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PHYSICS 231 Review problems for midterm 1 Some Housekeeping • The 1st exam will be Wednesday October 3. • The exam will take place right here in BPS 1410. • 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 1st 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 Clicker Quiz! A) This is an appropriate way to solve physics problems. B) This will probably not full credit. 3 Velocity (m/s) v(t)=v(0)+at x(t)=x(0)+v(0)t+0.5at2 Cut problem up in 1s pieces After 1 s: v(1)=0+0x1=0 x(1)=0+0x1+0.5x0X12=0 1 1 1.5 3 3 After 2 s: v(2)=v(1)+at=0+3x1=3 x(2)=x(1)+v(1)t+0.5at2 =0+0x1+0.5x3x12=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: PHY 231 After 3 s: v(3)=v(2)+at=3+2x1=5 x(3)=x(2)+v(2)t+0.5at2 =1.5+3x1+0.5x2x12=5.5 After 4 s: v(4)=v(3)+at=5-2x1=3 x(4)=x(3)+v(3)t+0.5at2 =5.5+5x1+0.5x(-2)x12=9.5 4 Cross straight a) Cross fast v? 6.50 m/s 3.30 m/s b) 6.50 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 v2+3.302=6.502 so v=(6.502-3.302)=5.6 m/s Time to cross river: t=x/v=262.0/5.6=46.8 s (use v=x/t) To cross fast: use picture b) V=6.5 m/s so t=x/v=40.3 s (but lands downstream) PHY 231 5 + a) b) c) d) Acceleration in vertical direction is ALWAYS g=–9.81 m/s2; TRUE See a) FALSE It is positive going up, negative going down: FALSE It is zero at the start and end and positive everywhere else: TRUE 6 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) B) C) D) E) 9:20 AM 9:10 AM 9:00 AM 8:55 AM None, I’m going to BPS 1415 for my exam vx(0)=29.0 m/s 2.19 m x(t)=x0+v0xt = 0+29t=29t vx(t)=v0x = 29 y(t)=y0+v0yt-0.5gt2 = 2.19+0t-0.5x9.8t2=2.19-4.9t2 vy(t)=v0y-gt = 0-9.8t=-9.8t When ball hits ground: y(t)=0 so: 2.19-4.9t2=0 t=0.67s Use in x(t) equation: x(0.67)=29 x 0.67 = 19.4 m PHY 231 8 1) Draw forces 2) Since the block is not moving, no acceleration, no net force mg T=15N 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… Tupper-mg-Tlower=0 so Tupper=4.3x9.81+15=42.2+15=57.2 N PHY 231 9 1) Draw forces 2) The block has constant velocity, so no acceleration, so no net force! n=FgL Fg//=Fgsin() Ffr=kn FgL=Fgcos() Fg In direction parallel to the slope: F-Fg//-Ffr=0 so F-Fgsin()-kn=0 and… F-mgsin()-Ffr=0 18.9-1.12x9.81xsin(36.5)- Ffr=0 Ffr=12.4 N PHY 231 10 n=mg Ffr=n= mg M T Fg=mg a) No acceleration: no net force. Ffr=T maximal Ffr=sn so T<= sn : TRUE b) k < s and acceleration will start if T> sn so FALSE c) At rest, so no net force: F=0: FALSE 11 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. 13 335 km/hr= 93.1 m/s Work-energy theorem: MEi-MEf=(PE+KE)i-(PE+KE)f=Wnc 14 15 L: length units T: time units 16 17 18 19 20 225o 1350 Decompose in ‘horizontal’ (west-east direction) and ‘vertical’ (northsouth) components Horizontal Vertical Day 1 15*cos(225)=-10.6 15*sin(225)=-10.6 Day 2 10*cos(135)=-7.07 10*sin(135)=7.07 Sum -17.7 -3.53 Total displacement: PHY 231 21 PHY 231 22 PHY 231 23 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. PHY 231 24 x(t)=x0+v0t+0.5at2 v(t)=v0+at X(t)=20-vot-0.5gt2 with g=9.81 m/s2 At t=1.5 s, x=0, so 0=20-1.5vo-0.5*9.81*(1.52) Solve for vo =-5.98 m/s magnitude: +5.98 m/s answer b) PHY 231 25 10 km/h 5 km/h The boat will move under an angle determined by tan=5/10 -> =26.57o The tan=0.5 also is equal to (distance d)/(width of river=1 km) = d/1 So d=0.5 km. PHY 231 26 x(t)=x0+v0xt vx(t)=v0x = v0cos() y(t)=y0+v0yt-0.5gt2 v(t)=v0y-gt g=9.81 m/s2 v0y=v0sin() x(t)=x0+v0xt = 40cos(40o)t we don’t now t… y(t)=y0+v0yt-0.5gt2 = 0.2+40sin(400)t-0.5*9.8*t2 =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(40o)5.239=161 m answer d) PHY 231 27 PHY 231 28 1 2 Object 1: Object 2: F=m1a, so F=m2a, so T-Ffr=m1a (moving to the right) Fg-T=m2a (note the – sign!!) m2g-T=m2a The frictional force Ffr=µkn=µkm1g (magnitude of normal force equals the gravitational force) So we have: Block 1 T-Ffr=m1a -> T-µkm1g=m1a -> T-0.1*40*9.8=40*a Block 2 m2g-T=m2a -> -T+10*9.8=10*a Sum to eliminate T 0-39.2+98=50a 29 So a=1.2 m/s2 The acceleration is the same for both masses 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) PHY 231 30 Newton’s law for motion parallel to the slope: F=ma -mgsin + Ffan =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 -> =300 answer a) PHY 231 31 PHY 231 32 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: KEi+PEi=KEf+PEf KEi : kinetic energy at top = 0 (starts from rest) PEi : potential energy at top = mgh = 70*9.8*200=137 kJ PEf : potential energy at bottom = mgh=0 (h=0) So, KEf = PEi=137 kJ this is also equal to the energy dissipated by friction after the skier comes to a full stop. 33 PHY 231 The masses are not accelerating since the two masses are equal. For either mass one can write Newton’s 2nd law: F=ma=0 T-mg=0 T=mg=1x9.8=9.8 N Answer: False PHY 231 34 True, see e.g. the drawing PHY 231 35 T 30o 200 N Tvertical The vertical component of the tension in the left cable must balance the weight of 200 N: So: Tvertical Tvertical makes an angle of 90-60=30 degrees with T (total tension in cable): Cos(30o)=Tvertical/T so T=Tvertical/cos(30o)=200/cos(300)=231 N PHY 231 36 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+0.5at2 at the end: x(t)=3 m So: 3=0+1.5t+0.5at2 use t=-1.5/a 3=-1.52/a+0.5a(-1.5/a)2 3=-2.25/a+1.125/a And 3=-1.125/a so a=-0.375 m/s2 F=ma = 40x-0.375 = -15 N (-, so the left) Method 2: Wnc=Ekin(initial)-Ekin(final) Ekin(final)=0 (at rest) so: (no change in height, so no change in potential energy) Wnc=0.5mvi2=0.5*40*1.52=45 J W=Fx and F=W/x=45/3=15 N Since object was moving to the right and slowing down, force must be 37 pointing to the left. PHY 231 n Ffriction T T M 1g M2g Velocity of the blocks is constant, so their acceleration is zero. For the block on the surface: F=ma=0 T-Ffriction=0 so T-kn=0 and T-kM1g=0 (1) For hanging block: F=ma=0 -T+M2g=0 (2) ------------------kM1g+M2g=0 Add (1) and (2) So k=M2/M1=2/6=0.33 PHY 231 + 38 Force in direction of motion: 20cos(60o)=10 N W=Fx=10 x 4 = 40 J PHY 231 39 1) x(t)=x0+v0xt 68 = 0+35t so t=1.95 s 2) vx(t)=v0x = v0cos() 3) y(t)=y0+v0yt-0.5gt2 4) vy(t)=v0y-gt g=9.81 m/s2 v0y=v0sin() Use 1) and note that =0o 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=18.51 PHY 231 40 Ffr n=-FgL Ffr=sn=smgcos Since the block is not moving: Ffr=Fg// so F =mgsin g// FgL=mgcos smgcos=mgsin s= sin/cos= tan=1.15 Fg=mg PHY 231 41 Acceleration of the total system: F=ma so T = Ma where M=1.0+2.0+3.0=6.0 kg and T=35 N, so a=5.8 m/s2 This is also the acceleration of the two blocks pulled by the string: F=ma Tension = (1.0+2.0) x 5.8=17.5 N PHY 231 42 1) v(t)=v(0)+at 2) x(t)=x(0)+v(0)t+0.5at2 Use 2) 950 = 0 + 0 +0.5 x 15.5 x t2 To find t=11.07 s for the time it takes to travel 950 m Use 1) v(t=11.07 s) = 0 + 15.5 x 11.07 = 171.6 m PHY 231 43 False: The component of the velocity that points across the river is largest for boat B, hence he crosses quicker. PHY 231 44 1) v(t)=v(0)+at 2) x(t)=x(0)+v(0)t+0.5at2 Use 2) with x(0)=3 m and v(0)=-6 m/s and a=+5 m/s2 X(t=4) = 3- (6 x 4)+(0.5x5x42) = +19 m PHY 231 45 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 = 1609.3 m 1 m = 1/1609.3 mile = 0.0006213 mile 1 s = 1/3600 hr = 0.000277 hr 331 m/s = 331 x 0.0006213 mile / 0.000277 hr = 742 mile/hr (mph) PHY 231 46 1) x(t)=x0+v0xt 2) vx(t)=v0x = v0cos() 3) y(t)=y0+v0yt-0.5gt2 4) vy(t)=v0y-gt g=9.81 m/s2 v0y=v0sin() Use 4) and realize that at the highest point vy=0 0= v0sin()-gt 0=100sin(35)-9.81t t=5.84 s Use 3) at t=5.84 s Y(t=5.84) = 0 + 100sin(35)x5.84 -0.5x9.81x5.842=170 m PHY 231 47 1) v(t)=v(0)+at 2) y(t)=y(0)+v(0)t-0.5gt2 Both rocks undergo 1D motion with a=-g=-9.81 m/s2 Use 2) for rock 1 (upward) and realize that y(t)=0 when the rock lands: 0 = 100 + 15t-0.5x9.81t2 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 -0.5x9.81t2 t=4.515 s. So if he drops rock 2 6.27-4.52=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 = 100 + 15t-0.5x9.81t2 you’ll find the same answer. PHY 231 48 Mass on surface: F=m1a, so T=15a (1) Hanging mass: F=m2a, so -T+5g=5a (2) Combine (1) and (2) -T+5g=5T/15 4T/3=5g T=15g/4=15x9.8/4=37N PHY 231 49 Initial kinetic energy: 0.5mv2=0.5*2000*302=900000 J Work done to stop the car: W=Fx=10000x Work done = change in kinetic energy 900000 = 10000 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+0.5at2 To get x(t)-x(0)= x 50 PHY 231