# PHY 131 Study Guide - Midterm Guide: Mississippi Highway 1, Arc Welding, Microwave OvenExamPremium

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**preview**shows pages 1-3. to view the full**12 pages of the document.**PHY131 Midterm 2 — Extra, Exam XXXXXXXXX (form: 1) 1

PHY131 Spring 2017, Midterm 2

Extra, Exam

XXXXXXXXX

Academic Integrity Statement

Please sign on side two of the answer sheet (over ”General Purpose Answer Sheet”). Your signature

acknowledges that your work on this exam has been done in accordance with an expectation of

academic honesty.

Fine Print

This is a closed book exam. You may use one sheet of notes - 81

2×11 inches, double sided, handwritten. ANY

manipulation of electronic equipment (iPod, iWatch, phone, laptop computer, taser, arc welder, microwave oven,

etc.) other than your calculator is strictly prohibited, and will result in a zero on this exam.

Your name and student identiﬁcation number are printed on this exam. Before you start, enter your name and

student number on the answer sheet. Mark 1 in the “Test Version” box. Answer all problems on the bubble scan

sheet. If you think there is more than one correct answer for a given question, bubble in all of the correct answers.

Some potentially useful expressions and constants are:

proton mass 1.67 ×10−27 kg neutron mass 1.67 ×10−27 kg

electron mass 9.11 ×10−31 kg electron charge 1.6×10−19 C

speed of light 3.0×108m·s−1gravity acceleration 9.8 m·s−2

derivative of a power d

dx (xn) = nxn−1integral of a power Rxndx =1

n+1 xn+1

Each question carries equal weight. Answer each problem with the answer closest to the correct answer. When

mistakes are found, the problematic question will be excluded from the score.

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PHY131 Midterm 2 — Extra, Exam XXXXXXXXX (form: 1) 2

1. An mb= 19 g bullet traveling vi= 3.5×102m·s−1penetrates an mB= 2.0 kg block of wood

sitting on a frictionless table and emerges going vf= 1.6×102m·s−1(this is an inelastic collision).

If the block is stationary when the bullet hits, how fast is it traveling as the bullet exits?

A. 1.8 m·s−1

B. 1.8×103m·s−1

C. 0.017 m·s−1

D. 19 m·s−1

E. 58 m·s−1

SOLUTION: Momentum is conserved, so Pi=Pfand we can write that

mbvi=mbvf+mBv↔v=mb

mB

(vi−vf)(1)

v=(19 g)(0.001 kg·g−1)

(2.0 kg) (3.5×102m·s−1)−(1.6×102m·s−1)= 1.8 m·s−1(2)

2. A ball with mass m= 1.83 kg, radius r= 0.100 m and moment of inertia I= 0.00800 kg·m2is

initially at rest and rolls without slipping down a ramp sloping at 25.0◦to the horizontal. What is

the magnitude of the acceleration?

A. 5.18 m·s−2

B. 7.58 m·s−2

C. 2.88 m·s−2

D. 4.14 m·s−2

E. 9.47 m·s−2

SOLUTION: The motion of the ball is described by

ma =X~

F(3)

Iα =X~τ (4)

and since the ball is rolling without slipping, α=ar/r, with a torque of τ=rFrwhere

arand Frare parallel to the ramp. Therefore

mar=mg sin θ−Fr(5)

Iar

r=rFr(6)

Solving for the acceleration gives

ar=mgr2sin θ

mr2+I=(1.83 kg)(9.8 m·s−2)(0.100 m)2(sin 25.0◦)

((1.83 kg)(0.100 m)2+ (0.00800 kg·m2)) = 2.88 m·s−2(7)

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PHY131 Midterm 2 — Extra, Exam XXXXXXXXX (form: 1) 3

3. Given ~

A= (0.56,0.56,2.8) and ~

B= (−5.0,−3.9,−1.7). What is the dot product of ~

Aand ~

B?

A. 2.9

B. 7.2

C. −9.7

D. (10,−13,0.62)

E. (−0.62,2.6,−2.0)

SOLUTION: The dot product is

~

A◦~

B=AxBx+AyBy+AzBz= (0.56)(−5.0) + (0.56)(−3.9) + (2.8)(−1.7) = −9.7(8)

4. Given ~

A= (−3.9,−5.0,−0.56) and ~

B= (−5.0,−0.56,2.8). What is the cross product, ~

A×~

B?

A. 21

B. 8.6

C. (−14,14,−23)

D. (17,4.4,−21)

E. −23

SOLUTION: The cross product is

[(−5.0)(2.8)−(−0.56)(−0.56)]ˆ

i+[(−0.56)(−5.0)−(−3.9)(2.8)]ˆ

j+[(−3.9)(−0.56)−(−5.0)(−5.0)]ˆ

k

(9)

(−14,14,−23) (10)

5. A crate is on a ramp tilted at θ= 40.2◦to the horizontal, with µk= 0.100. What is the

acceleration of the crate?

A. 6.85 m·s−2

B. 6.64 m·s−2

C. 7.07 m·s−2

D. 5.58 m·s−2

E. 6.33 m·s−2

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