Sample Problems PDF

Sample Problems 1. The mass of a sugar grain is 5 milligrams. Which one of the following statements indicates the correct mass of the sugar grain in grams? A. The sugar grain has a mass of 5 × 10² grams. B. The sugar grain has a mass of 5 × 10⁻² grams. C. The sugar grain has a mass...

Sample Problems 1. The mass of a sugar grain is 5 milligrams. Which one of the following statements indicates the correct mass of the sugar grain in grams? A. The sugar grain has a mass of 5 × 10² grams. B. The sugar grain has a mass of 5 × 10⁻² grams. C. The sugar grain has a mass of 5 × 10⁻³ grams. D. The sugar grain has a mass of 5 × 10¹ grams 2. In full-projectile motion, when does the projectile start? A. At the peak B. After the peak C.Before the peak D.At the end of the motion 3. Sarah is watching an object being thrown upwards into the air at an angle. It reaches a certain height and then falls back down due to gravity. What type of motion is Sarah witnessing? A. Linear motion B. Projectile motion C.Circular motion D.Free-fall motion 4. What happens to the horizontal acceleration of a projectile during its motion? A. It remains constant. B. It increases as the projectile rises. C.It decreases as the projectile rises. D.It is equal to zero. 5. A rectangular park has an area of 12.3 km². What is the area of the park in m²? (2 points) A. 1.23 × 10⁴ m² B. 1.23 × 10⁵ m² C. 1.23 × 10⁶ m² D. 1.23 × 10⁷ m² 6. What happens to the vertical acceleration of a projectile during its motion? A. It remains constant. B. It increases as the projectile rises. C.It decreases as the projectile rises. D.It fluctuates depending on the initial velocity. 7-8. An object is launched with an initial velocity of 20 m/s at an angle of 30 degrees above the horizontal. What is the horizontal component of its velocity? A. 10.0 m/s B. 17.32 m/s C.11.55 m/s D.20.0 m/s 7-8. Given: V0 = 20 m/s θ = 30° Vx = ? Formula: Vx = V0Cosθ Sol’n: Vx = 20 m/s(Cos 30°) Vx = 17.32 m/s 9-10. An object is launched with an initial velocity of 20 m/s at an angle of 30 degrees above the horizontal. What is the vertical velocity component of the projectile after 3 seconds? A. 39.4 m/s B. 19.4 m/s 5 MINUTE TIMER - COUNTDOWN TIMER (MINIMAL) C.-39.4 m/s This 5 MINUTE countdown timer is made for professional use and has some minimal sound effects in the last 5 seconds. D.-19.4 m/s 9-10. Given: V0 = 20 m/s θ = 30° t= 3s Vy = ? Formula: Vy = V0(Sinθ) + gt Solution: Vy = 20 m/s(Sin 30°) + (-9.8 m/s^2) (3s) Vy = 10 m/s + (-29.4 m/s) Vy = - 19.4 m/s 11-12. An object is projected vertically with an initial velocity of 9 m/s. What is the final vertical velocity of the projectile after 4 seconds? A. 25.2 m/s upward B. 39.2 m/s downward C. 30.2 m/s downward D. 0 m/s 11-12. Given: V0x = 9 m/s t=4s Vy = ? Formula: Vy = V0y + gt Sol’n: Vy = 9 m/s + (- 9.8 m/s^2) (4 s) Vy = 9 m/s - 39.2 m/s Vy = -30.2 m/s 13-14. Which one of the following expressions may be used to correctly find the angle θ in the drawing? A. θ = cos^-1(6/10) B. θ = sin^-1(6/10) C. θ = tan^-1(6/10) D. θ = tan^-1(10/6) 15. Which one of the following quantities is a vector quantity? A. The volume of a swimming pool B. The speed of a car C. The force exerted by the wind D. The population of a city 16-17. A hiker walks 2.00 km due south from a campsite. He then walks due east to a trailhead. If the magnitude of the hiker's total displacement vector is 2.40 km, what is the direction of his total displacement vector with respect to due south? A. 38° east of south B. 48° west of south C. 34° east of south D. 45° west of south 16-17. A hiker walks 2.00 km due south from a campsite. He then walks due east to a trailhead. If the magnitude of the hiker's total displacement vector is 2.40 km, what is the direction of his total displacement vector with respect to due south? Solution: cah cosƟ = a/h 2.40 km (h) 2.00 km (a) Ɵ = cos^-1(a/h) Ɵ = cos^-1(2.00/2.40) Ɵ = 34º East of South 19-21. If a projectile is launched from the ground with an initial velocity of 40 m/s at an angle of 30 degrees above the horizontal, what time does it take to reach its maximum height? A. 1.5 seconds B. 2.0 seconds C. 2.5 seconds D. 3.0 seconds 19-21. Given: V0 = 40 m/s θ = 30° t(maximum height) = ? Formula: t = Vy - V0(Sinθ) g t = 0 - 40 m/s (Sin 30°) - 9.8 m/s^2 t=2s 22-23. A motorcycle, starting from rest, accelerates in a straight-line path at a constant rate of 3.0 m/s². How far will the motorcycle travel in 10 seconds? A.150 m b. 200 m C. 300 m D. 30 m 22-23. Given: a = 3.0 m/s² t = 10 seconds Required: distance (d) Equation: X = ½ at^2 Solution: X = ½ (3.0 m/s^2) (10 s) ^2 X = ½ (3.0 m/s^2) 100 s^2 X = 150 m 24-25. A projectile is fired from a cannon with initial horizontal and vertical components of velocity equal to 25 m/s and 35 m/s, respectively. Determine the initial speed of the projectile. a. 30 m/s b. 43 m/s c. 50 m/s d. 60 m/s 24-25. Given: vy= 35m/s vx=25 m/s Required: V0 Equation: V0 = √(vx)^2 + (vy)^2 Solution: V0 = √(25 m/s)^2 + (35 m/s)^2 V0 = √(625 m^2/s^2) + (1225 m^2/s^2) V0 = √(1850 m^2/s^2) V0 = 43 m/s 24-25. At what angle is the projectile fired (measured with respect to the horizontal)? (Refer to the previous problem) a. 35° b. 40° c. 45° d. 54° 24-25. Given: Vy = 35 m/s Vx= 25 m/s Formula: θ = tan^-1 (o/a) Solution: θ = tan^-1 (35/25) θ = 54° Pointers: Unit Conversion Scientific Notation Right Triangle (soh cah toa) Vectors and Scalars Vector Addition (Component Method) Distance and Displacement Velocity and Acceleration Position- time graph Velocity- time graph Kinematic equations (X and Y Motion) Projectile Motion