Fall 2020 Final Exam Review Answers
Chapter 2
(1) and (2)
Equations (1) and (2) can be solved simultaneously to find d.
SOLUTION Solving Equation (1) for t and substituting into Equation (2), we find
Solving for d yields:
The minus sign indicates that the direction of the acceleration is opposite to the direction of motion, and the plane is slowing down.
33.
Solving Equation 2.9 for x shows that the second part of the stopping distance is
Here, the acceleration is assigned a negative value, because we have assumed that the car is traveling in the positive direction, and it is decelerating. Since it is decelerating, its acceleration points opposite to its velocity. The stopping distance, then, is
The speed of the plane as it enters the intersection can be found from Equation 2.9. Solving Equation 2.9 for v0 gives
The time required to traverse the intersection can then be found from Equation 2.4. Solving Equation 2.4 for t gives
Chapter 3.
The motion in the horizontal direction occurs at a constant velocity of and the displacement in the horizontal direction is . Thus, it follows that
Using this value for the time in the expression for y shows that
Since the displacement of the ball is 2.40 m downward, the ball is above the court when it leaves the racket.
with t equal to the time required for the water the reach its maximum vertical displacement. The time t can be found by considering the vertical motion. From Equation 3.3b,
When the water has reached its maximum vertical displacement, vy = 0 m/s. Taking up and to the right as the positive directions, we find that
And
Therefore, we have
The range of the leap is, therefore,
The first solution (t = 0.217 s) corresponds to the situation where the ball is moving upward and has a displacement of y = +5.50 m. The second solution represents the later time when the ball is moving downward and its displacement is also y = +5.50 m (see the drawing). This is the solution we seek, so t =.
Chapter 4.
15.
Substituting this result into Newton’s second law we obtain
The answer is negative, indicating that the force is directed away from the cushion.
SOLUTION The table lists the two vectors and their x and y components:
Vector | x component | y component |
T | +T cos 16.0° | –T sin 16.0° |
T¢ | –T¢ cos 16.0° | –T¢ sin 16.0° |
T + T¢ | +T cos 16.0° – T¢ cos 16.0° | –T sin 16.0° – T¢ sin 16.0° |
Since we are given that T = T¢ = 21.0 N, the sum of the x components of the forces is
SFx = +T cos 16.0° – T¢ cos 16.0° = +(21 N) cos 16.0° – (21 N) cos 16.0° = 0 N
The sum of the y components is
SFy = –T sin16.0° – T¢ sin 16.0° = -(21 N) sin 16.0° – (21 N) sin 16.0° = -11.6 N
The magnitude F of the net force exerted on the tooth is
We will take upward as the positive direction. If the acceleration of is a, then the acceleration of must be –a. From Newton’s second law, we have for that
(1) |
and for
(2) |
Block 2 (3.00 kg) has two ropes attached each carrying a tension T. Also, block 2 only travels half the distance that block 1 travels in the same amount of time so its acceleration is only half of block 1’s acceleration. Newton’s second law for block 2 is then (2)
Solving Equation (1) for a, substituting into Equation (2), and rearranging gives
Chapter 6.
Solving for the skier’s initial speed gives (1)
The work done by the kinetic frictional force is given by (Equation 6.1), where fk is the magnitude of the kinetic frictional force and s is the magnitude of the skier’s displacement. Because the kinetic frictional force points opposite to the displacement of the skier, q = 180°. According to Equation 4.8, the kinetic frictional force has a magnitude of , where is the coefficient of kinetic friction and is the magnitude of the normal force. Thus, the work can be expressed as
Substituting this expression for W into Equation (1), we have that (2)
Since the skier is sliding on level ground, the magnitude of the normal force is equal to the weight mg of the skier (see Example 10 in Chapter 4), so FN = mg. Substituting this relation into Equation (2) gives
Since the skier comes to a halt, vf = 0 m/s. Therefore, the initial speed is
In Equation (1), fk = 205 N (see the REASONING), the angle θ between the frictional force and the displacement is θ = 180º (see the REASONING), the final speed is vf = 0 m/s (the snowmobile coasts to a halt), and the initial speed is given as v0 = 5.50 m/s. Solving Equation (1) for s gives
Solving for the final speed gives
(6.8) |
Solving for the final speed gives
Chapter 7:
Here the subscripts “1” and “2” refer to the first log and lumberjack, respectively. Let the direction of motion of the lumberjack be the positive direction. Then, solving for v1f gives
The minus sign indicates that the first log recoils as the lumberjack jumps off.
In this expression, the subscripts “1” and “2” now represent the second log and lumberjack, respectively. Since the second log is initially at rest, . Furthermore, since the lumberjack and the second log move with a common velocity, . The statement of momentum conservation then becomes
Solving for vf, we have
The positive sign indicates that the system moves in the same direction as the original direction of the lumberjack’s motion.
Their common velocity after Miranda hops on is, therefore,
The common speed is the magnitude of this value or
Both the x and y components of the total momentum of the three-bullet system are zero after the collision, since the bullets form a stationary lump. The x component of the initial momentum of bullet 1 is positive and the y component is negative, because this bullet is fired to the right and downward in the drawing. The x and y components of the initial momentum of bullet 3 are negative, because this bullet is fired to the left and downward in the drawing. Either of the two equations presented above can be solved for the unknown mass m3. From the equation for the x component of the total momentum, we find that
Chapter 9:
Substituting the given values, we obtain
(1)
We will apply trigonometry to determine the lever arms and for the weight and the tension, respectively, and then calculate the magnitude T of the tension in the strap.
SOLUTION The lever arm of the crate’s weight is shown in the diagram at the right, and is given by , where d is the distance between the axis of rotation (lower edge of the crate) and the crate’s center of gravity, and q is the angle between that line and the bottom of the crate. The right triangle in the free-body diagram of the crate (drawing on the left) shows how we can use trigonometry to determine both the length d and the angle q. The length d is the hypotenuse of that right triangle, and the other sides are the half-height (H/2 = 0.20 m) and half-length (L/2 = 0.45 m) of the crate, so by the Pythagorean theorem (Equation 1.7) we find that
We can find the angle q from the inverse tangent function:
The lever arm of the tension force is illustrated in the drawing at the right, where we see that . Therefore, from Equation (1), the magnitude of the tension in the strap is
Table 9.1 indicates that the moment of inertia of a hoop is , while the moment of inertia of a disk is . The net external torques acting on the hoop and the disk are:
(1)
In Equation (1) we have used the fact that the marble starts at rest at the top of the ramp. Since the marble rolls without slipping, we know that (Equation 8.12), where r is the radius of the marble. Referring to Table 9.1, we also know that the marble’s moment of inertia is . Substituting these two expressions into Equation (1) gives
As applied to the cube, energy conservation gives
where we have used the fact that the cube starts at rest at the top of the ramp and does not rotate. The desired ratio of the center-of-mass speeds is
(1)
where IC is the moment of inertia of the carousel (without the person). Substituting and (Equation 9.6) for the person’s initial and final moments of inertia into Equation (1), we obtain
Solving for the mass m of the person yields
Chapter 10:
kx1 = m(a + µkg) (1)
We can find the acceleration using
In phase 2 of the block’s motion (constant speed) a = 0 m/s2, so the force exerted by the spring is
(2) |
so that
Using this expression for mk in Equation (1) we obtain
The block does not rotate, so the angular speeds wf and w0 are zero. Since the block comes to a momentary halt on the spring and is dropped from rest, the translational speeds vf and v0 are also zero. Because the spring is initially unstrained, the initial displacement y0 of the spring is likewise zero. Thus, the above expression can be simplified as follows:
The block was dropped at a height of h0 – hf above the compressed spring. Solving the simplified energy-conservation expression for this quantity gives
The block/bullet system now compresses the spring by an amount x. During the compression the total mechanical energy is conserved so that
Substituting the expression for V into this equation, we obtain
Solving this expression for v gives
The block does not rotate, so the angular speeds wf and w0 are zero. Since the block comes to a momentary halt on the spring and is released from rest, the translational speeds vf and v0 are also zero. Because the spring is initially unstrained, the initial displacement x0 of the spring is likewise zero. Thus, the above expression can be simplified as follows:
The term h0 – hf is the amount by which the spring has compressed, or h0 – hf = xf. Making this substitution into the simplified energy-conservation equation gives
Solving for xf, we find
Chapter 11.
In this result, hair is the height of a column of air that equals the height of the building. We can also express using the mercury barometer, which indicates that corresponds to a column of mercury that has a height of . Thus, we have
Equating the two expressions for , we obtain
Taking the density of mercury from Table 11.1 in the text and solving for hair gives
We want P = 2Patm, and we know h = hw + hm = 1.00 m. Using the above and rearranging gives
(1) |
The weight W of an object is equal to its mass m times the magnitude g of the acceleration due to gravity, or W = mg (Equation 4.5). The mass, in turn, is equal to the product of an object’s density r and its volume V, so m = r V (Equation 11.1). Combining these two relations, the weight can be expressed as W = r V g.
According to Archimedes’ principle, the magnitude FB of the buoyant force is equal to the weight of the cool air that the balloon displaces, so FB = mcool airg = (rcool airV)g. Since we are neglecting the weight of the balloon fabric and the basket, the weight of the balloon is just that of the hot air inside the balloon. Thus, m = mhot air = rhot airV and
W = mhot airg = (rhot airV)g .
Substituting the expressions FB = (rcool airV)g, m = rhot airV, and W = (rhot airV)g into Equation (1) gives
Solving Bernoulli’s equation for P2 and taking the elevation at the first floor to be , we have
Chapter 12:
We can use this equation to find the temperature of the water before the insertion of the thermometer.
SOLUTION Solving the equation above for , and using the value of from Table 12.2, we have
The temperature of the water before the insertion of the thermometer was
Solving for , we have
or
where and . The specific heats of mercury and water are given in Table 12.2, and the latent heat of vaporization of water is given in Table 12.3. Solving for the mass of the water that vaporizes gives
Chapter 14.
(14.1) |
The number of remaining moles of gas can be found from the ideal gas law, , so that
Equation 6.9b gives the principle of conservation of mechanical energy:
In this expression, we know that and that (since the atom comes to a halt at the top of its trajectory). Furthermore, we can take the height at the earth’s surface to be h0 = 0 m. Taking this information into account, we can write the energy-conservation equation as follows:
Using M to denote the molecular mass (in kilograms per mole) and recognizing that , where NA is Avogadro’s number and is the number of xenon atoms per mole, we have
Recognizing that kNA = R and that M = 131.29 g/mol = 131.29 × 10-3 kg/mol, we find
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