PHYSICS 231 Review problems for midterm 1

Transcription

1 PHYSICS 231 Review problems for midterm 1

2 Some Housekeeping The 1 st exam will be Wednesday October 3. The exam will take place right here in BPS We will have assigned seating, so show up early. You need to show up 15 min early to guarantee the full 50 min exam time. If you are sick, you MUST have a note from a doctor on the doctor s letterhead or prescription pad. RCPD forms cannot be turned in for the 1 st time on Oct 3. You CANNOT use cell phones during the exam. You can (should!) bring an equation sheet on a 8.5x11 sheet of paper. 2

3 Clicker Quiz! A) This is an appropriate way to solve physics problems. B) This will probably not full credit. 3

4 Velocity (m/s) v(t)=v(0)+at x(t)=x(0)+v(0)t+0.5at 2 Cut problem up in 1s pieces After 1 s: v(1)=0+0x1=0 x(1)=0+0x1+0.5x0x1 2 =0 After 2 s: v(2)=v(1)+at=0+3x1=3 x(2)=x(1)+v(1)t+0.5at 2 =0+0x1+0.5x3x1 2 =1.5 What is the displacement at t=4 s. By drawing: Derive v(t) diagram from a(t) diagram: red line x(t) is area under v(t) diagram: After 3 s: v(3)=v(2)+at=3+2x1=5 x(3)=x(2)+v(2)t+0.5at 2 =1.5+3x1+0.5x2x1 2 =5.5 After 4 s: v(4)=v(3)+at=5-2x1=3 x(4)=x(3)+v(3)t+0.5at 2 =5.5+5x1+0.5x(-2)x1 2 =9.5 4

5 Cross straight a) v? 6.50 m/s Cross fast b) 6.50 m/s To cross fast: use picture b) V=6.5 m/s so t=x/v=40.3 s (but lands downstream) 3.30 m/s 3.30 m/s 3.30 m/s To cross straight: use picture a) Velocity of water Velocity of boat needed to cancel motion of water Total velocity of the boat (I.e. available in still water) Velocity left over for crossing river v = so v= ( )=5.6 m/s Time to cross river: t=x/v=262.0/5.6=46.8 s (use v=x/t) 5

6 + a) Acceleration in vertical direction is ALWAYS g= 9.81 m/s 2 ; TRUE b) See a) FALSE c) It is positive going up, negative going down: FALSE d) It is zero at the start and end and positive everywhere else: TRUE 6

7 Clicker Quiz! What time should you arrive at BPS 1410 for your exam in order to guarantee that you ll have the full 50 minutes to take the exam? A) 9:20 AM B) 9:10 AM C) 9:00 AM D) 8:55 AM E) None, I m going to BPS 1415 for my exam

8 v x (0)=29.0 m/s 2.19 m x(t)=x 0 +v 0x t = 0+29t=29t v x (t)=v 0x = 29 y(t)=y 0 +v 0y t-0.5gt 2 = t-0.5x9.8t 2 = t 2 v y (t)=v 0y -gt = 0-9.8t=-9.8t When ball hits ground: y(t)=0 so: t 2 =0 t=0.67s Use in x(t) equation: x(0.67)=29 x 0.67 = 19.4 m 8

9 mg T=15N 1) Draw forces 2) Since the block is not moving, no acceleration, no net force Note that the tension in the lower block must be trying to pull The block down-the rope cannot support weight and so not Produce an upward force by itself T upper -mg-t lower =0 so T upper =4.3x = =57.2 N 9

10 F g// =F g sin( ) n=f gl 1) Draw forces 2) The block has constant velocity, so no acceleration, so no net force! F fr = k n F gl =F g cos( ) F g In direction parallel to the slope: F-F g// -F fr =0 so F-F g sin( )- k n=0 and F-mgsin( )-F fr = x9.81xsin(36.5)- F fr =0 F fr =12.4 N 10

11 F fr = n= mg M n=mg T F g =mg a) No acceleration: no net force. F fr =T maximal F fr = s n so T<= s n : TRUE b) k < s and acceleration will start if T> s n so FALSE c) At rest, so no net force: F=0: FALSE 11

12 Clicker Quiz! When you are passed the stack of exams, what should you do to identify your exam? A) Look for the exam with your name and face on it. B) Make sure the exam doesn t have someone else s name and face on it. C) Make sure you have only one exam booklet. D) Make sure you don t have another exam stuck to the staple on the back. E) All of the above.

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14 335 km/hr= 93.1 m/s Work-energy theorem: ME i -ME f =(PE+KE) i -(PE+KE) f =W nc 14

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16 L: length units T: time units 16

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21 225 o Decompose in horizontal (west-east direction) and vertical (northsouth) components Horizontal Vertical Day 1 15*cos(225)= *sin(225)=-10.6 Day 2 10*cos(135)= *sin(135)=7.07 Sum Total displacement: 21

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24 Initially, the velocity is pointing up, but is decreasing in magnitude (speed is decreasing) since the gravitational force is slowing it down. This goes on until it reaches the highest point, where the velocity/speed equals zero. The ball than moves down: the velocity becomes negative, but the speed (not a vector, just a positive number) increases. So answer c is correct. 24

25 x(t)=x 0 +v 0 t+0.5at 2 v(t)=v 0 +at X(t)=20-v o t-0.5gt 2 with g=9.81 m/s 2 At t=1.5 s, x=0, so 0=20-1.5v o -0.5*9.81*(1.5 2 ) Solve for v o =-5.98 m/s magnitude: m/s answer b) 25

26 10 km/h 5 km/h The boat will move under an angle determined by tan =5/10 -> =26.57 o The tan =0.5 also is equal to (distance d)/(width of river=1 km) = d/1 So d=0.5 km. 26

27 x(t)=x 0 +v 0x t v x (t)=v 0x = v 0 cos( ) y(t)=y 0 +v 0y t-0.5gt 2 v(t)=v 0y -gt g=9.81 m/s 2 v 0y =v 0 sin( ) x(t)=x 0 +v 0x t = 40cos(40 o )t we don t now t y(t)=y 0 +v 0y t-0.5gt 2 = sin(40 0 )t-0.5*9.8*t 2 =0 at landing Solve this for t (quadratic equation): t=5.239 s Plug this into the equation for x(t): x(5.239)= 40cos(40 o )5.239=161 m answer d) 27

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29 1 Object 1: F=m 1 a, so T-F fr =m 1 a (moving to the right) Object 2: F=m 2 a, so F g -T=m 2 a (note the sign!!) m 2 g-t=m 2 a The frictional force F fr =µ k n=µ k m 1 g (magnitude of normal force equals the gravitational force) 2 So we have: Block 1 T-F fr =m 1 a -> T-µ k m 1 g=m 1 a -> T-0.1*40*9.8=40*a Block 2 m 2 g-t=m 2 a -> -T+10*9.8=10*a Sum to eliminate T =50a 29 So a=1.2 m/s 2 The acceleration is the same for both masses

30 The reading of the scale equals the normal force provided by the scale. Write down Newton s law for the forces acting on you: F=ma n-mg=ma (normal force n is pointing up, gravitational force is pointing down) The elevator is accelerating upwards, so a>0 and thus: n>mg which means that the weight you read from the scale is larger than your nominal weight (i.e. when not accelerating), answer a) 30

31 Newton s law for motion parallel to the slope: F=ma -mgsin + F fan =ma (down the slope is negative) -0.5*9.81*sin +2.45=0.5*a (cart is at rest, so a=0) -0.5*9.81*sin +2.45=0, solve for -> =30 0 answer a) 31

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33 Since the skier eventually stops, all the kinetic energy he had at the bottom of the slope was taken (dissipated) by friction. So if we determine the kinetic energy at the bottom of the slope, we know the answer to the problem. To find the KE at the bottom of the slope we use conservation of energy: KE i +PE i =KE f +PE f KE i : kinetic energy at top = 0 (starts from rest) PE i : potential energy at top = mgh = 70*9.8*200=137 kj PE f : potential energy at bottom = mgh=0 (h=0) So, KE f = PE i =137 kj this is also equal to the energy dissipated by friction after the skier comes to a full stop. 33

34 The masses are not accelerating since the two masses are equal. For either mass one can write Newton s 2 nd law: F=ma=0 T-mg=0 T=mg=1x9.8=9.8 N Answer: False 34

35 True, see e.g. the drawing 35

36 T 30 o 200 N T vertical The vertical component of the tension in the left cable must balance the weight of 200 N: So: T vertical T vertical makes an angle of 90-60=30 degrees with T (total tension in cable): Cos(30 o )=T vertical /T so T=T vertical /cos(30 o )=200/cos(30 0 )=231 N 36

37 Method 1: (positive: motion to the right) v(t)=v(0)+at at the end v(t)=0 so 0=1.5+at t=-1.5/a x(t)=x(0)+v(0)t+0.5at 2 at the end: x(t)=3 m So: 3=0+1.5t+0.5at 2 use t=-1.5/a 3= /a+0.5a(-1.5/a) 2 3=-2.25/a+1.125/a And 3=-1.125/a so a= m/s 2 F=ma = 40x = -15 N (-, so the left) Method 2: W nc =E kin (initial)-e kin (final) E kin (final)=0 (at rest) so: (no change in height, so no change in potential energy) W nc =0.5mv i2 =0.5*40*1.5 2 =45 J W=F x and F=W/ x=45/3=15 N Since object was moving to the right and slowing down, force must be pointing to the left. 37

38 F friction n T T M 1 g M 2 g Velocity of the blocks is constant, so their acceleration is zero. For the block on the surface: F=ma=0 T-F friction =0 so T- k n=0 and T- k M 1 g=0 (1) For hanging block: F=ma=0 -T+M 2 g=0 (2) Add (1) and (2) - k M 1 g+m 2 g=0 So k =M 2 /M 1 =2/6=

39 Force in direction of motion: 20cos(60 o )=10 N W=F x=10 x 4 = 40 J 39

40 1) x(t)=x 0 +v 0x t 68 = 0+35t so t=1.95 s 2) v x (t)=v 0x = v 0 cos( ) 3) y(t)=y 0 +v 0y t-0.5gt 2 4) v y (t)=v 0y -gt g=9.81 m/s 2 v 0y =v 0 sin( ) Use 1) and note that =0 o so: 68 = 0+35t so t=1.95 s Use 3) at t=1.95, where y(t)=0, y(0)=h(eight) solved for: 0=h-0.5 x 9.81 x (1.95) 2 so h=

41 F fr n=-f gl F gl =mgcos F g =mg F g// =mgsin F fr = s n= s mgcos Since the block is not moving: F fr =F g// so s mgcos =mgsin s = sin /cos = tan =

42 Acceleration of the total system: F=ma so T = Ma where M= =6.0 kg and T=35 N, so a=5.8 m/s 2 This is also the acceleration of the two blocks pulled by the string: F=ma Tension = ( ) x 5.8=17.5 N 42

43 1) v(t)=v(0)+at 2) x(t)=x(0)+v(0)t+0.5at 2 Use 2) 950 = x 15.5 x t 2 To find t=11.07 s for the time it takes to travel 950 m Use 1) v(t=11.07 s) = x = m 43

44 False: The component of the velocity that points across the river is largest for boat B, hence he crosses quicker. 44

45 1) v(t)=v(0)+at 2) x(t)=x(0)+v(0)t+0.5at 2 Use 2) with x(0)=3 m and v(0)=-6 m/s and a=+5 m/s 2 X(t=4) = 3- (6 x 4)+(0.5x5x4 2 ) = +19 m 45

46 1 mile = 5280 ft = 5280 x 12 inch = 5280 x 12 x 2.54 cm= 5280 x 12 x 2.54 x 0.01 m = m 1 m = 1/ mile = mile 1 s = 1/3600 hr = hr 331 m/s = 331 x mile / hr = 742 mile/hr (mph) 46

47 1) x(t)=x 0 +v 0x t 2) v x (t)=v 0x = v 0 cos( ) 3) y(t)=y 0 +v 0y t-0.5gt 2 4) v y (t)=v 0y -gt g=9.81 m/s 2 v 0y =v 0 sin( ) Use 4) and realize that at the highest point v y =0 0= v 0 sin( )-gt 0=100sin(35)-9.81t t=5.84 s Use 3) at t=5.84 s Y(t=5.84) = sin(35)x x9.81x =170 m 47

48 1) v(t)=v(0)+at 2) y(t)=y(0)+v(0)t-0.5gt 2 Both rocks undergo 1D motion with a=-g=-9.81 m/s 2 Use 2) for rock 1 (upward) and realize that y(t)=0 when the rock lands: 0 = t-0.5x9.81t 2 t=6.27 s (quadratic equation, disregard negative answer) Use 2 for rock 2 (dropped) to find the time it takes to reach the ground 0=100 +0t -0.5x9.81t 2 t=4.515 s. So if he drops rock =1.75 s later, they will reach the ground at the same time Note: as an alternative, you can calculate the time it takes for rock 1 to return to the starting height: 100 = t-0.5x9.81t 2 you ll find the same answer. 48

49 Mass on surface: F=m 1 a, so T=15a (1) Hanging mass: F=m 2 a, so -T+5g=5a (2) Combine (1) and (2) -T+5g=5T/15 4T/3=5g T=15g/4=15x9.8/4=37N 49

50 Initial kinetic energy: 0.5mv 2 =0.5*2000*30 2 = J Work done to stop the car: W=F x=10000 x Work done = change in kinetic energy = x so x=90 m. Alternatively, you can use F=ma to calculate the acceleration And then use the equations v(t)=v(0) x(t)=x(0)+v(0)t+0.5at 2 To get x(t)-x(0)= x 50