Classical Mechanics - Test Banks and Practical Problems

Motion in Two Dimensions MULTIPLE CHOICE At t=0, a particle leaves the origin with a velocity of 9.0 m/s in the positive y direction and moves in the xy plane with a constant acceleration of (2.0i + 4.0j) m/s^2. At the instant the x coordinate of the particle is 15 m, what is the speed of the particle? x - axis y - axis a. 10 m/s b. 11 m/s c. 12 m/s d. 14 m/s e. 26 m/s V = sqrt(Vx^2 + Vy^2) ANS: A PTS: 2 DIF: Average Vx = Vx0 + at Vx = 0 + 2(2)(15) = 60m/s y = V0y t + 1/2 ayt^2 y = 9 + (2-4)t = m/s 15 = (1/2) 2 t^2, t = sqrt(15) A particle starts from the origin at t=0 with a velocity of 6.0i m/s and moves in the xy plane with a constant acceleration of (-2.0i + 4.0j) m/s^2. At the instant the particle achieves its maximum positive x coordinate, how far is it from the origin? a. 10 m b. 20 m c. 30 m d. 37 m e. 45 m ANS: B PTS: 2 DIF: Average at x max. positive x happen at Vx = 0, Vx = V0x + axt -> 0 = 6 - 2t -> t = 3s X = V0xt + 1/2 axt^2 = 6(3) + 1/2(-2)(9) = 9m. Y = V0y t + 1/2 ayt^2 = 0 + 1/2(4)(9) = 18m. r = sqrt(x^2 + y^2) = 20m A particle leaves the origin with a velocity of 7.2 m/s in the positive y direction and moves in the xy plane with a constant acceleration of (3.0i - 2.0j) m/s^2. At the instant the particle moves back across the x axis (y=0), what is the value of its x coordinate? a. 65 m b. 91 m c. 13 m d. 78 m e. 86 m ANS: D PTS: 2 DIF: Average y = V0y t + 1/2 ayt^2 -> y = 7.2t + 1/2(-2)t^2 y = 0. t^2 = 7.2t -> t = 7.2s x = V0xt + 1/2 axt^2 = 0 + 1/2(3)(7.2)^2 = 77.76m At t=0, a particle leaves the origin with a velocity of 5.0 m/s in the positive y direction. Its acceleration is given by a = (3.0i - 2.0j) m/s^2. At the instant the particle reaches its maximum y coordinate how far is the particle from the origin? Vy = V0 + at a. 11 m b. 16 m c. 22 m d. 29 m e. 19 m ANS: A PTS: 2 DIF: Average Similar to #3 Vy = V0 + ayt = 5 + (-2)t -> 0 = 5 - 2t -> t = 2.5s y = V0y t + 1/2 ayt^2 = 5(2.5) + 1/2(-2)(2.5)^2 = 6.25m x = V0x t + 1/2 axt^2 = 0 + 1/2(3)(2.5)^2 = 9.375m r = sqrt(x^2 + y^2) = 11.26m A particle moves in the xy plane with a constant acceleration given by a = -4.0j m/s^2. At t=0, its position and velocity are 10i m and (-2.0i + 8.0j) m/s, respectively. What is the distance from the origin to the particle at t = 2.0 s? a. 6.4 m X = X0 + V0xt + 1/2 axt^2 = 10 - 2(2) + 1/2(0)t^2 = 6m Y = Y0 + V0y t + 1/2 ayt^2 = 8(2) + 1/2(-4)(4) = 16 - 8 = 8m r = sqrt(X^2 + Y^2) a. 10 m b. 8.9 m c. 9.0 m d. 6.2 m PTS: 3 DIF: Challenging A particle starts from the origin at t = 0 with a velocity of (16i - 12j) m/s and moves in the xy plane with a constant acceleration of a = (3.0i - 6.0j) m/s^2. What is the speed of the particle at t = 2.0 s? a. 52 m/s b. 39 m/s c. 46 m/s d. 32 m/s e. 43 m/s V = V0 + at Vx = 16 - 3(2) Vx = 16 - 6 = 10 m/s Vy = -12 + (-6)(2) Vy = -12 - 12 = -24 m/s V = sqrt(Vx^2 + Vy^2) = sqrt(10^2 + (-24)^2) = sqrt(100 + 576) = sqrt(676) = 26 m/s ANS: D PTS: 2 DIF: Average At t = 0, a particle leaves the origin with a velocity of 12 m/s in the positive x direction and moves in the xy plane with a constant acceleration of (-2.0i + 4.0j) m/s^2. At the instant the x coordinate of the particle is 18 m, what is the x coordinate of the particle? a. 30 m b. 21 m c. 40 m d. 24 m e. 45 m X = X0 + V0xt + 1/2 axt^2 18 = 0 + 12t + 1/2(-2)t^2 18 = 12t - t^2 t^2 - 12t + 18 = 0 No real solution for t. This part is not clear. Y = Y0 + V0yt + 1/2 ayt^2 Y = 0 + 0 + 1/2(4)t^2 Y = 2t^2 X = V0xt + 1/2 axt^2 18 = 12t - t^2 t^2 - 12t + 18 = 0 Using quadratic formula: t = [ -(-12) +- sqrt((-12)^2 - 4(1)(18)) ] / 2(1) t = [ 12 +- sqrt(144 - 72) ] / 2 t = [ 12 +- sqrt(72) ] / 2 t = [ 12 +- 6sqrt(2) ] / 2 t = 6 +- 3sqrt(2) t1 = 6 - 3sqrt(2) approx 1.757 s t2 = 6 + 3sqrt(2) approx 10.243 s Y = 2t^2 At t = 1.757, Y = 2(1.757)^2 approx 6.176 m At t = 10.243, Y = 2(10.243)^2 approx 209.86 m ANS: C PTS: 2 DIF: Average The initial speed of a cannon ball is 0.20 km/s. If the ball is to strike a target that is at a horizontal distance of 3.0 km from the cannon, what is the minimum time of flight for the ball? a. 16 s b. 21 s c. 24 s d. 14 s e. 19 s X = V0x t 3.0 km = (0.20 km/s) t t = 3.0 / 0.20 = 15 s However, this is for projectile motion. Minimum time of flight usually implies a direct shot. Y = V0y t + 1/2 ayt^2 target V0sin(theta) = V0y V0cos(theta) = V0x X = V0cos(theta) * t Y = V0sin(theta) * t - 1/2gt^2 ANS: A PTS: 3 DIF: Challenging A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a horizontal distance given by X = (3.0i - 2.0j) m/s^2. At the instant the particle reaches its maximum y coordinate how far is the particle from the origin? a. 11 m b. 16 m c. 22 m d. 29 m e. 19 m Y = V0y t + 1/2 ayt^2 V0y = 0 0.10 km = 100 m 100 = 0 + 1/2 (9.8) t^2 t^2 = 200 / 9.8 approx 20.408 t = sqrt(20.408) approx 4.517 s X = V0x t X = 0 + V0x t X = (V0) * t There seems to be an error in the problem description, where the acceleration is given as (3.0i - 2.0j) m/s^2, but then it talks about a horizontal throw. If it's horizontally thrown, the initial vertical velocity is 0 and the horizontal acceleration should be 0 unless stated otherwise. If the acceleration is constant, then it's not just gravity. Assuming the problem meant a initial horizontal velocity and only gravity acting. Let's re-evaluate based on the provided values and handwritten notes. T = sqrt(2h/g) = sqrt(2 * 100 / 9.8) = sqrt(200/9.8) = sqrt(20.408) = 4.517 s V0x = X / T If the acceleration given was the initial velocity, this would be a different problem. From the handwriting, it appears they interpret V0 = 5.0 m/s for question 9. Y = V0 + at y = 0 + (9.8)t^2 t = sqrt(200/9.8) = 4.517 s X = Vx * t = sqrt(V0x^2 + Vy^2) ANS: B PTS: 2 DIF: Average A baseball is hit at ground level. The ball is observed to reach its maximum height above ground level: 3.0 m after being hit, and 2.5 s after reaching this maximum height, the ball is observed to barely clear a fence that is 97.5 m from where it was hit. How high is the fence? a. 8.2 m b. 13.8 m At t = 3.0 s, Y = 0 Y = V0y t - 1/2gt^2 0 = V0y(3) - 1/2(9.8)(3)^2 0 = 3V0y - 4.9(9) 3V0y = 44.1 V0y = 14.7 m/s Max height H = V0y^2 / (2g) = (14.7)^2 / (2 * 9.8) = 216.09 / 19.6 = 11.025 m The problem states max height 3.0 m which contradicts this calculation. Let's use the given "3.0 m after being hit" as the time to reach maximum height. V_y_final = V_y_initial - g * t 0 = V_y_initial - 9.8 * 3.0 V_y_initial = 29.4 m/s H_max = V_y_initial * t - 0.5 * g * t^2 = 29.4 * 3.0 - 0.5 * 9.8 * (3.0)^2 = 88.2 - 44.1 = 44.1 m This contradicts the "3.0 m" being the max height. It's likely "3.0 s" after being hit to reach max height. So, the maximum height is 44.1 m. The fence is at 97.5 m from where it was hit. Total time of flight T = 2 * (time to max height) = 2 * 3.0 = 6.0 s Range R = V0x * T = 97.5 m V0x = 97.5 / 6.0 = 16.25 m/s Now we need the height of the ball at X = 97.5 m. The ball clears the fence at 97.5 m from where it was hit. The question asks for the height of the fence. If the ball is at X = 97.5 m at the end of its flight (when it hits the ground), then the height is 0. This seems to be a misinterpretation. "3.0 m after being hit" could mean 3.0 m is the maximum height. If H_max = 3.0 m: V0y^2 = 2gH_max = 2 * 9.8 * 3.0 = 58.8 V0y = sqrt(58.8) approx 7.668 m/s Time to max height t_max = V0y / g = 7.668 / 9.8 approx 0.782 s Total flight time T = 2 * t_max = 1.564 s "and 2.5 s after reaching this maximum height" means that the time to reach the fence is 0.782 + 2.5 = 3.282 s. At t = 3.282 s, X = 97.5 m. V0x = X / t = 97.5 / 3.282 approx 29.707 m/s Now calculate the height Y at X = 97.5 m, or at t = 3.282 s. Y = V0y t - 0.5 g t^2 = 7.668 * 3.282 - 0.5 * 9.8 * (3.282)^2 Y = 25.176 - 4.9 * 10.7715 = 25.176 - 52.78 approx -27.6 m. This indicates the ball would have hit the ground much earlier. Let's re-read: "the ball is observed to reach its maximum height above ground level: 3.0 m after being hit, and 2.5 s after reaching this maximum height, the ball is observed to barely clear a fence that is 97.5 m from where it was hit." Interpretation: Maximum height H_max = 3.0 m. Time to reach max height t_Hmax = 2.5 s. This contradicts standard physics unless it's a relative time. Let's use the interpretation from the handwritten notes: V0y = 1/2 g T = 1/2(9.8)(5.5) = 26.95 Y = V0y t - 1/2gt^2 This seems to be from a problem where time to land is 5.5s. If the maximum height is 3.0 m, then t_peak = sqrt(2H/g) = sqrt(2*3/9.8) = sqrt(6/9.8) = sqrt(0.612) = 0.782 s. If "3.0 m after being hit" refers to the time to hit the maximum height, then t_peak = 3.0 s. In that case, H_max = 0.5 * 9.8 * (3.0)^2 = 44.1 m. This is likely the intended meaning given typical projectile motion problems. So, H_max = 44.1 m (occurs at t = 3.0 s). The ball clears a fence at 97.5 m horizontally. The time from the peak to clearing the fence is 2.5 s. So, the total time to reach the fence is 3.0 s + 2.5 s = 5.5 s. At t = 5.5 s, X = 97.5 m. V0x = X / t = 97.5 / 5.5 approx 17.727 m/s. Now, we need the height of the fence at t = 5.5 s. Y_fence = V0y(5.5) - 0.5 * g * (5.5)^2 We need V0y. Since the peak is at 3.0 s, V0y = g * t_peak = 9.8 * 3.0 = 29.4 m/s. Y_fence = 29.4 * 5.5 - 0.5 * 9.8 * (5.5)^2 Y_fence = 161.7 - 4.9 * 30.25 Y_fence = 161.7 - 148.225 Y_fence = 13.475 m This matches option (b) 13.8 m if rounded. ANS: B