Task 1: Vector components and free body diagrams QUESTION 1 a. π₯ πππππππππ‘ |→ πΉππ₯ | = πΉπ sin 2 5 |→ πΉππ₯ | = (40) sin 2 5 |→ πΉππ₯ | = 16.9π |→ πΉππ₯ | = 17π Y component |→ πΉππ¦ | = πΉπ cos 2 5 |→ πΉππ¦ | = (40) cos 2 5 |→ πΉππ¦ | = 36.252π |→ πΉππ¦ | = 36π b. X component → π£π₯ = (25π⁄π ) cos 7 5 → π£π₯ = 6.47π⁄π → π£π₯ = 6.5π⁄π Y component → π£π¦ = (25π⁄π ) sin 7 5 → π£π¦ = 24.148π ⁄π → π£π¦ = 24π⁄π c. X component → π£π₯ = (17π⁄π ) cos 6 5 → π£π₯ → π£π₯ = 7.185π ⁄π = 7.2π⁄π Y component → π£π¦ = (17π⁄π ) sin 6 5 → π£π¦ = 15.41π ⁄π → π£π¦ = 15π⁄π QUESTION 2 a. b. Task 2: Relative velocity and projectile motion problem solving QUESTION 1 a. Let helicopter represent H Let the ground represent G Let airspeed represent A Given: |→ π£π»π΄ | = 42.5π ⁄π [π] |→ π£π΄πΊ | = 25π ⁄π [πΈ 30°π] Solving for components of |→ π£π»π΄ | |→ π£π»π΄π₯ | =? |→ π£π»π΄π¦ | =? Vector falls on east – west horizontal line with no vertical components. Vector only has horizontal → components. Therefore, horizontal component of| → π£π»π΄ |, | π£π»π΄π₯ | is 42.5 m/s [W] and the vertical → component of | → π£π»π΄ | | π£π»π΄π¦ | is 0 m/s [S] Solving for components of | → π£π΄πΊ | |→ π£π΄πΊ | = 25.0π ⁄π [πΈ 30° π] |→ π£π΄πΊπ₯ | =? |→ π£π΄πΊπ¦ | =? Find the horizontal component using cosine formula |→ π£π΄πΊπ₯ | = π£πππ π |→ π£π΄πΊπ₯ | = (25π ⁄π ) cos 3 0° |→ π£π΄πΊπ₯ | = 21.65π ⁄π |→ π£π΄πΊπ₯ | = 22π ⁄π [πΈ] Find the vertical component using sine formula |→ π£π΄πΊπ¦ | = π£π πππ |→ π£π΄πΊπ¦ | = (25π ⁄π ) sin 3 0° |→ π£π΄πΊπ¦ | = 12.5π ⁄π |→ π£π΄πΊπ¦ | = 13π ⁄π [π] → Therefore, the horizontal component of | → π£π΄πΊ |, | π£π΄πΊπ₯ | is 22 m/s [E] and the vertical component of → |→ π£π΄πΊ | | π£π΄πΊπ¦ | is 13 m/s [S] b. Let west and south be positive directions Given: |→ π£π»π΄π₯ | = 42.5 π ⁄π [π] |→ π£π΄πΊπ₯ | = 22 π ⁄π [πΈ] rounded to 2 significant figures |→ π£π΄πΊπ¦ | = 13 π ⁄π [π] rounded to 2 significant figures → Required: | → π£π»πΊπ₯ | =? | π£π»πΊπ¦ | =? Solution: To determine | → π£π»πΊπ₯ |, and all given horizontal components. → → |→ π£π»πΊπ₯ | = | π£π»π΄π₯ | + | π£π΄πΊπ₯ | |→ π£π»πΊπ₯ | = 42.5π ⁄π + (−22.65π ⁄π ) |→ π£π»πΊπ₯ | = 19.85π ⁄π |→ π£π»πΊπ₯ | = 20π ⁄π [π] To determine | → π£π»πΊπ¦ |, add all given vertical components. → → |→ π£π»πΊπ¦ | = | π£π»π΄π¦ | + | π£π΄πΊπ¦ | |→ π£π»πΊπ¦ | = 0π ⁄π + 12.5π ⁄π |→ π£π»πΊπ¦ | = 12.5π ⁄π |→ π£π»πΊπ¦ | = 13π ⁄π [π] c. Solution: using Pythagorean theorem to find magnitude of resultant velocity 2 2 → → |→ π£π»πΊ | = √ | π£π»πΊπ₯ | + | π£π»πΊπ¦ | 2 2 |→ π£π»πΊ | = √ (19.85π ⁄π ) + (12.5π ⁄π ) |→ π£π»πΊ | = √ 550.2725π ⁄π |→ π£π»πΊ | = 23.458π ⁄π |→ π£π»πΊ | = 23π ⁄π Using tangent function to find direction π‘πππ = π = tan−1 |→ π£π»πΊπ¦ | |→ π£π»πΊπ₯ | (12.5π⁄π ) (19.85π⁄π ) π = 32° Therefore, the resultant velocity of the helicopter relative to the ground is 23 m/s [π32°π] QUESTION 2 Solution: first, determine horizontal and vertical components of initial velocity to help find final velocity which we will use to find the time of flight. Given: → π = 9.8 π⁄π 2 [πππ€π] β³ ππ¦ = 4.3 π[πππ€π] → π£π = 30 π⁄π π = 35° Solving for horizontal components cos 3 5° =→π£ππ₯ ⁄ → π£π → π£ππ₯ → π£ππ₯ → π£ππ₯ =→ π£π cos 3 5° = (30π⁄π ) sin 3 5° = 24.575π⁄π [πππβπ‘] Solving for vertical components sin 3 5° =→ π£ππ¦ ⁄ → π£ππ¦ = → π£π → π£π sin 3 5° → π£ππ¦ = (30π⁄π ) sin 3 5° → π£ππ¦ = 17.208π⁄π [π’π] a. Required: β³ π‘ =? → . let up be positive direction Using motion equation 5 to determine π£ππ¦ → 2 π£ππ¦ → π£ππ¦ = → 2 π£ππ¦ + 2→ π β³ ππ¦ = √(17.208π⁄π )2 + 2(−9.8π⁄π )(−4.3) → π£ππ¦ = √380.395264 → π£ππ¦ = ±19.50π ⁄π Negative root chosen since the ball is travelling downward at this time. Therefore, → = −19.50π ⁄π [π’π] π£ππ¦ Solving for time of flight using motion equation 1. → π£ππ¦ → =→ π£ππ¦ +π β³ π‘ (−19.50π⁄π ) = (17.208π⁄π ) + (−9.8π⁄π 2 β³ π‘ ) β³π‘= (−36.708π ⁄π ) (−9.8π⁄π 2 ) β³ π‘ = 3.746π β³ π‘ = 3.7π Therefore, the time of flight of the golf ball is 3.7s b. Required: β³ ππ₯ =? Using constant equation for velocity to find range, β³ ππ₯ → π£ππ₯ = β³ ππ₯ β³π‘ → π£ππ₯ β³ ππ₯ = β³π‘ β³ ππ₯ = (24.575π⁄π )(3.746π ) β³ ππ₯ = 92.06π β³ ππ₯ = 92π Therefore, the range for the golf ball is 92m. c. Required: → π£π =? Using Pythagorean theorem to solve for final velocity 2 2 → → |→ π£π | = √| π£ππ₯ | + | π£ππ¦ | 2 2 |→ π£π | = √(24.575π ⁄π ) + (19.50π ⁄π ) |→ π£π | = ±31.371π ⁄π Choose a positive root since magnitude is always positive |→ π£π | = 31π ⁄π Using tangent function to determine direction π‘πππ = π£ππ¦ ⁄ → π£ππ₯ → π = tan−1 (19.50π⁄π ) (24.575π⁄π ) π = 38° Therefore, the velocity of the golf ball the instant before the ball impacts the ground is 31 π⁄π [πππβπ‘ 38° πππ€π] Task 3: Newton’s laws of motion and uniform circular motion problem-solving QUESTION 1 a. b. → πΉπππ‘π¦ = → πΉπ + (−→ πΉππ¦ ) All force is happening on x-axis, therefore movement on the y-axis. → πΉπππ‘π¦ = 0 because acceleration is 0 since there is no 0 = πΉπ − ππ cos 3 0° πΉπ = ππ cos 3 0° πΉπ = (120ππ)(9.8π⁄π 2 ) cos 3 0° πΉπ = 1018.45ππ [π’π] πΉπ = 1.0 × 103 ππ [π’π] c. Solution → πΉπ = ππ πΉπ → (0.10)(1018.45ππ) πΉπ = → πΉπ = 101.845π → πΉπ = 1.0 d. Solution for πΉπππ‘π₯ × 102 π πΉπππ‘π₯ = πΉππ₯ + (−πΉπ) πΉπππ‘π₯ = (ππ) sin 3 0° − 101.845π πΉπππ‘π₯ = (120ππ)(9.8π⁄π 2 ) sin 3 0° − 101.845π πΉπππ‘π₯ = 588π − 101.845π πΉπππ‘π₯ = 486.155π Solving for acceleration → πΉπππ‘π₯ = ππ→π₯ 486.155π = (120ππ)ππ₯ ππ₯ = 486.155π 120ππ ππ₯ = 4.05π⁄π 2 [πππ€π π‘βπ ππππ] Therefore, the acceleration of the piano down the ramp along the axis parallel to the ramp is 4.05π ⁄π 2 [πππ€π π‘βπ ππππ] QUESTION 2 a. Given: m=100kg, r=0.25m Required: v=? Solution: using equation to determine minimum speed required to maintain the path of circular motion v=√ππ π£ = √(0.25)(9.8π⁄π 2 ) π£ = ±1.565π ⁄π Choose positive root since magnitude is always positive π£ = 1.57π⁄π (Rounded to 3 significant figures) b. Given: r = 0.25m, m=0.100kg, v= 1.57m/s g= 9.8m/s^2 Required: πΉπ =? πΉπΆ = πΉπ − πΉπ π£2 π ( ) = πΉπ − ππ π πΉπ = (0.100ππ)(1.57π⁄π 2 ) + ππ 0.25π πΉπ = 0.97969 + 0.98 πΉπ = 1.959π πΉπ = 2.0π Therefore, thew magnitude of the tension in the string at the bottom of the circle is 2.0N QUESTION 3 a. Given: πΉπΎ = 5.0π π1 = 2.1ππ π2 = 3.3ππ Solving for acceleration using Newton’s Second Law πΉπππ‘ = ππ π= π= πΉπππ‘ π πΉπ − πΉπ + π2 π − πΉπ (π1 + π2 ) π= π= π2 π − πΉπ (π1 + π2 ) 32.34π − 5.0π 5.4ππ π = 5.06π⁄π 2 To solve for πΉπ Solve for π1 πΉπππ‘1 = πΉπ − πΉπ π1 π = πΉπ − πΉπ πΉπ = (2.1ππ)(5.06π ⁄π 2 ) + 5.0π πΉπ = 15.626π πΉπ = 16π Therefore πΉπ = 16π . For verification, solve for π2 to see if πΉπ will have same outcome has π1 b. Solving for π2 πΉπππ‘2 = πΉπ2 − πΉπ π2 π = π2 π − πΉπ πΉπ = π2 π − π2 π πΉπ = (3.3ππ)(9.8π⁄π 2 ) − (3.3ππ)(5.06π⁄π 2 ) πΉπ = 32.34π − 16.698π πΉπ = 15.642π πΉπ = 16π Therefore, the magnitude of the tension in the string is 16N
