Physics 3A: Midterm C Solutions - Useful Equations and Constants, Exams of Physics

Download Physics 3A: Midterm C Solutions - Useful Equations and Constants and more Exams Physics in PDF only on Docsity! Physics 3A: Basic Physics I Shoup – Midterm C : Solutions Useful Equations Ax A cos
 Ay A sin
  tan  1
AyAx  A
Ax2 Ay2  A Ax i Ay j Az k A B  A  B  cos   A B  A  B  sin ! A "#B Ax Bx Ay By Az Bz A $#B
Ay Bz % Az By  i
Az Bx % Ax Bz  j
Ax By % Ay Bx  k v d& t 'vx ( & x& t vx ( lim) t * 0 & x& t dx dt xf + x i , vx t -a ( & v& t ax ( lim) t * 0 & vx& t d v x d t vxf + v xi , ax t xf x i 12
vxi vxf  t xf x i vx t 12 ax t 2 v f 2 v i2 2a
x f % x i  .r x /i y 0 j vxf  vi cos  i  constant vyf  vi sin  i 21 g t x f  x i 3 vi cos  i  t y f yi vi cos
 i  t % 12 g t 2 ac v2 r at d 4 5v 4 d t y f
tan
 i 6 x f %
g2 vi2 cos2
7 i   x f 2 T 2 8 r v h vi 2 sin2
 i  2 g R vi 2 sin
2  i  g ar % ac % v 2 r 9vPO ' ;:vPO %<vO ' O = >F 0 ?F12 @1@AF21 BF g m -g = >F m -a m1m2 ( a2 a1 Useful Constants g 9.80 m C s2 32.0 ft C s2 c 2.99 x108m C s Physics 3A: Basic Physics I Shoup – Midterm C : Solutions Name:XXXXXXXXXXXXXXXXXX_____ Student ID #:_________________________ Discussion Section:____________________ Date:________________________________ Signature:____________________________ (circle the letter of your answer) 1. (5 pts) In the computation
47.787 % 42.1 
5999.6 % 874.326  $ 0.00375 what is the correct answer, giving the proper number of significant figures (at each stage of the calculation, apply significant figures)? a.) 0.000004125 b.) 0.000004161 c.) 0.00000413 d.) 0.0000041 e.) 0.0000042 2. (5 pts) After a long night of studying for your midterms, you leave a stack of three books on your desk, Physics on the bottom, Chemistry in the middle, and Biology on top. Order these books by the magnitude of their vertical normal force, smallest magnitude first: a.) Chemistry, Biology, Physics b.) Biology, Chemistry, Physics c.) Physics, Chemistry, Biology d.) All have the same vertical normal force. e.) Not enough information to tell. 3. (5 pts) Consider a particle moving along the postive x-axis (where positive x is right, and positive y is up). If the particle's acceleration in y is positive and its acceleration in x is zero, then a.) the particle's speed will increase. b.) the particle's velocity will “turn” downward c.) the particle's velocity will “turn” upward d.) the particle's speed will decrease. e.) None of the above. 11.(5 pts) If
A % #B  " C 0, and A #B , then which of the follow statements is true? a.) The component of A in the direction of C is equal to the component of B in the direction of
C . b.) The magnitudes of A and B are equal. c.) A is perpendicular to B d.) all of the above are true e.) None of the above are true. 12.(5 pts) A ball is attached to one end of a string. The other end of the string is attached to a spool (like a spool of thread). If the ball is going around the spool with a constant speed so that the string is winding onto the spool, then the acceleration of the ball is a.) increasing b.) decreasing c.) stays the same d.) not enough information to tell e.) none of the above. 13.(20 pts) Consider the situation in the figure below, where the inclined plane is frictionless, the rope is massless and non-stretching, and the pully is frictionless and massless. Assume m1 is 12.5 kg, and theta is 45.0o. a.) What should m2 be so that m1 does not move? b.) What is the tension in the rope with this value of m2? c.) Would your result for a.) change if the inclined plane were located on the moon? Explain why or why not. d.) Finally, assume the rope breaks. What is the acceleration of m1 and m2? (circle your final answers and show all work) b.) From a.) above: T m2g
m2a m2g  9.80   8.84  86.6N c.) No, because g does not enter into the equation for m2. d.) When the rope breaks, m2 would be in free fall, so: a  g  9.80m  s2 for m1: m1g sin  m1a so a  g sin   9.80sin 45  6.93m  s2   14.(20 pts) At a pistol shooting range a policeman fires a gun, aiming horizontally, directly at the center of a block of wood 18.2 m away (see diagram below). The block is nailed to a table so it can't move. The bullet strikes the wood block and comes to rest inside the block, 12.5 cm from the front. If the initial speed of the bullet when it left the gun was 500. m/s and the bullet's mass is 75.0 g, then compute a.) the stopping acceleration of the bullet, b.) the average force of the wooden block on the bullet, c.) the time the bullet was in the air before hitting the block, and d.) how far below the center of the front of the wood block did the bullet strike. (circle your final answers and show all work) a.) concept: constant acceleration (de-acceleration). Knowns: x
12.5cm
0.125m , v i
500.0 m  s , v f
0.0 m  s , so use: v f 2
v i2  2a x a
v f 2  v i2 2 x
0  500.2 2  0.125 
 1000000
 1.00 x 106 m  s2 b.) Force which causes above a? Use F = ma: F  ma   0.075 kg  1.00 x106  75000 N c.) concept: projectile motion, in x direction, no acceleration during flight use: xf
x i  v ix t or x
v ix t so t
x v ix
18.2 500.
0.0364 s d.) concept: projectile motion, in y direction: use: y f  y i  v iy t  12 g t 2 or  y  v iy t  12 g t 2   12 g t 2   12  9.80   0.0364  2  0.00649 m. ! #"$&%'(*),+