review for 2nd midterm - university of colorado boulderjcumalat/phys1110/lectures/lec23.pdf ·...

Review for 2nd Midterm

•  Midterm on Thursday at 7:30pm! Old exams available on website. Chapters 6–9 are covered. Go to same room as last time. – You are allowed one calculator and one double-

sided sheet of paper with hand written notes – Multiple choice plus two long answer questions – Test time from 7:30pm – 9:00pm. If you have a

conflict, then contact Professor Daniel Dessau. – 2 Sample exams have been posted on D2L!

Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/

Announcements:

Clicker Score Update •  Clickers are posted on D2L up to March 9.

00153C29 0CCDF332 0E8D199A 0F8C6AE9 26657437 32AD801F 33505A39 33506B08 33B10587 33BC9817

These clickers have scores, but are not registered to any students in the class. Please check D2L and if your clicker score is 0 and you have been using your clicker, then you need to register it.

Clicker Score Update

Have not yet thrown out lowest 5 scores!

How to Study for the Exam •  Read the Chapters in H+R; look at the Review &

Summary pages at the end of the Chapter. •  Review the Lectures and Concept Questions •  Try the two Sample Exams on D2L •  If you feel this is not enough – then find a study

partner and look at odd-numbered problems at the end of each chapter in H+R – don’t do the problems, rather see if you know how to start them.

•  Review CAPA problems, but these are often harder, than what you would see on an exam.

Chapter 6 – Force and Motion

•  Inclined planes, pulleys, and hanging weights –  Tension in a massless string is the same everywhere –  Acceleration on either side of taut string is the same – Remember to choose coordinate systems intelligently

•  Friction –  Static friction applies when no relative motion –  Kinetic friction when sliding is occurring

•  Circular motion. Since, •  Section 6.4 Drag Force and Terminal Speed (will not

be on this exam)

Clicker question 1 Set frequency to BA

A block slides down a plane inclined at an angle of 15 degrees. Its speed remains constant. What is the coefficient of kinetic friction?

A) sin 15° B) cos 15° C) tan 15° D) 1/sin 15° E) 1/tan 15°


A block slides down a plane inclined at an angle of 15 degrees. Its speed remains constant. What is the coefficient of kinetic friction?

A) sin 15° B) cos 15° C) tan 15° D) 1/sin 15° E) 1/tan 15°

N

mg

Ff

€

Ff = µkN

€

µk =Ff

N=mgsinθmgcosθ

= tanθ

Chapter 7 – Work and kinetic energy •  Work is force times displacement

–  where is the angle between and

– Can obtain from area under the curve of force versus displacement

•  Kinetic energy: •  Work–energy theorem: •  Power: •  Work done by gravitational force up = -mgh


An object is pushed 9 m along a path and the force varies as shown in the figure. The scale is 1 N/division vertically and 1 m/division horizontally. What is the work done by this force?

A) 12 J B) 26 J C) 32 J D) 42 J E) None of these


An object is pushed 9 m along a path and the force varies as shown in the figure. The scale is 1 N/division vertically and 1 m/division horizontally. What is the work done by this force?

A) 12 J B) 26 J C) 32 J D) 42 J E) None of these

If you plot force versus distance the area under the curve is work . Just count the number of boxes. Number = 9x2 + ½(4x3)+1/2(2x2) = 26


A 200 watt electric winch hauls a 10 kg load straight up with steady speed. If there is no friction, approximately how fast will the load move up?

A) 9.8 m/s B) 20 m/s C) 200 m/s D) 0.05 m/s E) None of these


A 200 watt electric winch hauls a 10 kg load straight up with steady speed. If there is no friction, approximately how fast will the load move up?

A) 9.8 m/s B) 20 m/s C) 200 m/s D) 0.05 m/s E) None of these

If we allow the force to be in any direction, then the instantaneous power is:

€

v =Pmg

=200W

10kg•9.8m /s2≈ 2m /s

Chapter 8 – Conservation of Energy •  For conservative forces the work done depends

only on position so can consider potential energy – Gravitational potential energy – Elastic potential energy

•  Conservation of energy •  Mechanical Energy Emech= K+U within it. •  The force associated with a potential energy:

– One dimension: – Three dimensions

Instantaneous Power = dE/dt


Tarzan, mass M, uses a rope to swing across a gap. The rope length is 100 m. He starts at rest, and loses 50 m in elevation while swinging in a circular path. What is the tension in the rope at the lowest point ?

A)  M B) 1.5 M g C) 2 M g D) 3 M g E) None of these.


Tarzan, mass M, uses a rope to swing across a gap. The rope length is 100 m. He starts at rest, and loses 50 m in elevation while swinging in a circular path. What is the tension in the rope at the lowest point ?

A)  M B) 1.5 M g C) 2 M g D) 3 M g E) None of these.

€

T −Mg = Ma =Mv 2

l

T =Mv 2

l+ Mg

At the bottom of the swing

€

Mgh + 0 =12Mv 2 = Mg l

2∴v 2 = gl

€

T = 2mg


A spring (height h1 off the ground) is initially compressed by amount x. It launches a mass m horizontally (from rest.) The mass flies through the air and hits a table at height h2. What is the final kinetic energy of the mass just before it hits?

A) mgh1 B) mg(h1-h2) C) mg(h1-h2) - (1/2)kx2 D) mg(h1-h2) + (1/2)kx2 E) None of these is the correct expression for the kinetic energy.


A spring (height h1 off the ground) is initially compressed by amount x. It launches a mass m horizontally (from rest.) The mass flies through the air and hits a table at height h2. What is the final kinetic energy of the mass just before it hits? A) mgh1 B) mg(h1-h2) C) mg(h1-h2) - (1/2)kx2 D) mg(h1-h2) + (1/2)kx2 E) None of these is the correct expression for the kinetic energy.

Initial Energy is all potential

Mgh1+(1/2)kx2

Final Energy is

Mgh2+ K.E. Solve for KE

Chapter 9 – Momentum •  Momentum . Newton’s 2nd law: •  Impulse: •  Impulse–momentum theorem: •  If no external forces act on a system then the total momentum

of the system is conserved –  The two objects involved in a collision can be considered a

system so total momentum is conserved –  Elastic Collisions conserve kinetic energy. –  Completely inelastic collisions, the objects have the same

final velocity after the collision. •  Can use the center-of-mass to determine the motion of a

system – understand examples in section 9-2 •  Thrust Equation: Ma=VreldM/dt if varying mass (ie.

rocket ejecting gases)


A ball bounces off the ceiling, as shown. (Neglect gravity and friction in this problem)

What is the direction of the impulse, ∆p, on the ball?

A) left B) right C) down D) up E) zero


A ball bounces off the ceiling, as shown. (Neglect gravity and friction in this problem)

What is the direction of the impulse, ∆p, on the ball?

A) left B) right C) down D) up E) zero

Vector subtraction:

€

Δ p = p 2 −

p 1

The horizontal component of momentum (px) does not change, while the vertical component goes from up to down so the change is down


A large (ordinary) ball, mass M=5m, speed v, strikes a smaller (ordinary) ball, mass m, which is initially at rest.

Could the big ball come to a complete stop, and the small ball takes off with speed 5v?

A) Yes, this could occur. B) No, it cannot occur because it would violate conservation of momentum.

C) No, it cannot occur because it would violate conservation of mechanical energy.


A large (ordinary) ball, mass M=5m, speed v, strikes a smaller (ordinary) ball, mass m, which is initially at rest.

Could the big ball come to a complete stop, and the small ball takes off with speed 5v?

A) Yes, this could occur. B) No, it cannot occur because it would violate conservation of momentum.

C) No, it cannot occur because it would violate conservation of mechanical energy.

€

5mv + 0 = 0 + 5mv125mv 2 ≠ 1

2m(5v)2

Conservation of Momentum

Conservation of Energy


A 2.0 kg firecracker slides across a frictionless ice rink with velocity v= 20. m/s . At time t=0, BOOM, it explodes into two identical pieces (each of mass 1.0 kg). One of the two chunks continues straight ahead with final velocity v’= 30. m/s . How much total energy did the chemical explosion add to the two pieces?

A) 900 J B) 500 J C) 450 J D) 400 J E) 100 J


A 2.0 kg firecracker slides across a frictionless ice rink with velocity v= 20. m/s . At time t=0, BOOM, it explodes into two identical pieces (each of mass 1.0 kg). One of the two chunks continues straight ahead with final velocity v’= 30. m/s . How much total energy did the chemical explosion add to the two pieces?

A) 900 J B) 500 J C) 450 J D) 400 J E) 100 J

€

mivi =mi

2v1 +

mi

2v2

20m /s =v12

+30m /s2

v1 =10m /s

Initial Energy =

€

12(2kg)(20m /s)2 = 400J

Final Energy =

€

12(1kg)(10m /s)2 +

12(1kg)(30m /s)2

Final – Initial Energy = 500J - 400J = 100J

review for 2nd midterm - university of colorado boulderjcumalat/phys1110/lectures/lec23.pdf ·...

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