, 2, 1 . The train’s speed stays the same, so it will follow uniform circular motion around these curves.(13) The magnitude of the radial acceleration is a = v2/r. The speed is the same for all four cases, so the acceleration depends only on the radius of curvature. So the order from greatest to least acceleration is the order of least to greatest radius of curvature: 2, 1 and 4 tied, 3 . Problems The cart has acceleration components ax = 4.0 m/s 2 and ay = −2.0 m/s2, and initial velocity(15) components v0x = 8.0 m/s and v0y = 12 m/s. We want the final velocity when the cart reaches its maximum y position. The final position occurs when vy = 0, which is part of the final answer. It also provides extra information in the y direction and suggests the use of eq. 2-15, vy = v0y + ayt. Solving for the time: t = vy − v0y ay = 0− 12 m/s −2.0 m/s2 = 6.0 s. Using this time in the x direction with eq. 2-15 gives vx = v0x+axt = (8.0 m/s) + (4.0 m/s 2)(6.0 s) = 32 m/s. Expressing the final velocity in unit-vector notation: vx = (32 m/s)̂i . The displacements ∆x = 77.0 m and ∆y = 0. The direction of the initial velocity is θ = 12.0◦(23) above horizontal. We can ignore air resistance, so the accelerations are ax = 0 and ay = −9.8 m/s2. We’re asked for the initial speed. The motion in the vertical direction determines the time of flight, with initial speed v0y = v0 sin θ. From eq. 2-11 vy = v0y + at. The symmetry of the vertical motion, or the use of eq. 2-16 with ∆y = 0, gives vy = −v0y, so eq. 2-11 becomes −2v0y = ayt or t = −2v0y/ay. In the horizontal direction, eq. 2-15 gives ∆x = v0xt + 1 2 axt 2 = v0xt. Substituting for t from the vertical motion and then v0x = v0 cos θ and v0y = v0 sin θ gives ∆x = v0xt = v0x ( −2v0y ay ) = (v0 cos θ) ( −2(v0 sin θ) ay ) = ( −2 cos θ sin θ ay ) v20 . Solving for the initial velocity, v0 = √√√√(−2 cos θ sin θ ay∆x ) = √√√√( −2 cos 12◦ sin 12◦ (−9.8 m/s2)(77 m) ) = 43.1 m/s . 4-6 The initial velocity v0 = 10 m/s horizontal, so in the x and y directions v0x = 10 m/s and v0y = 0.(25) The time t = 0.19 s. Ignoring air resistance, the accelerations are ax = 0 and ay = −9.8 m/s2. (a) We’re asked for the distance the dart fell, i.e. the magnitude of the vertical displacement, ∆y. From eq. 2-15, ∆y = v0yt+ 1 2 ayt 2 = 0 + 1 2 (−9.8 m/s2)(0.19 s)2 = −0.18 m. The distance PQ is the magnitude, 0.18 m . (b) Now we’re asked for the horizontal displacement. Again from eq. 2-15, ∆x = v0xt+ 1 2 axt 2 = (10 m/s)(0.19 s) + 0 = 1.9 m . The initial velocity v0 = 25.0 m/s at θ = 40.0 ◦ above horizontal. The x displacement is the distance(38) to the wall, ∆x = 22.0 m. We can find the initial x and y velocities: v0x = v0 cos θ = (25.0 m/s)(cos 40.0 ◦) = 19.15 m/s and v0y = v0 sin θ = (25.0 m/s)(sin 40.0 ◦) = 16.07 m/s. As usual, we know the accelerations: ax = 0 and ay = −9.8 m/s2. (a) The final point is set by crossing the horizontal distance to the wall, so we start with the x motion, which we can use to find the time. From eq. 2-15, ∆x = v0xt + 1 2 axt 2 = v0xt (using ax = 0), so the time t = ∆x v0x = 22.0 m 19.15 m/s = 1.149 s. Using this time in the y direction along with eq. 2-15: ∆y = v0yt+ 1 2 ayt 2 = (16.07 m/s)(1.149 s) + 1 2 (−9.8 m/s2)(1.149 s)2 = 12.0 m . (b) The horizontal velocity doesn’t change, so the final value vx = v0x = 19.2 m/s . (c) Eq. 2-11 gives the final vertical velocity: vy = v0y + ayt = (16.07 m/s) + (−9.8 m/s2)(1.149 s) = 4.81 m/s . (d) At the wall, the ball is still moving upward (vy is positive), so it has not passed its highest point. The satellite has a height h = 640 km above the Earth’s surface and a period T = 98.0 min.(60) The radius of the satellite’s orbit is the radius of the Earth plus the height above the surface: r = RE + h = 6370 + 640 km = 7010 km. (a) The speed equals the distance traveled divided by the time for one orbit (i.e. using eq. 4-35): v = 2πr T = 2π(7010 km) 98.0 min × ( 1000 m 1 km )( 1 min 60 s ) = 7490 m/s . (b) The magnitude of the centripetal acceleration a = v2 r = (7490 m/s)2 7.01× 106 m = 8.00 m/s2 . 4-7 The circle has radius r = 1.5. For the motion after the string breaks, the initial height y0 = 2.0 m and(67) the final position y = 0, so ∆y = −2.0 m. Horizontal displacement ∆x = 10 m. Because the circular motion is horizontal, the initial velocity when the string breaks is horizontal, so v0 = v0x = vc, the circular velocity, and the initial y velocity v0y = 0. We also know the accelerations while the stone falls: ax = 0 and ay = −9.8 m/s2. We have enough information to fully describe the motion as the stone falls. Starting with the y direction, we can use eq. 2-15, ∆y = v0yt+ 1 2 ayt 2, to find the time. Setting v0y = 0, ∆y = 1 2 ayt 2, so solving for t gives t = √ 2∆y ay = √√√√2(−2.0 m) −9.8 m/s2 = 0.6388 s. Turning to the x direction, eq. 2-15 gives ∆x = v0xt + 1 2 axt 2 = v0xt (with ax = 0). With the time from the y direction, the initial velocity v0x = ∆x t = 10 m 0.6388 s = 15.65 m/s. Since this is the circular velocity before the string broke, we can now find the centripetal acceleration: a = v2 r = (15.65 m/s)2 1.5 m = 160 m/s2 . The radius of the circular motion r = 5.0 m and the centripetal acceleration a = 7.0g.(104) (a) From eq. 4-34, the centripetal acceleration a = v2/r, so the speed v = √ ar = √ (7.0g)(5.0 m) =√ (7.0)(9.8 m/s2)(5.0 m) = 18.52 m/s ' 19 m/s . (c) From eq. 4-35, the period T = 2πr v = 2π(5.0 m) 18.52 m/s = 1.7 s . (b) The period is the time for 1 revolution, so the number of revolutions per minute (called the frequency) is f = 1 T = 1 rev 1.7 s × ( 60 s 1 min ) = 35 rpm .