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For every question also consider

**(e) none of the above.**

Be sure to put your name on the "scantron sheet" or "bubble sheet". Be sure to fill in the circles for your name. Be sure to fill in all the circles for your answers. Be sure to make any erasures complete.

There are no intentionally "tricky" questions but arithmetic
round-off errors can always cause problems. In my solutions I have
used g = 10 m/s^{2}. If you get an answer like 123.4 and find
that the only thing close is 123.5, choose that one.

Possibly useful information:

s = s

_{i}+ v_{i}t + (^{1}/_{2}) a t^{2}v = v

_{i}+ a tv

^{2}= v_{i}^{2}+ 2 a (s - s_{i})sin = opp/hyp

cos = adj/hyp

tan = opp/adj

F = m a

F

_{12}= - F_{21}f

_{smax}= µ_{s}nf

_{k}= µ_{k}ng = 9.8 m/s

^{2}= 10 m/s^{2}

Questions 1 through 8 count 10 points each.

Questions 9 through 12 count 5 points each.

This change in points will not show up on the "tear sheet" that is returned from Academic Testing Services next week.

A hot air balloon is moving up at 5.0 m/s and is at an altitude of 10.0 m when a sandbag breaks loose and falls to the ground below

**1. How long is it before the sandbag hits the ground?**

y = y

_{i}+ v_{i}t + (1/2) a_{y}t^{2}y = y

_{i}+ v_{i}t + (1/2) (- g) t^{2}0 = 10 m + (5 m/s) t + (1/2) (- 10 m/s

^{2}) t^{2}0 = 10 + 5 t + (1/2) (- 10) t

^{2}0 = 10 + 5 t - 5 t

^{2}5 t

^{2}- 5 t - 10 = 0t

^{2}- t - 2 = 0t = { - (-1) ± SQRT[ ( - 1)

^{2}- 4 (1) (-2)] }/2(1)t = { 1 ± SQRT[ 1 + 8)] }/2(1)

t = { 1 ± SQRT[ 9] }/2

t = { 1 ± 3 }/2

t

_{1}= - 2/2 = - 1 = - 1 s (extaneous; throw out)t

_{2}= + 4/2 = 2 = 2 s(a) t = 1 s

(b) t = 2 s(c) t = 4 s

(d) t = 9.8 s

** **

**2. How fast is the sandbag moving when it hits the ground?**

v

^{2}= v_{i}^{2}+ 2 a (s - s_{i})v

_{y}^{2}= v_{yi}^{2}+ 2 a_{y}(y - y_{i})v

_{y}^{2}= (5 m/s)^{2}+ 2 ( - 10 m/s^{2}) (0 - 10 m)v

_{y}^{2}= 5^{2}+ 2 ( - 10) ( - 10 )v

_{y}^{2}= 225v

_{y}= 15 m/s(a) 5 m/s

(b) 10 m/s

(c) 15 m/s(d) 20 m/s

A blue Jeep traveling at a constant
velocity of 15 m/s (that's about 30 mi/h) passes through an
intersection just as the light turns green. A
red Mustang is sitting at the
intersection at rest when the light turns green. Just as the
Jeep passes the
Mustang, the
Mustang starts to accelerate at a
constant acceleration of 2 m/s^{2}. The
Jeep easily goes ahead of the
Mustang. But the
Mustang is accelerating so it eventually
catches up with the Jeep and then
overtakes it.

**3. How far from the intersection does the Mustang catch up with
the Jeep?**

s = s

_{i}+ v_{i}t + (1/2) a t^{2}

s

_{m}= s_{mi}+ v_{mi}t + (1/2) a_{m}t^{2}s

_{j}= s_{ji}+ v_{ji}t + (1/2) a_{j}t^{2}s

_{mi}= 0; v_{mi}= 0; a_{m}= 2 m/s^{2}s

_{ji}= 0; v_{ji}= 15 m/s; a_{j}= 0s

_{m}= 0 + 0 + (1/2) (2 m/s^{2}) t^{2}s

_{j }= 0 + (15 m/s) t + 0The Mustang catches up with the Jeep when s

_{m}= s_{j}s

_{m}= s_{j}(1/2) (2 m/s

^{2}) t^{2}= (15 m/s) tt

^{2}= 15 t

t

_{1}= 0; true but uninterestingt

_{2}= 15 = 15 sec; this is the one we want!Now we know when the Mustang catches the Jeep; but we want to know where this happens.

s

_{j}= (15 m/s) ts

_{j}= 225 m(a) 20 m

(b) 150 m

(c) 225 m(d) 980 m

In trying to catch the Roadrunner Wyle E Coyote speeds off a cliff
with an **horizontal** velocity of 25
m/s. The cliff is 45 m above the canyon floor below.

**4. How far from the cliff does our hapless hero hit the canyon
floor?**

First we find how long the coyote is in the air.

v

_{yi}= 0y

_{i}= 45 my = y

_{f}= 0y = y

_{i}+ v_{yi}t + (1/2) a_{y}t^{2}0 = 45 m + 0 + (1/2) ( - 10 m/s

^{2}) t^{2}5 t

^{2}= 45t

^{2}= 9t = 3 = 3 s

How far horizontally does he travel in this time?

x = x

_{i}+ v_{xi}t + (1/2) a_{x}t^{2}x = 0 + (25 m/s) (3 s) + 0

x = 75 m

(a) 25 m

(b) 75 m(c) 150 m

(d) 300 m

An airplane travels along the following displacement vectors:

r_{1}= 300 km 30° E of N

r_{2}= 400 km 60° W of N

r_{3}= 100 km due West

**5. Where does the airplane end up? What is its resultant
displacement R?**

R=r_{1}+r_{2}+r_{3}But this means

X = x

_{1}+ x_{2}+ x_{3}and Y = y_{1}+ y_{2}+ y_{3}Look closely at the diagram. In keeping with aviation tradition, these directions or angles are measured from

north(or the positive y-axis).

x

_{1}= r_{1}sin 30^{o}y

_{1}= r_{1}cos 30^{o}x

_{1}= (300 km)(0.500) = 150 kmy

_{1}= (300 km)(0.866) = 259.8 km

Always look at your diagram and ask if it is "reasonable".Clearly, Ifailedto do that initially. When Iincorrectlywrote down x_{1}= 259.8 km and y_{1}= 150 km, the diagram itself should have shown me the errors in my calculations. Take the diagrams seriously.

x

_{2}= - (400 km) (0.866) = - 346.4 kmy

_{2}= (400 km) (0.500) = 200 kmThese values

dolook "reasonable" when the diagram is consulted.

x

_{3}= - 100 kmy

_{3}= 0Now, we are ready for

X = x

_{1}+ x_{2}+ x_{3}Y = y

_{1}+ y_{2}+ y_{3}X = (150 - 346.4 - 100) km

Y = (259.8 + 200 + 0) km

X = - 296.4 km

Y = 459.8 km

Now that we have the components of

R, we need to find the magnitude and direction of resultant R,R = SQRT [ X

^{2}+ Y^{2}] = SQRT [ (- 296.4)^{2}+ (459.8)^{2}] km = 547.1 km= tan-1 [opp/adj] = tan-1 [X/Y] = 296.4 / 459.8 = 0.645

= 32.8°

R= 547 km, at 32.8° West of North(a) 800 km, 53° West of North

(b) 397 km, 28° West of North

(c) 567 km, 23° East of North

(d) 247 km, 64° East of North

(e) none of the above

A 1.0 kg block slides down a plane inclined at 30° from the horizontal. The coefficient of sliding friction between the block and the plane is 0.2.

**6. What is the acceleration of the block?**

First, a free-body diagram of all the forces on the block,

F_{net}=w+n+f_{k}

F

_{net,x}= w_{x}+ n_{x}+ f_{k,x}F

_{net,y}= w_{y}+ n_{y}+ f_{k,y}F

_{net,x}= - w sin 30° + 0 + f_{k}F

_{net,y}= - w cos 30° + n + 0F

_{net,x}= - w (0.500) + µ nF

_{net,y}= - w (0.866) + n + 0 = 0 = m a_{y}F

_{net,x}= - m g (0.500) + 0.2 nn = 0.866 w = 0.866 m g

F

_{net,x}= - m g (0.500) + 0.2 (0.866 m g)F

_{net,x}= - 0.500 m g + 0.173 m gF

_{net,x}= - 0.500 m g + 0.173 m gF

_{net,x}= - 0.327 m g = m a_{x}a

_{x}= - 0.327 g = - 0.327 ( 10 m/s^{2})a

_{x}= - 3.27 m/s^{2}

or, a

_{x}= 3.27 m/s^{2}, down and to the left, along the plane(a) a = 0.500 m/s

^{2}

(b) a = - 3.27 m/s^{2}(c) a = - 4.90 m/s

^{2}(d) a = 9.6 m/s

^{2}

A 0.250 kg block is released from rest on a horizontal plane. The coefficient of friction between the block and the plane is 0.20. The block is attached, by a lightweight string, to a hanging 0.150-kg mass.

**7. What is the acceleration of the block?**

**8. What is the tension in the string?**

This should look familiar; it is the same arrangement we had in this week's lab.

As always, begin with good free-body diagrams.

First, consider the forces on M, the 0.250-kg block.

F

_{x}= T - f_{k}= M a_{x}= M aF

_{y}= n - Mg = M a_{y}= 0n = M g

f

_{k}= µ n = µ M gT - µ M g = M a

T = M a + µ M g

Now consider the forces on m, the 0.150-kg hanging mass

Down is taken as

positive.F

_{y}= m g - T = m a_{y}= m aT = m g - m a

With two expressions for the tension T,

set these two expressions equal to each other.

M a + µ M g = m g - m a

M a + m a = m g - µ M g

(M + m) a = (m - µ M) g

a = [(m - µ M) / (M + m) ] g

a = [(0.150 - {0.2}{0.25} ) / (0.250 + 0.150) ] (9.8 m/s

^{2})a = [(0.150 - 0.05) / 0.4 ] (9.8 m/s

^{2})a = [ 0.10 / 0.4 ] (9.8 m/s

^{2})a = (0.25) (9.8 m/s

^{2})a = 2.45 m/s

^{2}Now, for the tension,

T = m g - m a = m (g - a)

T = (0.150 kg) (9.80 - 2.45) (m/s

^{2})T = 1.1025 N

T = 1.1 N

** **

**7. What is the acceleration of the block?**

(a) a = 1.1 m/s

^{2}

(b) a = 2.45 m/s^{2}(c) a = 4.90 m/s

^{2}(d) a = 9.80 m/s

^{2}

** **

**8. What is the tension in the string?**

(a) T = 1.1 N(b) T = 1.5 N

(c) T = 2.5 N

(d) T = 4.0 N

Questions 1 through 8 count 10 points each.

Questions 9 through 12 count 5 points each.

This change in points will not show up on the "tear sheet" that is returned from Academic Testing Services next week.

** **

**9. Projectile motion is a combination of**

(a) horizontal motion at constant velocity and vertical motion at constant velocity

(b) horizontal motion at constant velocity and vertical motion at constant, non-zero acceleration(c) horizontal motion at constant, non-zero acceleration and vertical motion at constant velocity

(d) horizontal motion at constant, non-zero acceleration and vertical motion at constant, non-zero acceleration

** **

**10. The force of friction depends upon the normal force. The
normal force is always given by**

(a) n = mg

(b) n = T

(c) n = mg sin

(d) n = mg tan

(e) none of the above-- We have no idea what angle is. The normal force can vary considerably so it is certainlynotalways given by any of the expressions above.

** **

** **

**11. A block of mass m sits on a plane inclined at angle
**** from the
horizontal. The normal force of the plane on the block is**

(a) n = mg

(b) n = mg cos(c) n = mg sin

(d) n = mg tan

** **

**12. In the absence of a net force, a body in motion**

(a) will gradually come to rest.

(b) will continue in motion with constant, non-zero velocity.(c) will continue in motion with constant, non-zero acceleration

(d) will fall along a parabolic trajectory.

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