1

ANSWER A

MIDTERM #2, PHYS 1211 (INTRODUCTORY PHYSICS), October 30, 2017 NAME: STUDENT ID: This exam book has 5 pages including an equation sheet on the last page PART I: MULTIPLE CHOICE QUESTIONS (question 1 to 4) For each question circle the correct answer (a,b,c,d or e).

1. (2.5 points) Diagram below shows three projectiles. Select the one correct statement. a) Since they reach the same maximum height, they all have the same initial speed v0 . b) Since Projectile 1 covers the least distance, it spends the least amount of time in the air. c) Projectile 1 has the greatest initial speed, since it spends the least amount of time in the air.

d) Since they reach the same maximum height, the y-component of the initial velocity v0yare the same for all projectiles. e) Projectile 3, which travels the farthest horizontally, spends the longest time in the air.

2. (2.5 points) A boy on the edge of a 20 m high cliff throws a stone horizontally with a

speed of 20 m/s. It strikes the ground at what distance from the foot of the cliff? Use

!!g=10m/ s2 .

a) 10 m b) 40 m c) 50 m d) !!50 5m e) none of these Since the initial velocity is horizontal, the time the stone takes to fall to the ground is

!! 20m= 1/2( ) 10mi s−2( )t2→ t =2s . The horizontal component of the velocity is

constant, which gives a distance !!d =20m/ s ×2s = 40m . ANSWER B 3. (2.5 points) A 1000-kg plane moves in a straight line at constant speed. The force of air

friction is 1800 N. The net force on the plane is:

A) 0N B) 11600 N C) 1800 N D) 9800 N E) None of these 4. (2.5 points) mid2 2015 A crate rests on a horizontal surface and a woman pulls on it with

a 10-N force. No matter what the orientation of the force, the crate does not move. Rank the situations shown below according to the magnitude of the frictional force of the surface on the crate, least to greatest

A) 1, 2, 3 B) 2, 3, 1 C) 2, 1, 3 D) 1, 3, 2 E) 3, 2, 1

ANSWER E

ANSWER D

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PART II: FULL ANSWER QUESTIONS (question 5 to 7) Do all four questions on the provided space. Show all work.

5. (10 points) In the figure, you throw a ball toward a wall at speed 24.0 m/s and at angle θ0 = 43.0˚ above the horizontal. The wall is d = 16.0 m from the release point of the ball.

(a) Calculate the initial horizontal (x) and vertical (y) components of its velocity. Hence calculate how far above the release point does the ball hit the wall?

vox = 24

mscos43! = 17.55m

s, voy = 24

mssin 43! = 16.37m

s. (1 point)

The time to the wall can be calculated from the horizontal component by the equation

d = voxt→ t = 16m

17.55m i s−1= 0.91s (2 points). Taking up as positive, and launch point as

yo = 0 the y-component of the position is

y = voyt −

12gt 2 = 16.37m i s( ) 0.91s( )− 1

29.8 m

s2⎛⎝⎜

⎞⎠⎟ 0.91s( )2 = 10.84m (2 points)

So it hits the wall 10.84 m above the release point.

(b) What are the horizontal and vertical components of its velocity as it hits the wall?

x-component is constant vox = 17.55ms

(1 point)

y-component vy = voy − gt = 16.37ms− 9.8 m

s2⎛⎝⎜

⎞⎠⎟ 0.91s( ) = 7.45m

s (2 point)

(c) Based on the results of part b, plot a motion diagram of the trajectory that labels the x- and y- coordinate system, as well the direction of the velocity when the ball hits the wall. Since vy > 0 , the ball has not reached its maximum height, so the trajectory is as below: +y

!v

!v (2 points)

!v

3

6. (10 points) In the diagram below block A ( mA = 2.0kg ) is on a frictionless surface. Block B ( mB = 1.5kg ) rests on block A with the coefficient of friction (between block A and B) beingµk = 0.33 (kinetic) and µs = 0.45 (static). Block B is being pulled by a force F =10N that causes an acceleration, a.

F = 10N a

a) Assume for this part that block B does not slip and fall off block A. Draw a free body diagram of all forces acting on the composite block A + B system. Hence determine the acceleration, a, of block B (and A).

!FN a

(2 points) F = 10N F = mA +mB( )a→ a = 10N ÷ 3.5kg( ) = 2.86m i s−2 (1 point)

!Fg,A+B

b) Draw a free-body diagram of all forces acting on block A. Assume that the system accelerates with the acceleration of part a). Using the diagram calculate the direction and magnitude of the force of friction acting on block A. Is this friction force static or kinetic? Explain your answer. FN ,ground ≡ normal force of ground on A (2 points) a = 2.86m i s−2 fS

AB ≡ static friction on A due to surface B second law fs

AB = mAa = 2kg × 2.86m i s−2 = 5.72N Fg,A = mAg (1 point) Fg,B = mBg Since the ground is frictionless, the only force that can accelerate block A is friction, and if we assume that B do not slip on A, then the friction must be static. (1 point) The normal force associated with the surface between A and B is the weight of B Fg,B = mBg = 14.7N . Hence the maximum possible static friction is

fSAB,Max = Fg,Bµs = 14.7N × 0.45 = 6.615N , which is greater than the actual static friction

fsAB = 5.72N . Hence there will be no slipping, and the friction is indeed static.

Finally note that the 10N force acts on B, and not on A. (1 point) c)Using Newton’s third law, calculate the magnitude and direction of the force of friction acting on block B. Since the surface of B exert a force to the right on block A of the magnitude fs

AB = 5.72N . Hence using Newton’s third law the surface of A must exert a force on block B to the left of magnitude fs

BA = fsAB = 5.72N (2 points)

A

B

A + B

A

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HINT: 1) Assume that the acceleration is down incline; 2) Note that the F = 10 N force is in the same direction as gravity (vertical)

7. (10 points) Below a downward force, F = 10 N is applied to a 5 kg crate moving up a 30! incline. The crate is connected to a hanging 2kg box by a rope through a frictionless pulley. Coefficient of friction between crate and incline are µk = 0.1 and µs = 0.2.

F = 10N v 30°

A) Draw free-body diagram (FBD) that includes all forces on the 5kg crate. Calculate the normal force, and the friction force on the crate. Use Newton’s second law to find an equation that relates the acceleration, a, and the tension, T, in the rope. +y T Gravity Fg1 = 5kg × 9.8 m•s-2 = 49 N a Fg1,x = Fg1 sin 30°=24.5N v fk Fg1,y = −Fg1 cos 30°=−42.4N(0.5point) F = 10 N Force

+x down incline Fx = 10 N sin 30° = 5 N (2 points) θ Fy = −10N sin 30° = −8.7 N (0.5 point) Y-component θ = 30° Fy

Net = Fy +Fg1,y +FN =0

F θ FN − 42.4 N − 8.7 N = 0 FN = 51.1 N (0.5 point)

Kinetic Friction down (+x) to oppose motion: fk = FNµk =5.11N (0.5 point)

X-component:Fxnet = Fg1,x +Fx + fk −T =m1a

24.5N+5N+5.11N − T = (5kg) a, 34.61N − T = (5kg) a (1) (1 point) B) Draw a FBD on the hanging 2kg box. Use Newton’s second law to find a equation that relates the acceleration, a, and the tension, T, in the rope. Combine this equation with the one found in part a, and solve to find a and T.

FBD of 2 kg box Since acceleration of 5 kg crate is down incline, the acceleration of +y 2 kg box must be up incline

Use 2nd law, FyNet =T −m2g=m2a

T a T −19.6N = 2kg( )a (2), (1point) Add equation 1 by 2, (1) +(2)

(1point) 15N =7kg×a→a=2.14mi s−2 (1point)

Substituting into (2), T =19.6N + 2kg( )a=23.9N (or use equation 1)

Fg2 = m2g = 2kg×9.8 m•s-2 = 19.6 N (1 point) C) The crate is initially traveling at 5 m/s up incline. What is its velocity after 2 seconds? Use v = v0 +at = −5mi s−1 +2.14mi s−2 2s( ) = −0.72mi s−1 (up incline) (1 point)

Fy Fg1

Fx

2 kg

5 kg

Fg1,y

Fg1,x

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USEFUL EQUATIONS

!A = Ax

⌢i + Ay

⌢j + Az

⌢k , unit vector notation.

Kinematics x = x0 + v0xt +12axt

2 , vx = v0x + axt , vx2 = v0x

2 + 2ax x − x0( )

2/8.9 smg = , Fg = mg Centripetal acceleration arad =v2

r

Free Fall g= 9.8m/ s2 , with +y up, y = y0 + v0t −

12gt 2 , v = v0 − gt , v2 = v02 − 2g y − y0( )

Solution of quadratic equation, ax2 + bx + c = 0→ x =−b ± b2 − 4ac

2a

Newton’s Laws 0=∑= FF net

!! (Object in equilibrium)

amF net !!= (Nonzero net force); third law, an applied force on object A by object B will induce

an equal (in magnitude) and opposite (in direction) force on object B by object A. Friction fs ≤ µsFN , fk = µkFN .