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Midterm 1 Practice Exam 1 Solutions

Question 1 Solution

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 hangs from the rope, 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 $\theta$.

$m_{1}a=m_{1}g-T$

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

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

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

For parts C-G consider a case where $m_{1}=$4 kg and $m_{2}$=1 kg, $\mu_{s}$=0.2 and $\mu_{k}$=0.15.

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?

$a=\frac{4+1\sin 30^{o}-0.15\cos30^{o}}{4+1}\times9.8=8.57\mathrm{ ms^{-2}}$

$v=at=8.57\times0.5=4.28\mathrm{ms^{-1}}$ down

$y=\frac{1}{2}at^{2}=0.5\times8.57\times0.5^{2}=1.07m$ down

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

$T=m_{1}g-m_{1}a=4\times(9.8-8.57)=4.92\mathrm{N}$

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

$W_{g}=m_{1}gd+m_{2}gd\sin 30^{o}=4\times9.8\times1.07+1\times9.8\times1.07\times\sin 30^{o}=47.187\mathrm{J}$

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

$W_{Fr}=-\mu_{k}m_{2}g\cos\theta d=-0.15\times 1 \times 9.8\times\cos30^{o}\times 1.07=-1.36\mathrm{J}$

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 Solution

A 200 kg rocket is fired straight up from the ground with an initial velocity of 100 $\mathrm{ms^{-1}}$ after which it is subject to a constant gravitational acceleration of 9.8 $\mathrm{ms^{-2}}$ down. A 15kg projectile is fired from a cannon 1 km away at the same time as the rocket is launched. 10 seconds later the projectile hits the rocket.

A. (5 points) At what height above the ground does the collision take place?

Solve using rocket equation of motion

$y=v_{0}-\frac{1}{2}gt^{2}=100\times10-0.5\times9.8\times10^{2}=510\mathrm{m}$

B. (5 points) At what angle $\theta$ above the horizontal must the cannon be fired in order to hit the rocket?

Equations of motion of projectile

$x=v_{0}\cos\theta t$

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

These give

$1000=v_{0}\cos\theta \times 10$

$1000=v_{0}\sin\theta \times 10$

So

$\theta=45^{o}$

C. (5 points) What is the initial velocity of the projectile launched from the cannon?

$v_{0x}=100\mathrm{ms^{-1}}$

$v_{0}=\frac{v_{0x}}{\cos45^{o}}=141\mathrm{ms^{-1}}$

D. (5 points) What is the kinetic energy of the rocket when it is hit by the projectile?

Can use either conservation of mechanical energy

$KE=\frac{1}{2}mv_{0}^{2}-mgh=0.5\times200\times100^{2}-200\times9.8\times510=400\mathrm{J}$

of find velocity of rocket when it is hit

$v=100-9.8\times10=2\mathrm{ms^{-1}}$

$KE=\frac{1}{2}mv^{2}=400\mathrm{J}$

E. (5 points) What is the kinetic energy of the projectile when it hits the rocket?

$KE=\frac{1}{2}mv_{0}^{2}-mgh=0.5\times15\times100^{2}\times2-15\times9.8\times510=75030\mathrm{J}$

F. (10 points) If instead of starting with the velocity found in part B the projectile was fired with a velocity of 500 ms$^{-1}$ find the required angle $\theta$ above the horizontal for the projectile to the hit the rocket and the height at which the collision occurs in this case.

Equations of motion for projectile are now:

$x=500\cos\theta t$

$y=500\sin\theta t-\frac{1}{2}gt^{2}$

The angle can be found by considering when the y coordinates of the two objects are the same

$500\sin\theta t-\frac{1}{2}gt^{2}=100t-\frac{1}{2}gt^{2}$

$\sin\theta=\frac{1}{5}$

$\theta=11.54^{o}$

The time at which the collision takes place can now be found using the equation for the x motion of the projectile.

$t=\frac{1000}{500\cos\theta}=2.04\mathrm{s}$

The height can now be found by substituting back in the y equation of motion

$y=500\times\frac{1}{5}\times2.04-0.5\times9.8\times2.04^{2}=183.6\mathrm{m}$

Question 3 Solution

A 500 kg car is traveling with a constant speed of 50 $\mathrm{ms^{-1}}$ in a circular path around the inside edge of a cylinder. The radius of the curve is 40 m. The coefficient of static friction between the road and the tires is $\mu=0.2$.

A. (5 points) How long does it take for the car to go around the cylinder once?

$T=\frac{2\pi r}{v}=\frac{2\pi \times40}{50}=5.03\mathrm{s}$

B. (5 points) What is the magnitude of the normal force exerted by the wall of the cylinder on the car.

$F_{N}=\frac{mv^{2}}{r}=\frac{500\times50^{2}}{40}=31250\mathrm{N}$

C. (5 points) What is the magnitude and direction of the frictional force on the car. Give the direction as one of the following, up, down, towards the center of the cylinder, radially outwards, in the direction the car is moving, opposite to the direction the car is moving.

$F_{Fr}=mg=500\times9.8=4900\mathrm{N}$ up

D. (5 points) How much work is done by the normal force in one complete circuit of the track?

$O \mathrm{J}$

E. (5 points) How much work is done by the frictional force in one complete circuit of the track?

$O \mathrm{J}$

F. (5 points) The car starts to slow down. At what speed does it start to slip down the wall of the cylinder?

The car starts to slip when the maximum frictional force is less than the weight of the car. The minimum velocity for this not to happen can be found from

$\mu\frac{mv^{2}}{r}=mg$

$v^{2}=\frac{rg}{\mu}=\frac{40\times9.8}{0.2}$

$v=44.3\mathrm{ms^{-1}}$

phy131studiof15/lectures/m1p1sol.txt · Last modified: 2015/09/30 17:23 by mdawber
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