Download Midterm I Physics 231

FORM B
Solutions to Midterm II Physics 231
November 10th, 2010
1. Block A has mass 1.0 kg and is moving on a straight line with a velocity 2.0 m/s on
a frictionless track. It collides with block B, which has mass 3.0 kg and velocity 1.0
m/s and is moving in the same direction as block A. After the collision the blocks
stick together. Which of the following statements is correct?
a) The final velocity of the blocks is 0.5 m/s.
b) The final velocity of the blocks is 0.8 m/s.
c) The final velocity of the blocks is 2.0 m/s.
d) The final velocity of the blocks is 1.3 m/s.
Conservation of linear momentum in one dimension:
pi= 1x2+3x1=5 kgm/s; pf= (1+3)v. Since pi= pf then 5=4v and v=1.3 m/s
2. A student throws a 120 g snow ball at the side of the schoolhouse. The ball has a
velocity of 7.5 m/s only in the horizontal direction when it hits the wall and sticks.
What is the magnitude of the average force on the wall if the duration of the collision
is 0.15 s?
a) 135 N
b) 60 N
c) 24 N
d) 6 N
The variation of momentum corresponds to the impulse of the force. We know that
the variation of momentum of the ball is going to be pf-pi=0-0.120x7.5=-0.9 kgm/s.
This corresponds to the impulse of the force I=F t; -0.9=F x 0.15 from which we
conclude that F=-6 N.
3. A ball on a string moves in a circular path, once a second, on a frictionless
horizontal table. The tension in the string is measured to be 6.0 N. What would be the
tension if the ball went round in only half a second, while the orbital radius is kept
constant?
a) 24 N
b) 12 N
c) 6 N
d) 3 N
If the period of rotation of the ball is reduced to half, this means the angular speed
is doubled. The tension of the string has to equal the net force which is centripetal:
T=mv2/R. Using the relation between linear velocity and angular velocity v=R we
can write the tension in terms of the angular velocity: T=m2R. This means that for
a fixed radius, if we double the angular speed, then the tension is 22=4 quadrupled.
The correct answer is 6x4=24N.
4. What is the average angular speed of the tip of the minute hand on a clock in rad/s?
a) 3.14 rad/s
b) 6.28 rad/s
c) 1.75x10-3 rad/s
d) 1.05x10-3 rad/s
/period. It takes 60x60 s for the minute hand on the clock to complete one
circle. /3600=1.75x10-3 rad/s
5. A 1500 kg car drives around a flat 200-m-radius circular track at a constant linear
speed of 20 m/s. Which of the following statements is correct?
a) The net force on the car is zero because the speed is constant.
b) The net force on the car is 3000 N along the direction of the velocity.
c) The net force on the car is 3000 N pointing radially outward.
d) The net force on the car is 3000 N pointing radially
inward.
For the car to drive at constant speed, the tangential force has to be zero. But the
net force cannot be zero because the car is changing direction, so the velocity vector
is changing. The net force on the car has to be centripetal, pointing radially inward.
Check the magnitude: F=mv2/R=1500x202/200=3000 N.
6. A canonical pendulum is formed by attaching a 500 g ball to a 1.0-m-long string,
then allowing the mass to move in a horizontal circle of radius 20 cm (note that this
means the angle the string makes with the y-axis is 11.34 degrees). What is the
tension on the string?
y-axis
a) 25 N
b) 4.8 N
c) 4.9 N
d) 5.0 N
T
Fg
The two forces represented in the figure add up to produce the net force for the
system which is a centripetal force pointing radially inward. The forces along the
vertical need to cancel to zero: Tcos(x)=mg; T=mg/cos(x)=0.5x9.8/cos(11.34)=5 N.
7. Planet X orbits the star Omega with a period that is 200 earth-days long. Planet Y,
is twice as heavy as planet X and circles around Omega at four times the distance.
How many earth-days does it take for planet Y to complete a full orbit?
a) 200
b) 800
c) 1600
d) 12800
According to Kepler’s 3rd law, the cube of the radius R3 of a planet’s circular orbit is
proportional to the square of the period T2 and does not depend on the mass of the
planet. This means that T is directly proportional to R3/2. If the radius of Y is 4 times
that of X then the period is 43/2= 8 times larger. The period is then 8x200=1600
earth-days long.
8. A nut bolt needs to be tightened with a wrench. Which will be the most effective
way to do this?
a) A
b) B
c) C
d) D
The torque is responsible for producing
rotation of the wrench: torque=rFsin(theta). The torque on C is maximum because r
is maximum and the angle between r and F is 90 maximizing the sin as well.
9. Two balls are connected by a very light rod of length L. The mass on the right is 3
times the mass on the left. What is the distance between the center of gravity of the
system and the ball on the left?
a) Zero
b) L/4
c) L/2
d) 3L/4
x
If we put the origin of our reference system on the ball on the left then;
Xcm=(3mL)/4m=3/4L.
10. A thin 2.00 kg box rests on a 6.00 kg board that hangs over the end of the table as
shown in the figure. The thin box is placed 5 cm from the edge of the board and 15
cm from point O at the edge of the table. Which
statement is correct?
a) The torque produced by the weight of
the box relative to O is larger than that
O
by the weight of the board, thus the
systems rotates.
Fboard
Fbox
b) The torque produced by the weight of
the board is larger than that by the weight of the box, thus the system rotates.
c) The net torque on the system is zero and thus the
system stays in equilibrium.
d) None of the above.
We need to calculate the net torque relative to O. The two relevant forces are the
gravitational force of the board and the gravitational force of the box. Net torque is
6x9.8x0.05-2x9.8x0.15=0. If the net torque is zero then the system stays in
equilibrium.
11. How much force does it take to stretch a 10-m-long, 1.0-cm-diameter steel cable
by 5.0 mm? The Young’s modulus for steel is 20x1010 N/m2.
a) 16x102 N
b) 79x102 N
c) 31x103 N
d) 16x105 N
The force is related to the linear deformation by F=Y L A/L0. We need to first
determine the area A= r2= (0.005)2=7.854X10-5 m2. Then
F=20x1010x5x10-3x7.854X10-5/10=79x102 N.
12. Three uniform blocks are placed in water according to the figure. When they reach
equilibrium, 2/3 of the volume of block A is submerged in water, 1/3 of the volume of
block B is submerged and ½ of the volume of block C is submerged. Which of the
following statements is correct?
a) The density of block A is smaller than the
density of block B.
C
B
b) The density of block B is
smaller than the density of the
block C.
A
c) The density of block C is larger than the
density of block A.
d) The densities of all three blocks are the
same.
Each block is in equilibrium therefore the gravitational force of the block has to be
equal in magnitude to its buoyancy force: Fg=bVb g and B=wVw g. If Fg=B then
bVb=wVw and therefore b=wVw/ Vb. This means that the density of the block is
directly proportional to the volume fraction that is submerged: A>C>B .