Tutorial work - chapters 2, 4, 5 practice PDF

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Physics 121

Chapter 2- Practice Questions

1) A car is making a 12-mile trip. It travels the first 6.0 miles at 30 miles per hour and the last 6.0 miles at 60 miles per hour. What is the car's average speed for the entire trip? A) 35 mph B) 40 mph C) 45 mph D) 50 mph 2) The position of a particle as a function of time is: x (t) = (3.1 m/s) t - (4.2 m/s2) t2. What is the average velocity of the particle between t = 1.0 s and t = 2.0 s? A) -11.7 m/s B) 11.7 m/s C) -9.5 m/s D) 9.5 m/s 3) A car is moving with a constant acceleration. At t = 5.0 s its velocity is 8.0 m/s and at t = 8.0 s its velocity is 12.0 m/s. What is the distance traveled in that interval of time? A) 10 m B) 20 m C) 30 m D) 40 m 4) Car A is traveling at 18.0 m/s and car B at 25.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with an acceleration of 1.80 m/s2. How long does it take car A to overtake car B? A) 5.50 s B) 12.6 s C) 22.6 s D) Car A never overtakes car B. 5) Two objects are dropped from a bridge, an interval of 1.00 s apart. What is their separation 1.00 s after the second object is released? A) 4.90 m B) 9.80 m C) 14.7 m D) 19.8 m 6) From the edge of a roof top you toss a green ball upwards with initial velocity v0 and a blue ball downwards with the same initial velocity. When they reach the ground below, A) the green ball will be moving faster than the blue ball. B) the blue ball will be moving faster than the green ball. C) the two balls will have the same speed. D) None of the previous choices is correct.

Answers: 1- B Average velocity is total distance divided by total time. Find time for first part 6/30 = 0.2hr and for the second part 6/60= 0.1 hr, then average velocity is 12/0.3 = 40 mph 2- C x (t) = (3.1 m/s) t - (4.2 m/s2) t2 , x(1) = 3.1 – 4.2 = -1.1 m, x(2) = 3.1(2) – 4.2 (2)2 = -10.6m, Vave = [-10.6 – (-1.1)] / (2s-1s) = -9.5 m/s 3- C xf = xi + vi Δt + ½ a(Δt)2 , but first we need to find a = Δv/Δt = 4/3 m/s 2. Then total x traveled = vi Δt + ½ a (Δt)2 = 8 (3) + ½ (4/3) (3)2 = 30 m. 4- C To overtake they both should be at the same x at the same t. We can choose the origin to be at the location of car A: xA = xi + vi Δt + ½ a(Δt)2 = 0 + 18 t + ½ (1.8) t2 xB = xi + vi Δt = 300 + 25 t xA = xB , or, 18 t + 0.9 t2 = 300 + 25 t, rearranging: 0.9 t2 - 7 t - 300 = 0, solving this quadratic equation gives: t = 22.6 s (acceptable answer) or -14. 8 s (not acceptable) Using this time we can also find x traveled during this time (from either one of the equations): xB = xi + vi Δt = 300 + 25 t = 300 + 25 (22.6) = 865 m 5- C One second after 2nd stone is released, it has traveled: xA = ½ a(Δt)2 = 4.9 m, at this time the first stone has been traveling for 2 seconds and had traveled: 4.9 (4) = 19.6 m. Their separation is therefore: 19.6 – 4.9 = 14.7 m. 6- C

Physics 121

Chapter 2 – Practice 2

1) A speeding motorist traveling 30 m/s passes a stationary police officer. The officer immediately begins pursuit at a constant acceleration of 3 m/s2. a) How much time will it take for the police officer to reach the speeder? b) What is the velocity of officer when he catches the speeder? c) Plot two graphs, show the displacement vs. time for both motorist and police in one graph and velocity vs. time for both of them in a the other graph.

2) At a certain instant, a ball is thrown downward with a velocity of 8.0 m/s from a height of 40 m. At the same instant, another ball is thrown upward from ground level directly in line with the first ball with a velocity of 12 m/s. a) Find the time when the balls collide. b) Find the height at which they collide. c) In what direction the second ball is traveling when they collide.

3) An object starts from rest (at the origin) and moves in one dimension, along the x-axis. Below graph shows the velocity of the object as a function of time. Sketch two graphs, showing the acceleration (first graph) and position of the object as functions of time. You graphs should show the magnitudes for acceleration for the three regions shown in velocity graph (t < 3s, 3 s < t < 5 s, and t > 5s), and also the position of the object at 3, 5, 8 and 11 seconds.

Solution: 1- For motorist with constant speed: x= v t, For officer moving with constant acceleration: x = ½ a t2 a) Police officer reaches the speeder when their x is equal: v t = ½ a t 2 solve for t, t = 0 in one answer ( when the motorist passes the officer initially) and t = 20 sec, is the answer we are looking for. b) V = a x t = 60 m/s for police officer when they meet.

2) a) x = ½ a t2+ v0 t + x0 , for the first one v0 = -8 m/s ( downward) , and x0 is +40m, if the origin is assumed at the ground, for the second ball, x0 = 0, and v0 = 12 m/s. Also a = -9.8m/s2 . At the point they meet x for both of them should be the same: -4.9 t2 - 8 t + 40 = - 4.9 t2 + 12 t , or 20 t = 40 and t = 2 sec. b) Put time = 2 sec in any of the equations you wrote in part a for balls, and you would get x about 4 m. c) Find velocity of the second ball at t = 2 sec , using, v= at + v0 , or, v ~ -8 m/s, from its sign we can say it is moving downward. (or, find the time for the second ball to get to maximum height, it will be less than 2sec, which means at 2 sec it had reached it maximum height already and is moving downward) 3) From the data given on velocity graph, you can find the slope of each segment (magnitude of acceleration) and the area under the line (magnitude of position) and plot a and x graph accurately:

Physics 121

Chapter 4 – Practice

1- Pictured below are two possible trajectories of a golf shot. (a)State which shot will stay in the air the longest and (b) which will have the higher launch speed. Completely explain your reasoning for full credit.

2- A ball is initially 1.5m above the horizontal ground and 10m horizontally from a brick wall. The ball is thrown toward the wall at an angle of 40° with respect to the ground. It takes 1.0 second for the ball to reach the wall. a. What is the initial speed of the ball? b. At what distance above the ground does the ball hit the wall? c. What is its velocity (magnitude and direction) when it hits the wall?

3- A ball is kicked from the ground such that it spends 4.00s in the air and travels 40.0m horizontally at landing. Find its initial velocity of the ball (magnitude and direction).

4- A stone is thrown from a cliff with an initial velocity of 2.0 m/s at 30° below the horizontal. It lands 1.0m out from the base of the cliff. Find the height of the cliff.

5- A ball of mass 0.2 kg on the end of a string is revolved at a uniform rate of 4.0 m/s in a vertical circle of radius 50 cm. Find the difference between the tensions in the string when the ball is (a) at the top of its path, and (b) at the bottom of its path.

Solution: 1- (a)Both balls will remain in the air the same amount of time. They both go up to the same

height and the vertical motion of a projectile is independent of the horizontal motion. The time in the air doesn’t depend upon the horizontal motion, only the vertical motion. (b)Ball B had the higher launch speed. Since they go to the same height, they both must have had the same vertical component of the initial velocity. Since ball B goes the furthest horizontally, it must have had the higher horizontal component. Adding the components of these vectors will give a higher launch speed for B. 2- a. Form x direction equations: 10m = V0 cos 40 (1s), which gives V0 = 13m/s. b.In y direction: y = (13m/s) sin 40 (1s) + ½ ( -9.81 m/s2) ( 1 s) 2 = 3.5 m ,from the initial point which is 3.5 m + 1.5 m = 5 m from the ground. c. Vx= V0 cos 40 = 10 m/s, Vy = at + V 0y = (-9.81 m/s2) (1 s) + 13 sin 40 = - 1.45 m/s, From vector sum of these components: V = 10.1 m/s, at an angle 8º below the horizontal which means it is coming down. 3- 40m = V0x (4s), V0x = 10 m/s. At the point of landing y = 0 = 1/2 (-9.8) (4)2 + V0y (4), gives V0y = 19.6 m/s, therefore V0 = √(102 + 19.62) = 22 m/s, at tan-1 (19.6/10) = 63°. 4- V0x = 2.0cos 30° = 1.73 m/s, V0y = -2 sin 30° = -1.0 m/s x = V0x t, or, t = 1/1.73 = 0.58s y = (-1)(0.58) – (1/2) (9.8)(0.58)2 = - 2.2 m. Height of the cliff is 2.2m. 5- At the top both T and mg are toward center: Ttop+mg= mv2/r, or, Ttop= mv2/r - mg At the bottom Tbottom-mg= mv2/r, or, Tbottom= mv2/r + mg The difference is Tbottom – Ttop = mv2/r + mg – (mv2/r – mg) = 2mg= 2(0.2kg) (9.8 m/s2) = 3.92 N.

Physics 121

Chapter 5- Practice

1- A 1.0 kg box on a 30º incline is connected to a 3.0 kg box on a horizontal surface. The coefficient of kinetic friction between the boxes and the surfaces is 0.15. The pulley is frictionless and massless. A 4.0 N horizontal force (F) is pushing the 3.0 kg box to the right, as shown below. Find the acceleration of the boxes and the tension in the string.

2- A roller-coaster vehicle has a mass of 500 kg when fully loaded with passengers. a) If the vehicle has a speed of 20.0 m/s at point A, what is the force exerted by the track on the vehicle at this point? b) What is the maximum speed the vehicle can have at B and still remain on the track?

3- A 10 kg block is attached to a wall and also placed on top of a 40 kg block. The coefficients of static friction are 0.30 between the floor and the 40 kg block and 0.15 between two blocks. Find the minimum force F needs to move the 40 kg block.

4- A 2.5 kg mass on a 16.3° ramp is connected to a hanging 2 kg mass using massless pulley and string. The coefficient of kinetic friction between the 2.5 kg mass and the plane is 1/6. Find (a) the acceleration of the blocks and (b) the tension in the string.

Solution: 1-

We need to write forces in the direction of motion for each box, and find friction from its formula using normal force. Pulley has no weight, tensions in the string connecting the boxes are the same (T). For the box on the incline: m1g sin30˚-T-f1 = m1a, N1= m1g cos30˚, f1 = µN1= µ m1g cos30˚ Combining these we have: m1g sin30˚-T- µ m1g cos30˚ = m1a (I) For the box on the horizontal surface: F + T – f2 = m2a, N2= m2g, f2 = µN2= µ m2g, Or: F + T - µ m2g = m2a (II) (I)+ (II) gives: F+ m1g sin30˚-- µ m1g cos30˚- µ m2g = (m1+ m2) a a= (F+ m1g sin30˚-- µ m1g cos30˚- µ m2g)/ (m1+ m2) = (4+ 1x 9.8 x0.5-0.15x 1x 9.8x 0.87- 0.15x3x9.8)/4 = 0.8m/s2. T can be found by putting a in one of (I) or (II): F+T- µ m2g = m2a, T= m2a +µ m2g –F = 3x0.8 + 0.15x3x9.8- 4= 2.8N. 2- a) At point p, The forces on mass m, the roller-coaster vehicle, are its weight mg pointing down and the force the track exerts N pointing up. Therefore, N-mg = mv2/r , or, N= mv2/r + mg = 500 kg [((20m/s)2/10)+ 9.8m/s2]= 24900N (which is about five times of its weight).

b) To "still remain on the track" means the normal force has just gone to zero, N = 0. The only force toward center is weight: mg = mv2/r, gives v = √(rg) = 12 m/s. 3) There are two frictional forces f1 (between m1 and floor), and f2 (between m1 and m2) both in opposite direction with respect to F: F – f1 –f2 = ma Minimum force means overcoming static friction and start to move, a=0. F = f 1 + f 2 = µ 1 N 1 + µ 2N 2 N1 is normal force on the floor, both masses are pushing to the floor: N 1 = (m1 + m2) g, N2 is normal force on m1, which is weight of m2: N2 = m2 g. F = µ 1(m1 + m2) g + µ 2m2 g = 0.3(50kg)(9.8m/s2) + 0.15(10 kg)(9.8 m/s2) = 162 N.

4) a) First isolate m1, write forces in two axis, one in the direction of motion and one perpendicular: T – m1g sin16.3˚ – f = m1a, or, N – m1g cos 16.3˚ = 0, f= µm1g cos 16.3˚ Gives: T – m1g sin 16.3˚ – µm1g cos16.3˚ = m1a (I) Now isolate m2: m2g – T = m2a (II) Add both sides of (I) and (II): m2g – m1g sin 16.3˚ – µm1g cos16.3˚ = (m1 + m2) a or, a = (19.6 N - 6.88 N – 3.92N)/ 4.5 kg = 1.96 m/s2. Put acceleration in (I) or (II) to find T: 19.6N – T = 2 (1.96 m/s2), T = 15.68 N....