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Fall 2015 Midterm 1 Solutions

Question 1

Two blocks of mass $m_{1}$ and $m_{2}$ are connected by a rope which runs over a frictionless pulley of negligible mass. One block sits on a flathorizontal surface, while the other rests on a plane with coefficients of static and kinetic friction $\mu_{s}$ and $\mu_{k}$ inclined at an angle of $\theta$ to the horizontal.

A. (5 points) Add arrows indicating the direction of all of the forces acting on both $m_{1}$ and $m_{2}$ 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 $\theta$.

$m_{1}a=T-\mu_{k}m_{1}g$

$m_{2}a=-T+m_{2}g\sin\theta-\mu_{k}m_{2}g\cos\theta$

$(m_{1}+m_{2})a=m_{2}g\sin\theta-\mu_{k}m_{2}g\cos\theta-\mu_{k}m_{1}g$

$a=\frac{m_{2}g\sin\theta-\mu_{k}m_{2}g\cos\theta-\mu_{k}m_{1}g}{m_{1}+m_{2}}$

For parts C-G consider a case where $m_{1}=$2 kg and $m_{2}$=5 kg, $\mu_{s}$=0.25 and $\mu_{k}$=0.2.

C. (5 points) If $\theta=30^{o}$ what is the velocity and displacement of $m_{1}$, 0.5s after the system has been released from rest?

$a=\frac{5\sin 30^{o}-0.2\times5\times\cos30^{o}-0.2\times 2}{7}\times9.8=1.73\mathrm{ ms^{-2}}$

$v=at=1.73\times0.5=0.84\mathrm{ms^{-1}}$ right

$y=\frac{1}{2}at^{2}=0.5\times1.73\times0.5^{2}=0.22m$ right

D. (5 points) What is the magnitude of the tension in the rope during the motion in part (C)?

$T=m_{1}a+\mu_{k}m_{1}g=2\times1.73+0.2\times2\times9.8)=3.78\mathrm{N}$

E. (5 points) How much work is done by gravity on the two block system during the motion in part (C)?

$W_{g}=m_{2}gd\sin 30^{o}=5\times9.8\times0.22\times\sin 30^{o}=5.39\mathrm{J}$

F. (5 points) How much work is done by friction during the motion in part (C)?

$W_{Fr}=(-\mu_{k}m_{1}g-\mu_{k}m_{2}g\cos\theta) d=(-0.2\times2\times9.8-0.2\times 5 \times 9.8\times\cos30^{o})\times 0.22=-2.73\mathrm{J}$

G. (5 points) How much work is done by the normal force on $m_{2}$ during the motion in part (C) ?

$W_{N}=0\mathrm{J}$

Question 2

A catapult is designed to fire a rock at a fixed angle of 45$^{o}$ above the horizontal. A knight begins to charge from a distance of 100m towards the catapult at a constant speed of 10m/s. If the catapult is fired at the same instant as the knight begins to charge then:

A. (5 points) Write equations for both components ($x$ and $y$) of the rock displacement as a function of time, taking t=0\,s to be the time the catapult is fired.

$x=v_{0}\cos\theta\, t$ $y=v_{0}\sin\theta t-\frac{1}{2}gt^2$

B. (5 points) Write equations for both components ($x$ and $y$) of the knight's displacement as a function of time.

$x=100-10t$ $y=0$

C. (10 points) What initial velocity must the rock fired from the catapult have to hit the knight?

At time of impact $t_{i}$

$0=v_{0}\sin\theta t_{i}-\frac{1}{2}gt_{i}^{2}$

$t_{i}=\frac{2v_{0}\sin\theta}{g}$

$100-10t_{i}=v_{0}\cos\theta\, t_{i}$

$(v_{0}\cos\theta+10)\frac{2v_{0}\sin\theta}{g}=100$

$\frac{2v_{0}^{2}\sin\theta\cos\theta}{g}+\frac{20v_{0}\sin\theta}{g}=100$

$v_{0}^{2}+\frac{20}{\sqrt{2}}v_{0}=100g$

$v_{0}^{2}+\frac{20}{\sqrt{2}}v_{0}+50=100g+50$

$(v_{0}+\frac{10}{\sqrt{2}})^2=100g+50$

$v_{0}=\frac{-10}{\sqrt{2}}\pm \sqrt{100g+50}$

$v_{0}=25.0 \mathrm{ms^{-1}}$ or $v_{0}=-39.2\mathrm{ms^{-1}}$ but we discard the negative solution.

D. (5 points) How much work did gravity do on the rock while it was in the air?

0J

E.(5 points) Why is the catapult designed to fire at a fixed angle of 45$^{o}$ above the horizontal?

To maximize the horizontal range

Question 3

Typed solution to appear soon!

phy131studiof16/lectures/phy131f15m1solutions.txt · Last modified: 2016/10/05 11:38 by mdawber
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