# Fall 2012 Midterm 1 Solutions

## Midterm 1 Q1 solution (Ave Score: 33.3/35)

Two blocks 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 frictionless plane inclined at an angle of $\theta$ to the horizontal.

A. Add arrows indicating the direction of all of the forces acting on both $m_{1}$ and $m_{2}$ to the diagram.

B. Find an expression for the acceleration, $a$, of $m_{1}$ as a function of $m_{1}$, $m_{2}$, $g$ and $\theta$. Specify which direction is positive (i.e up or down).

Taking up positive.

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

$m_{2}a=m_{2}g \sin\theta-T$

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

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

For parts C-G consider a case where $m_{1}=1\mathrm{kg}$ and $m_{2}=3\mathrm{kg}$.

C.For which value of the angle $\theta$ would the system remain at rest after it has been released?

This is when the numerator of the expression found in B is zero.

$3\sin\theta=1$

$\sin\theta=\frac{1}{3}$

$\theta=19.47^{o}$

D. If $\theta=30^{o}$ what is the velocity and displacement of $m_{1}$, 0.5 s after the system has been released from rest? Specify which direction is positive (i.e up or down).

$a=\frac{\frac{3}{2}-1}{4}g=\frac{g}{8}=1.225\mathrm{m\,s^{-2}}$

$v=1.225\times0.5=0.6125\mathrm{m\,s^{-1}}$

$x=0.5\times1.225\times0.5^{2}=0.15\mathrm{m}$

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

$T=m_{1}g=9.8\mathrm{kg}$

OR

What is the magnitude of the tension in the rope during the motion in part (D)?

$T=(1+\frac{1}{8}g)=11.025\mathrm{N}$

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

$0 \mathrm{J}$

OR

How much work is done by gravity on the two block system during the motion in part (D)?

$\frac{1}{2}(m_{1}+m_{2})v^{2}=\frac{1}{2}(4)0.6125^{2}=0.75\mathrm{J}$

G. How much work is done by the normal force on m$_{2}$ during the motion in part (C) or part (D)?

$0 \mathrm{J}$

## Midterm 1 Q2 Solution (Ave Score: 29.9/35)

A freedom loving American in the back of a stationary pickup truck fires his rifle in to the air at an angle $\theta$ to the horizontal. Assume that the bullet is not subject to drag forces. At the same time as the gun is fired the driver of the pickup truck begins to accelerate the truck to the right with a constant acceleration $a$.

A. If after 10s the pickup truck is 250m down the road from where it started, what is the value of the acceleration $a$?

$\frac{1}{2}at^{2}=250$

$a=\frac{250\times 2}{100}=5\mathrm{m\,s^{-2}}$

B. How fast is the pickup truck going at this point?

$5\times 10= 50 \mathrm{m\,s^{-1}}$

C. At this time the freedom loving American is hit by the bullet he fired into the air. What is the initial velocity of the bullet? Express your answer using unit vector notation, using $\hat{i}$ for the direction the truck moves in and $\hat{j}$ for vertically upward.

For the bullet

$x=v_{x0}t$

$y=v_{y0}t-\frac{1}{2}gt^{2}$

At t=10s the bullet must be at (250,0).

$250=v_{x0}10$

$v_{x0}=25\mathrm{m\,s^{-1}}$

$0=v_{y0}-\frac{1}{2}g\times100$

$v_{y0}=\frac{\frac{1}{2}\times9.8\times100}{10}$

$v_{y0}=49\mathrm{m\,s^{-1}}$

$\vec{v_{0}}=25\hat{i}+49\hat{j}\,\mathrm{m\,s^{-1}}$

D. What is the angle $\theta$ at which the rifle is fired?

$\tan^{-1}(\frac{49}{25})=62.97^{o}$

E. What is the maximum height that the bullet reaches before it turns around?

$y=49\times5-\frac{1}{2}\times9.8\times5^{2}=122.5\mathrm{m}$

F. If the bullet fired has mass 100g find the value of it's total mechanical energy at 3 positions: when it is first fired, when it is at the top of it's trajectory, and just before it hits the freedom loving American.

Mechanical energy is conserved. It is always equal to the initial kinetic energy, which is

$\frac{1}{2}0.1\times(49^{2}+25^{2})=151.3\mathrm{J}$

## Midterm 1 Q3 Solutions (Ave Score: 20.8/30)

A 500kg car is traveling with constant speed along a road which follows a circular path on the face of a cone. The radius of the curve is 100m. The road is 10m lower in height than the top of the cone. The coefficient of static friction between the road and the tires is $\mu=0.2$

A. What is the angle of the slope, $\theta$?

$\tan^{-1}(\frac{10}{100}=5.71^{o})$

B. What is the maximum speed that the car can have without sliding off the road?

Forces in horizontal direction

$F_{Fr}\cos\theta-F_{N}\sin\theta=\frac{mv^{2}}{r}$

Forces in vertical direction

$F_{Fr}\sin\theta+F_{N}\cos\theta=mg$

The maximum value of the frictional force is $F_{Fr}=\mu F_{N}$

so

$\mu F_{N}\cos\theta-F_{N}\sin\theta=\frac{mv^{2}}{r}$

$\mu F_{N}\sin\theta+F_{N}\cos\theta=mg$

Divide one equation by the other to get

$\frac{\mu\cos\theta-\sin\theta}{\mu\sin\theta+\cos\theta}=\frac{v^{2}}{rg}$

Subbing in gives

$v^{2}=100\times9.8\times\frac{0.2\cos 5.71^{o}-\sin 5.71^{o}}{0.2\sin 5.71^{o}+\cos 5.71^{o}}=96.09$

$v=9.8\mathrm{m\,s^{-1}}$

C.What is the magnitude of the normal force exerted by the road on the car when it is traveling at this maximum speed?

This can be determined from either of the force equations

$F_{N}=\frac{mg}{\mu\sin\theta+\cos\theta}=\frac{9.8\times500}{0.2\sin 5.71^{o}+\cos 5.71^{o}}=4827\mathrm{N}$

D. How much work does the frictional force do on the car while it is driving around the path?

Friction is perpendicular to motion so work done by friction is $0 \mathrm{J}$

E. How much work does gravity do on the car while the car is driving around the path?

Gravity is perpendicular to motion so work done by gravity is $0 \mathrm{J}$ 