Honors Physics Examination I Review Questions #1-#11 - Vectors & Measurements vector quantity is specified by magnitude and direction (examples: displacement, velocity, acceleration, momentum, and weight). lthough a ll forces are vector quantities not all vectors are forces. Contact forces include tension, air resistance, and friction. Tension can be measured using a spring balance. The negative of a vector is a vector of the same magnitude, but in the opposite direction. Two vectors are equal if and only if they have the same magnitude and direction. The tip-to-tail method of vector addition consists of moving the vectors parallel to their original position until the tail of one vector is at the tip of the next vector. The resultant vector is drawn from the tail of the first vector to the head of the last vector. + - Scalar quantities are described by magnitude only (examples: distance, speed, temperature, energy, and mass). The base SI units include length (meters), time (seconds), temperature (Kelvin) and mass (kg). n example of a derived unit is a Newton (kg-m/s 2 ). The surface area of an object equals its length multiplied by its width. For example, the surface area of a rectangular plate measuring 10cm by 15cm is 150cm 2 or 0.015m 2. Precision is an indication of accuracy only if there are no systematic errors. Large random errors yield neither precision nor accuracy. Measurements can be precise (with small random errors) but not necessarily accurate (agreement with accepted value). For example, if measuring the length of a wire yields results of 3.52cm, 3.41cm, 3.55cm, 3.45cm and the actual length is 3.50cm, the results were both precise and accurate. system may be placed in equilibrium if a vector of equal magnitude but opposite direction to the sum of the vectors (resultant) is added to the set of vectors. The x-component of a vector (v) is vcosθ and the y-component is vsinθ. 2 2 R = Σ ( v cos θ ) + Σ ( v sin θ ) θ = tan 1 v v y x What is the magnitude and direction of a vector with the components x = 9m and y = 12m? R= (9 2 +12 2 )=15m and θ=tan -1 (12/9)=53 o Questions #12-#30 - Kinematics n object with a constant acceleration increases its velocity by the same during each time interval. The amount of time required for an object to reach its maximum height is equal to the time required for that object to return to ground level relative to its starting position. The y or vertical component of velocity at the maximum height of a projectile equals zero. The horizontal or x-component of a projectile is constant and non-zero throughout its entire path. The rate of change of the velocity is the same during the entire time of projectile motion due to the constant downward acceleration of gravity (-9.8m/s 2 ). The vertical and horizontal components of a projectile s velocity are independent of each other. The maximum range is achieved at a 45 o angle.
The slope of a position-time graph yields velocity, while the slope of a velocity-time graph yields the object s acceleration. The displacement of an object is a straight-line vector from its origin to its end point. The distance traveled is a scalar quantity which depends upon the actual path taken by the object. x=v i t+½at 2 v f =v i +at x=(v f2 -v i2 )/(2a) x=[(v i +v f )/2]t n object with an initial velocity of 5m/s accelerates at 3m/s 2 for 4s. What is the displacement and final velocity of the object? x=(5m/s)+½(3m/s 2 )(4s) 2 =29m v f=(5m/s)+(3m/s 2 )(4s)=17m/s car is moving at a constant velocity when it is involved in a collision. The car comes to rest after 0.45s with an average acceleration of 65m/ s 2 in the direction opposite that of the car s velocity. What was the speed of the car before the collision? [nswer: 29.2m/s] car is initially traveling at 13.8m/s. The brakes are applied and the car stops over a distance of 35m. What was magnitude of the car's acceleration while it was braking? [nswer: -2.8m/s 2 ] v y =v i sinθ+gt (velocity) y=v isinθt+½gt 2 (height) R=v i cosθt (range) n object with an initial velocity of 20m/s is launched at a 60 o angle. What is the maximum height and range of the object? 0=v y=v isinθ+gt at maximum height so 0=(20m/s)sin60 o +(-9.8m/s 2 )t therefore t=1.7s y=(20m/s)sin60 o (1.7s)+½(-9.8m/s 2 )(1.7s) 2 = 15.3m Note that the time is doubled to reach the ground for the range: R=(20m/s)cos60 o (1.7s x 2)=34m ball is thrown straight upward from the top of a 35m tall cliff with an initial velocity of 15 m/sec. ssuming the ball just misses the edge of the cliff on its way down, what is the ball s maximum height and the velocity before it hits the ground? v f=v i + gt Note that v y=0 at maximum height t=(0-15m/s)/-9.8m/s 2 =1.5s y=v it+ ½gt 2 y=(15m/s)(1.5s)+½(-9.8m/s 2 )(1.5s) 2 y=11.5m The total distance traveled by the ball from the maximum height is 11.5m+35m=46.5m where its initial velocity is 0. y=v it+ ½gt 2 Note y is negative as the ball is moving downward -46.5m=(0)(t)+½(-9.8m/s 2 )(t) 2 solving for t yields 3.1s v f=v i + gt v f=(0)(3.1s)+(-9.8m/s 2 )(3.1s)=-30.2m/s (negative indicates downward velocity of ball relative to maximum height) ball is dropped from rest from a tower and strikes the ground 125 m below. pproximately how many seconds does it take the ball to strike the ground after being dropped from the tower? [nswer: 5.05s] ball rolls off a frictionless table top 4m above the ground with a constant horizontal velocity of 2m/s. How far from the table does the ball strike the floor? [nswer: 1.8m]
Questions #31-#50 - Translational Equilibrium & Newton s Laws of Motion Newton s first law of motion (law of inertia) states that an object at rest will remain at rest or an object in motion will remain in motion in a straight line with a constant velocity unless acted upon by an external unbalanced force. No force is required to keep an object in motion. ccording to Newton's second law (F=ma), whenever a net external unbalanced force acts upon a body, it will cause that body to accelerate directly proportional to the applied force and inversely proportional to the mass of the body in a direction in keeping with the direction of the applied force. Inertia describes a body s resistance to change in motion. Mass (kg) is a measure of an object s inertia and is independent of gravity. n object s weight (W=mg) measured in Newtons is equal to the force of gravity acting on the body that varies with location with respect to the Earth s gravitational field. The weight of an object only equals the Normal force on a flat surface where no other forces except those parallel to the surface act on the object. The Normal force of an object on an incline with no other forces acting upon it equals the y-component of the object s weight. The apparent weight of an object equals zero when the object moves vertically downward with an acceleration equal to gravity. Whenever an object is in translational equilibrium, the body is either at rest or moving at a constant speed, since the sum of the forces acting on the object equals zero. Newton's third law of motion states that for every acting force, there exists a reacting force of equal magnitude and opposite direction. (Example: For a box resting on a surface, the force of the earth pulling on the box and the box pulling on the earth are the action-reaction pair). Friction is a dissipative force caused by the irregularities present between surfaces. The direction of friction is always opposed to the direction of the actual or the impending relative motion of the body. The frictional force is proportional to the normal force (F N) which may not necessarily be equal to the object s weight. [F k=μ kf N]. The frictional force is dependent upon the nature/conditions of the surfaces, but nearly independent of the area of contact between the surfaces and the velocity of the object in contact with the surface. More force is required to initiate motion than to maintain motion for a given set of surfaces. The angle at which an object will start to move down an incline at a constant speed depends upon the coefficient of friction rather than the mass. ll objects regardless of their mass experience the same acceleration (-9.8m/s 2 ) when in a state of free fall. Newton s Law of Universal Gravitational Force states that gravitational force is directly proportional to the product of the masses and inversely proportional to the square of the distance between them. F=ma m=w/g where g = 9.8 What is the acceleration of a 245N block on a flat surface pushed horizontally by a 147N force if the coefficient of kinetic friction is 0.20 for the surface? For the 245N block, Fy=F N-245N=0 since there is no vertical motion, so the normal F N = 245N pplying F=ma where F= Fx yields: (note that the weight of 245N is divided by 9.8m/s 2 to yield mass in the solution) Fx=147N-(0.20)(245N)=(245N/9.8m/s 2 )a so a=3.92m/s 2 What is the acceleration of the system and the tension in the cable if the coefficient of friction for the surface is 0.1 in this case?
F x =0 and F y =0 F k =μ k N [nswer: 2.6m/s 2 and T=14.4N] 18kg box is moved up a 60 o incline plane at a constant speed by a horizontal force of 160N. What is the coefficient of friction? Note that 18kg mass must be converted to weight (18kg)(9.8m/s 2 )=180N Draw a free-body force diagram and label the forces. Fx=160N-μ sf N+180Ncos210 o =0 Fy=180sin210 o +F N=0 solving for F N yields F N=90N then substituting gives Fx=160N-μ s(90n)+180cos210 o =0, solving for μ s yields 0.046 What is the tension in each cable? [nswer: 300N, 260N] dditional Practice Problems 1. Which one of the following choices is equivalent to 2.0 m 2? (a) 2.0 10 4 cm 2 (b) 2.0 10 2 cm 2 (c) 2.0 10 3 cm 2 (d) 2.0 10 4 cm 2 (e) 2.0 10 2 cm 2 2. Two vectors and are added together to form a vector C. The relationship between the magnitudes of the vectors is given by + = C. Which one of the following statements concerning these vectors is true? (a) and must be displacements. (b) and must have equal lengths. (c) and must point in opposite directions. (d) and must point in the same direction. (e) and must be at right angles to each other. 3. student adds two displacement vectors with magnitudes of 3.0 m and 4.0 m, respectively. Which one of the following could not be a possible choice for the resultant? (a) 1.3 m (b) 5.0 m (c) 7.8 m (d) 3.3 m (e) 6.8 m 4. Which expression is false concerning the vectors shown in the sketch?
(a) C = + (b) C < + (c) C + = (d) 2 + 2 = C 2 (e) + + C = 0 5. force, F1, of magnitude 2.0 N and directed due east is exerted on an object. second force exerted on the object is F2 = 2.0 N, due north. What is the magnitude and direction of a third force, F3, which must be exerted on the object so that the resultant force is zero? (a) 1.4 N, 45 north of east (b) 2.8 N, 45 north of east (c) 4.0 N, 45 east of north (d) 1.4 N, 45 south of west (e) 2.8 N, 45 south of west 6. bird flies 25.0 m in the direction 55 east of south to its nest. The bird then flies 75.0 m in the direction 55 west of north. What are the northward and westward components of the resultant displacement of the bird from its nest? (a) 29 m, 41 m (b) 41 m, 29 m (c) 35 m, 35 m (d) 81 m, 57 m (e) 57 m, 81 m 7. During the first 18 minutes of a 1.0-hour trip, a car has an average speed of 11 m/s. What must the average speed of the car be during the last 42 minutes of the trip be if the car is to have an average speed of 21 m/s for the entire trip? (a) 21 m/s (b) 25 m/s (c) 29 m/s (d) 23 m/s (e) 27 m/s 8. n object moving along a straight line is decelerating. Which of the following statements concerning the object s acceleration is necessarily true? (a) The value of the acceleration is positive. (b) The direction of the acceleration is in the same direction as the displacement. (c) n object that is decelerating has a negative acceleration. (d) The direction of the acceleration is in the direction opposite to that of the velocity. (e) The acceleration changes as the object moves along the line. 9. car is stopped at a red traffic light. When the light turns to green, the car has a constant acceleration and crosses the 9.10m intersection in 2.47 s. What is the magnitude of the car s acceleration? (a) 1.77 m/s 2 (b) 3.60 m/s 2 (c) 9.80 m/s 2 (d) 2.98 m/s 2 (e) 7.36 m/s 2 10. 5.0-kg rock is dropped from rest down a vertical mine shaft. How long does it take for the rock to reach a depth of 79 m? (a) 2.8 s (b) 4.9 s (c) 4.0 s (d) 9.0 s (e) 1.0 s 11a. football is kicked with a speed of 18 m/s at an angle of 65 to the horizontal. How long is the football in the air? (a) 1.1 s (b) 2.0 s (c) 4.0 s (d) 1.6 s (e) 3.3 s 11b. How far does the football travel horizontally before it hits the ground? (a) 18 m (b) 36 m (c) 72 m (d) 25 m (e) 48 m 12a. 600N box rests on a 45 o inclined surface (u k=0.5). What is the normal force? (a) 424N (b) 600N (c) 325N (d) 255N (e) 732N 12b. What would be the frictional force if the surface were level? (a) 1200N (b) 300N (c) 450N (c) 625N (d) 180N 12c. If a 450N weight were attached by a horizontal cable passing over a frictionless pulley to the 600N block described above, what would be the acceleration of the system? (a) 3.1m/s 2 (b) 1.4m/s 2 (c) 8.0m/s 2 (d) 2.6m/s 2 (e) 9.8m/s 2 12d. What would be the tension in the cable when the 450N weight is attached to the 600N block? (a) 127N (b) 450N (c) 600N (d) 285N (e) 385N nswers: 1D, 2D, 3C, 4, 5E, 6, 7, 8D, 9D, 10C, 11a E, 11b D, 12a, 12b, 12c, 12d E