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phy131studiof17:lectures:phy131f16m1solutions [2017/09/27 09:15]
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phy131studiof17:lectures:phy131f16m1solutions [2017/09/27 09:16] (current)
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 ====== Fall 2016 Midterm 1 Solutions ====== ====== Fall 2016 Midterm 1 Solutions ======
  
 +===== Question 1 Solutions (Ave Score:​29.6/​35) =====
  
-Question ​Solutions (Ave Score:29.6/35)+Two blocks of mass $m_{1}$ and $m_{2}$ are connected by a rope which runs over a frictionless pulley of negligible mass. Initially, $m_{1}$ rests on a surface with an incline of $\theta_{1}$ to the horizontal, while $m_{2}$ rests on a plane inclined at an angle of $\theta_{2}$ to the horizontal. Both surfaces have coefficients of static and kinetic friction $\mu_{s}$ and  $\mu_{k}$.
  
-Two blocks ​of mass m1m1 and m2m2 are connected by a rope which runs over a frictionless pulley ​of negligible mass. Initially, m1m1 rests on a surface with an incline of θ1θ1 to the horizontal, while m2m2 rests on a plane inclined at an angle of θ2θ2 to the horizontal. Both surfaces have coefficients of static and kinetic friction μsμs and μkμk.+A. (5 points) Add arrows indicating the direction ​of all of the forces acting ​on both $m_{1}$ and $m_{2}$ to the diagram.
  
-A. (5 points) ​Add arrows indicating ​the direction ​of all of the forces acting on both m1m1 and m2m2 to the diagram.+B. (5 points) ​Assuming that the incline is sufficiently steep that the blocks move when they are released from rest, find an expression for the downwards acceleration,​ $a$ in terms of $m_{1}$, $m_{2}$, $g$,  $\mu_{s}$ or  $\mu_{k}$ ​and $\theta$.
  
-B. (5 points) Assuming that the incline is sufficiently steep that the blocks move when they are released from rest, find an expression for the downwards acceleration,​ aa in terms of m1m1, m2m2, gg, μsμs or μkμk and θθ.+$m_{1}a=T+m_{1}g\sin\theta_1-\mu_{k}m_{1}g\cos\theta_1$
  
-m1a=T+m1gsinθ1−μkm1gcosθ1m1a=T+m1gsin⁡θ1−μkm1gcos⁡θ1+$m_{2}a=-T+m_{2}g\sin\theta_2-\mu_{k}m_{2}g\cos\theta_2$
  
-m2a=−T+m2gsinθ2−μkm2gcosθ2m2a=−T+m2gsin⁡θ2−μkm2gcos⁡θ2+$(m_{1}+m_{2})a=m_{2}g\sin\theta+m_{1}g\sin\theta_1-\mu_{k}m_{2}g\cos\theta-\mu_{k}m_{1}g\cos\theta_1$
  
-(m1+m2)a=m2gsinθ+m1gsinθ1−μkm2gcosθ−μkm1gcosθ1(m1+m2)a=m2gsin⁡θ+m1gsin⁡θ1−μkm2gcos⁡θ−μkm1gcos⁡θ1+$a=\frac{m_{2}g\sin\theta_2+m_{1}g\sin\theta_1-\mu_{k}m_{2}g\cos\theta_2-\mu_{k}m_{1}g\cos\theta_1}{m_{1}+m_{2}}$
  
-a=m2gsinθ2+m1gsinθ1−μkm2gcosθ2−μkm1gcosθ1m1+m2a=m2gsin⁡θ2+m1gsin⁡θ1−μkm2gcos⁡θ2−μkm1gcos⁡θ1m1+m2+For parts C-G consider ​case where $m_{1}=$kg and $m_{2}$=5 kg, $\mu_{s}$=0.25 and $\mu_{k}$=0.2.
  
-For parts C-G consider a case where m1=m1=2 kg and m2m2=5 kgμsμs=0.25 and μkμk=0.2.+C.  (5 points) If $\theta_{1}=15^{o}$ ​and $\theta_{2}=30^{o}$, what is the speed of $m_{1}$, ​ 0.5\,s after the system has been released from rest? How far has $m_{1}$ travelled ​0.5\,s after the system has been released from rest?
  
-C. (5 points) If θ1=15oθ1=15o and θ2=30oθ2=30o,​ what is the speed of m1m1, 0.5\,s after the system has been released from rest? How far has m1m1 travelled 0.5\,s after the system has been released from rest?+$a=2.47 \mathrm{m\,s^{-2}}$
  
-a=2.47ms−2a=2.47ms−2+$v=at=2.47\times0.5=1.24 \mathrm{m\,​s^{-1}}$
  
-v=at=2.47×0.5=1.24ms−1v=at=2.47×0.5=1.24ms−1 +$x=\frac{1}{2}at^{2}=0.5\times2.47\times0.5^2=0.31 \mathrm{m}$
- +
-x=12at2=0.5×2.47×0.52=0.31mx=12at2=0.5×2.47×0.52=0.31m+
  
 D. (5 points) What is the magnitude of the tension in the rope during the motion in part (C)? D. (5 points) What is the magnitude of the tension in the rope during the motion in part (C)?
  
-Using either+Using either ​
  
-m1a=T+m1gsinθ1−μkm1gcosθ1m1a=T+m1gsin⁡θ1−μkm1gcos⁡θ1+$m_{1}a=T+m_{1}g\sin\theta_1-\mu_{k}m_{1}g\cos\theta_1$
  
 or or
  
-m2a=−T+m2gsinθ2−μkm2gcosθ2m2a=T+m2gsin⁡θ2−μkm2gcos⁡θ2+$m_{2}a=-T+m_{2}g\sin\theta_2-\mu_{k}m_{2}g\cos\theta_2$
  
-with a=2.47ms−2a=2.47ms−2+with $a=2.47 \mathrm{m\,​s^{-2}}$
  
-gives+gives 
  
-T=3.66NT=3.66N+$T=3.66\mathrm{N}$
  
 E. (5 points) How much work is done by gravity on the two block system during the motion in part (C)? E. (5 points) How much work is done by gravity on the two block system during the motion in part (C)?
  
-Wg=m2gdsinθ2+m1gdsinθ1=9.14JWg=m2gdsin⁡θ2+m1gdsin⁡θ1=9.14J+$W_{g}=m_{2}gd\sin\theta_2+m_{1}gd\sin\theta_1=9.14\mathrm{J}$
  
-F. (5 points) How much work is done by friction on the two block system during the motion in part (C)?+F. (5 points) How much work is done by friction on the two block system ​ during the motion in part (C)?
  
-WFr=−μkm2gdcosθ2−μkm1gdcosθ1=3.79JWFr=−μkm2gdcos⁡θ2−μkm1gdcos⁡θ1=−3.79J+$W_{Fr}=-\mu_{k}m_{2}gd\cos\theta_2-\mu_{k}m_{1}gd\cos\theta_1=-3.79\mathrm{J}$
  
-G. (5 points) How much work is done by the normal force on the two block system during the motion in part (C) ?+G. (5 points) How much  work is done by the normal force on the two block system during the motion in part (C) ?
  
-0J0J+$0\mathrm{J}$
  
-Edit 
-Question 2 (Ave score: 21/30) 
  
-A cannon fires a tennis ball with an initial velocity v0v0 at an angle of θθ above the horizontal at time t=0t=0. A robot bucket is supposed to catch the ball. Once it starts moving the bucket moves to the left with a constant speed of vBvB. The bucket is initially xB0xB0 m to the right of the cannon and can only move to the left.+===== Question 2 (Ave score: 21/30)=====
  
-A. (5 points) Write down equations for the xx and yy coordinates ​of the tennis ​ball as function ​of time in terms of v0v0, θθ and gg.+cannon fires a tennis ball with an initial velocity $v_{0}$ at an angle of $\theta$ above the horizontal at time $t=0$. A robot bucket is supposed to catch the ball. Once it starts moving the bucket moves to the left with constant speed of $v_{B}$. The bucket is initially $x_{B0}$ m  to the right of the cannon ​and can only move to the left.
  
-x=v0cosθtx=v0cos⁡θt ​y=v0sinθt−12gt2y=v0sin⁡θt−12gt2+A. (5 points) Write down equations for the $x$ and $y$ coordinates of the tennis ball as a function of time in terms of $v_{0}$, $\theta$ and $g$.
  
-B. (5 points) From the equations you wrote down in part A, derive an expression for the horizontal distance RR from the cannon at which the tennis ball will hit the ground in terms of v0v0, θθ and gg.+$x=v_{0}\cos\theta t$ 
 +$y=v_{0}\sin\theta t-\frac{1}{2}gt^{2}$
  
-When the ball returns to the ground ​y=0y=so+B. (5 points) From the equations you wrote down in part A, derive an expression for the horizontal distance $R$ from the cannon at which the tennis ​ball will hit the ground ​in terms of $v_{0}$, $\theta$ and $g$.
  
-v0sinθt=12gt2v0sin⁡θt=12gt2+When the ball returns to the ground $y=0$ so
  
-t=2v0sinθgt=2v0sin⁡θg+$v_{0}\sin\theta ​t=\frac{1}{2}gt^{2}$
  
-Substituting for tt in the xx equation gives+$t=\frac{2v_{0}\sin\theta}{g}$
  
-x=R=2v20sinθcosθgx=R=2v02sin⁡θcos⁡θg+Substituting for $t$ in the $x$ equation gives
  
-C. (5 points) If the bucket starts moving at a time t=tBt=tB write an expression for the xx position of the bucket for times t>​tBt>​tB in terms of xB0xB0, vBvB and tBtB.+$x=R=\frac{2v_{0}^2\sin\theta\cos\theta}{g}$
  
-x=xB0−vB(t−tB)x=xB0−vB(t−tB)+C. (5 points) If the bucket starts moving at a time $t=t_{B}$ write an expression for the $x$ position of the bucket for times $t>t_{B}$ in terms of $x_{B0}$, $v_{B}$ and $t_{B}$.
  
-D. (5 points) If θ=30∘θ=30∘ ​and xB0=40xB0=40m what is the maximum tennis ball launch velocity ​v0v0 for which the robot bucket can successfully catch the ball?+$x=x_{B0}-v_{B}(t-t_{B})$ 
 + 
 +D. (5 points) If $\theta=30^{\circ}$ ​and $x_{B0}=40$m what is the maximum tennis ball launch velocity ​$v_{0}$ ​for which the robot bucket can successfully catch the ball?
  
 If the ball overshoots the bucket, the bucket cannot catch it so If the ball overshoots the bucket, the bucket cannot catch it so
  
-2v20sinθcosθg<​402v02sin⁡θcos⁡θg<40+$\frac{2v_{0}^2\sin\theta\cos\theta}{g}<40$
  
-v20<20gsinθcosθv02<​20gsin⁡θcos⁡θ+$v_{0}^2<\frac{20g}{\sin\theta\cos\theta}$
  
-v0<21.29ms−1v0<​21.29ms−1+$v_{0}<21.29\mathrm{\,​m\,​s^{-1}}$
  
-E. (10 points) Find the time tBtB that the bucket should start moving so that it catches the ball if v0=20v0=20m/s, θ=30∘θ=30∘vB=4vB=4m/s and xB0=40xB0=40m.+E. (10 points) Find the time $t_{B}$ ​that the bucket should start moving so that it catches the ball if $v_{0}=20$m/s, $\theta=30^{\circ}$$v_{B}=4$m/s and $x_{B0}=40$m.
  
 The bucket needs to reach the point at which the ball hits the ground at the same time as it hits the ground. ie. The bucket needs to reach the point at which the ball hits the ground at the same time as it hits the ground. ie.
  
-x=R=2v20sinθcosθgx=R=2v02sin⁡θcos⁡θg 
  
-and+$x=R=\frac{2v_{0}^2\sin\theta\cos\theta}{g}$ ​
  
-t=2v0sinθgt=2v0sin⁡θg+and  
 + 
 +$t=\frac{2v_{0}\sin\theta}{g}$
  
 Putting this in to the equation of motion for the bucket gives Putting this in to the equation of motion for the bucket gives
  
-2v20sinθcosθg=xB0−vB(2v0sinθg−tB)2v02sin⁡θcos⁡θg=xB0−vB(2v0sin⁡θg−tB)+$\frac{2v_{0}^2\sin\theta\cos\theta}{g}=x_{B0}-v_{B}(\frac{2v_{0}\sin\theta}{g}-t_{B})$
  
-which can be solved for+which can be solved for 
  
-tB=0.87stB=0.87s+$t_{B}=0.87\mathrm{s}$
  
-Edit +===== Question 3 (Ave Score: 22/​35) ​=====
-Question 3 (Ave Score: 22/35)+
  
-The diagram shows the motion of a block of mass 100 grams. The block is given an initial velocity ​v0v0 by a spring which was previously compressed by a distance ​x0x0 from it’s equilibrium length. The block slides along a flat surface, goes through a loop and then continues up a ramp that has an angle of 45∘∘ ​to the horizontal, eventually coming to rest at height ​hh before turning around and sliding back down the slope. All of the surfaces that the block slides along are frictionless. The diameter of the loop is 40cm.+The diagram shows the motion of a block of mass 100 grams. The block is given an initial velocity ​$v_0$ by a spring which was previously compressed by a distance ​$x_0$ from it’s equilibrium length. The block slides along a flat surface, goes through a loop and then continues up a ramp that has an angle of 45$^{\circ}$ ​to the horizontal, eventually coming to rest at height ​$h$ before turning around and sliding back down the slope. All of the surfaces that the block slides along are frictionless. The diameter of the loop is 40cm.
  
 A. (5 points) Add arrows to the pictures of the block to indicate the directions of the forces acting on the block at each of the 5 positions shown in the diagram. Label each arrow to indicate which force it corresponds to. A. (5 points) Add arrows to the pictures of the block to indicate the directions of the forces acting on the block at each of the 5 positions shown in the diagram. Label each arrow to indicate which force it corresponds to.
  
-B. (10 points) What is the minimum value of v0v0 required such that the block does not fall off the track at the top of the loop?+B. (10 points) What is the minimum value of $v_{0}$ ​required such that the block does not fall off the track at the top of the loop? 
 + 
 +$\frac{mv^{2}}{r}=mg$ 
 + 
 +$\frac{1}{2}mv^{2}=\frac{1}{2}mv_{0}^{2}-2mgr$
  
-mv2r=mgmv2r=mg+$\frac{1}{2}mgr=\frac{1}{2}mv_{0}^{2}-2mgr$
  
-12mv2=12mv20−2mgr12mv2=12mv02−2mgr+$\frac{1}{2}mv_{0}^{2}=\frac{5}{2}mgr$
  
-12mgr=12mv20−2mgr12mgr=12mv02−2mgr+$v_{0}^{2}=5gr=5\times9.8\times0.2=9.8$
  
-12mv20=52mgr12mv02=52mgr+$v_{0}=3.13\mathrm{\,​m\,​s^{-1}}$
  
-v20=5gr=5×9.8×0.2=9.8v02=5gr=5×9.8×0.2=9.8+C(5 points) If the block has this value of $v_{0}$ what is the maximum height $h$ reached by the block?
  
-v0=3.13ms−1v0=3.13ms−1+$\frac{1}{2}mv_{0}^{2}=mgh$
  
-C(5 points) If the block has this value of v0v0 what is the maximum height hh reached by the block?+$h=\frac{v_{0}^{2}}{2g}=\frac{3.13^2}{2\times9.8}=0.50\mathrm{m}$
  
-12mv20=mgh12mv02=mgh+D. (5 points) If the spring constant $k$ of the spring is 30N/m by what distance $x_{0}$ was the spring compressed initially to provide the value of $v_{0}$ you found earlier?
  
-h=v202g=3.1322×9.8=0.50mh=v022g=3.1322×9.8=0.50m+$\frac{1}{2}mv_{0}^{2}=\frac{1}{2}kx_{0}^{2}$
  
-D(5 points) If the spring constant kk of the spring is 30N/m by what distance x0x0 was the spring compressed initially to provide the value of v0v0 you found earlier?+$x_{0}^{2}=\frac{mv_{0}^{2}}{k}=\frac{0.1\times9.8}{30}=0.0326$
  
-12mv20=12kx2012mv02=12kx02+$x_{0}=0.18\mathrm{m}$
  
-x20=mv20k=0.1×9.830=0.0326x02=mv02k=0.1×9.830=0.0326+E(5 points) How much work did gravity do on the block between the time the spring was released and the time when it reached height $h$?
  
-x0=0.18mx0=0.18m+$-\frac{1}{2}mv_{0}^{2}=-mgh=W_{g}=-0.49\mathrm{J}$
  
-E. (5 points) How much work did gravity ​do on the block between the time the spring was released and the time when it reached height ​hh?+F. (5 points) ​The ramp exerts a normal force on the block at all times. ​How much work does the normal force do on the block between the time the spring was released and the time when it reached height ​$h$?
  
-−12mv20=−mgh=Wg=−0.49J−12mv02=−mgh=Wg=−0.49J+$W_{N}=0\mathrm{J}$
  
-F. (5 points) The ramp exerts a normal force on the block at all times. How much work does the normal force do on the block between the time the spring was released and the time when it reached height hh? 
  
-WN=0J 
phy131studiof17/lectures/phy131f16m1solutions.txt · Last modified: 2017/09/27 09:16 by mdawber
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