Question 1

A small ball is tied to one end of a light 2.5-m wire, and the other end of the wire is hooked to the ceiling. A person pulls the ball to the side until the wire makes an angle of 35° with the plane of the ceiling and then gently releases it. What is the angular speed of the ball, in rad/s, as it swings through its lowest point?

Accepted Answer

The answer of A small ball is tied to one...

Question 2

A wheel having a moment of inertia of 5.00 kg ∙ m² starts from rest and accelerates under a constant torque of 3.00 N ∙ m for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s?

Accepted Answer

The rotational kinetic energy (KE_rot) of the wheel can be calculated using the formula KE_rot = 0.5 * I * ω^2, where I is the moment of inertia and ω is the angular velocity. First, we need to find the angular acceleration (α) using the torque (τ) and moment of inertia (I) with the formula τ = I * α, giving α = τ / I = 3.00 N∙m / 5.00 kg∙m^2 = 0.60 rad/s^2. Then, we find the angular velocity (ω) after 8.00 s using ω = α * t = 0.60 rad/s^2 * 8.00 s = 4.80 rad/s. Finally, KE_rot = 0.5 * 5.00 kg∙m^2 * (4.80 rad/s)^2 = 0.5 * 5.00 * 23.04 = 57.6 J.

Question 3

In a certain nuclear reactor, neutrons suddenly collide with deuterons, which have twice the mass of neutrons. In a head-on elastic collision with a stationary deuteron, what fraction of the initial kinetic energy of a neutron is transferred to the deuteron?

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Question 4

A 1200-kg pick-up truck traveling south at 15 m/s suddenly collides with a 750-kg car that is traveling east. The two vehicles stick together and slide along the road after colliding. A highway patrol officer investigating the accident determines that the final position of the wreckage after the collision is 25 m, at an angle of 50° south of east, from the point of impact. She also determines that the coefficient of kinetic friction between the tires and the road at that location was 0.40. What was the speed of the car just before the collision?

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Question 5

Two simple pendulums of equal length l = 0.45 m are suspended from the same point. The pendulum bobs are small solid steel spheres. The first bob is drawn back to make a 35° angle with the vertical, while the other one is left hanging at rest. If the first bob has a mass of 0.25 kg and the second has a mass of $0.54 \mathrm {~kg} \text {, }$ how high will the second bob rise above its initial position when struck elastically by the first bob after it is released?

Accepted Answer

Explanation:Momentum is conserved in the collision, so the total momentum before the collision is equal to the total momentum after the collision. Since the second bob is initially at rest, the total momentum before the collision is equal to the momentum of the first bob. After the collision, the first bob has zero velocity, so the total momentum is equal to the momentum of the second bob. Therefore, the momentum of the second bob after the collision is equal to the momentum of the first bob before the collision.The momentum of an object is equal to its mass times its velocity. Therefore, the momentum of the first bob before the collision is equal to 0.25 kg * v1, where v1 is the velocity of the first bob before the collision. The momentum of the second bob after the collision is equal to 0.54 kg * v2, where v2 is the velocity of the second bob after the collision.Since the momentum of the second bob after the collision is equal to the momentum of the first bob before the collision, we have:0.25 kg * v1 = 0.54 kg * v2Solving for v2, we get:v2 = 0.463 v1The kinetic energy of an object is equal to 1/2 * m * v^2, where m is the mass of the object and v is its velocity. Therefore, the kinetic energy of the first bob before the collision is equal to 1/2 * 0.25 kg * v1^2. The kinetic energy of the second bob after the collision is equal to 1/2 * 0.54 kg * v2^2.Since the collision is elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Therefore, we have:1/2 * 0.25 kg * v1^2 = 1/2 * 0.54 kg * v2^2Substituting v2 = 0.463 v1, we get:1/2 * 0.25 kg * v1^2 = 1/2 * 0.54 kg * (0.463 v1)^2Solving for v1, we get:v1 = 1.41 m/sThe height to which the second bob rises above its initial position is equal to the maximum height reached by the first bob. The maximum height reached by the first bob is equal to the height from which it was released, which is 0.45 m * sin(35°) = 0.26 m.Therefore, the second bob will rise to a height of 0.26 m above its initial position.

Question 6

To drive a typical car at 40 mph on a level road for one hour requires about 3.2 × 10⁷ J of energy. Suppose we tried to store this much energy in a spinning, solid, uniform, cylindrical flywheel. A large flywheel cannot be spun too fast or it will fracture. If we used a flywheel of diameter 1.2 m and mass 400 kg, what angular speed would be required to store 3.2 × 10⁷ J?

Accepted Answer

The moment of inertia for a solid, uniform, cylindrical object is given by $I=\frac{1}{2}mr^2$, where $m$ is the mass and $r$ is the radius. Therefore, the moment of inertia for the given flywheel is $I=\frac{1}{2}(400\;m{kg})(0.6\;m{m})^2=72\;m{kg\cdot m^2}$. 
The kinetic energy of a rotating object is given by $K=\frac{1}{2}I\omega^2$, where $\omega$ is the angular velocity. Setting $K$ equal to the energy required to drive the car, we have:
$$3.2	imes 10^{17}\;m{J}=\frac{1}{2}(72\;m{kg\cdot m^2})\omega^2$$
Solving for $\omega$, we get $\omega=940\;m{rad/s}$. Thus, the answer is choice C.

Question 7

A uniform solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with an angular speed of 33.4 rpm. What is its kinetic energy?

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Question 8

What is the kinetic energy of a 120-cm thin uniform rod with a mass of 450 g that is rotating about its center at 3.60 rad/s?

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The answer of What is the kinetic energy of a...

Question 9

On a frictionless horizontal surface, a 1.50-kg mass traveling at 3.50 m/s suddenly collides with and sticks to a 3.00-kg mass that is initially at rest, as shown in the figure. This system then runs into an ideal spring of force constant (spring constant)50.0 N/cm.
(a)What will be the maximum compression distance of the spring?
(b)How much mechanical energy is lost during this process? During which parts of the process (the collision and compression of the spring)is this energy lost?

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The answer of On a frictionless horizontal surface, a 1.50-kg...

Question 10

As shown in the figure, two blocks are connected by a light string that passes over a frictionless pulley having a moment of inertia of 0.0040 kg ∙ m² and diameter 10 cm. The coefficient of kinetic friction between the table top and the upper block is 0.30. The blocks are released from rest, and the string does not slip on the pulley. How fast is the upper block moving when the lower one has fallen 0.60 m?

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Question 11

The L-shaped object shown in the figure consists of three small masses connected by thin uniform rods, each rod of mass 3.00 kg. Assume that the masses shown are accurate to three significant figures. How much work must be done to accelerate the object from rest to an angular speed of 3.25 rad/s about the y-axis?

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Question 12

A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill, starting from rest. If the angular speed of the disk is 4.27 rad/s at the bottom, how high did it start on the hill?

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Question 13

The L-shaped object shown in the figure consists of three small masses connected by extremely light rods. Assume that the masses shown are accurate to three significant figures. How much work must be done to accelerate the object from rest to an angular speed of 3.25 rad/s (a)about the x-axis, (b)about the y-axis?

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Question 14

A solid uniform ball with a mass of 125 g is rolling without slipping along the horizontal surface of a table with a speed of 4.5 m/s when it rolls off the edge and falls towards the floor, 1.1 m below. What is the rotational kinetic energy of the ball just before it hits the floor?

Accepted Answer

We can use the conservation of energy to solve this problem. Initially, the ball has both translational and rotational kinetic energy given by:
KE_initial = (1/2)mv^2 + (1/2)Iω^2
where m is the mass of the ball, v is the speed with which it is rolling, I is the moment of inertia of the ball (which depends on its shape and mass distribution), and ω is its angular velocity (which is related to its speed by ω = v/r, where r is its radius).

Since the ball is rolling without slipping, we can also relate its translational and rotational motion by v = ωr, so we can rewrite the expression for KE_initial as:
KE_initial = (1/2)mv^2 + (1/2)(I/m)v^2    [Using I = (2/5)mr^2 for a solid sphere]
KE_initial = (7/10)mv^2

When the ball falls off the table, its gravitational potential energy is converted into kinetic energy. At the instant it hits the floor, all of its initial kinetic energy has been converted into translational kinetic energy, so we have:
KE_final = (1/2)mv_final^2

We can relate v_final to the height from which the ball falls by the conservation of energy:
PE_initial = KE_final
mgh = (1/2)mv_final^2
v_final = (2gh)^0.5

Substituting this expression into the expression for KE_final, we get:
KE_final = (1/2)mv_final^2 = (1/2)m(2gh) = mgh

Finally, we can compute the change in kinetic energy as the difference between KE_final and KE_initial:
ΔKE = KE_final - KE_initial = mgh - (7/10)mv^2

Substituting the given values, we get:
ΔKE = (0.125 kg)(9.81 m/s^2)(1.1 m) - (7/10)(0.125 kg)(4.5 m/s)^2 ≈ 0.51 J

Therefore, the answer is A) 0.51 J.

Question 15

While spinning down from 500 rpm to rest, a flywheel does $3.9 \mathrm {~kJ}$ of work. This flywheel is in the shape of a solid uniform disk of radius 1.2 m. What is the mass of this flywheel?

Accepted Answer

The correct answer is A.The work done by the flywheel is equal to the change in its rotational kinetic energy. The rotational kinetic energy of a solid uniform disk is given by:$$K = \frac{1}{2} I \omega^2$$where:* K is the rotational kinetic energy* I is the moment of inertia* ω is the angular velocityThe moment of inertia of a solid uniform disk is given by:$$I = \frac{1}{2} MR^2$$where:* M is the mass* R is the radiusSubstituting the given values into the equation for rotational kinetic energy, we get:$$K = \frac{1}{2} \left(\frac{1}{2} MR^2ight) \omega^2 = \frac{1}{4} MR^2 \omega^2$$The change in rotational kinetic energy is:$$\Delta K = K_f - K_i = \frac{1}{4} MR^2 \omega_f^2 - \frac{1}{4} MR^2 \omega_i^2$$where:* ω_i is the initial angular velocity* ω_f is the final angular velocitySubstituting the given values into the equation for the change in rotational kinetic energy, we get:$$\Delta K = \frac{1}{4} M(1.2 	ext{ m})^2 (0 	ext{ rad/s})^2 - \frac{1}{4} M(1.2 	ext{ m})^2 (500 	ext{ rpm})^2$$Converting the initial angular velocity from rpm to rad/s:$$\omega_i = 500 	ext{ rpm} \left(\frac{2\pi 	ext{ rad}}{60 	ext{ s}}ight) = 52.36 	ext{ rad/s}$$Substituting the converted value into the equation for the change in rotational kinetic energy, we get:$$\Delta K = \frac{1}{4} M(1.2 	ext{ m})^2 (0 	ext{ rad/s})^2 - \frac{1}{4} M(1.2 	ext{ m})^2 (52.36 	ext{ rad/s})^2$$$$\Delta K = -3.9 	ext{ kJ}$$Solving for the mass, we get:$$M = \frac{4\Delta K}{R^2 \omega_i^2} = \frac{4(-3.9 	ext{ kJ})}{(1.2 	ext{ m})^2 (52.36 	ext{ rad/s})^2} = 4.0 	ext{ kg}$$Therefore, the mass of the flywheel is 4.0 kg.

Question 16

An object initially at rest suddenly explodes in two fragments of masses $2.6 \mathrm {~kg}$ and $3.3 \mathrm {~kg}$ that move in opposite directions. If the speed of the first fragment is $3.6 \mathrm {~m} / \mathrm { s } \text {, }$ find the internal energy of the explosion.

Accepted Answer

The internal energy of the explosion is equal to the kinetic energy of the fragments. The kinetic energy of the first fragment is:$$K_1 = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} (2.6 \mathrm{~kg}) (3.6 \mathrm{~m/s})^2 = 18.48 \mathrm{~J}$$The kinetic energy of the second fragment is:$$K_2 = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} (3.3 \mathrm{~kg}) v_2^2$$Since the fragments move in opposite directions, the total kinetic energy is:$$K = K_1 + K_2 = 18.48 \mathrm{~J} + \frac{1}{2} (3.3 \mathrm{~kg}) v_2^2$$The internal energy of the explosion is:$$U = K = 18.48 \mathrm{~J} + \frac{1}{2} (3.3 \mathrm{~kg}) v_2^2$$We don't know the value of $v_2$, but we can use the conservation of momentum to find it:$$m_1 v_1 = m_2 v_2$$$$v_2 = \frac{m_1 v_1}{m_2} = \frac{(2.6 \mathrm{~kg})(3.6 \mathrm{~m/s})}{3.3 \mathrm{~kg}} = 2.7 \mathrm{~m/s}$$Substituting this into the equation for the internal energy, we get:$$U = 18.48 \mathrm{~J} + \frac{1}{2} (3.3 \mathrm{~kg}) (2.7 \mathrm{~m/s})^2 = 30 \mathrm{~J}$$Therefore, the internal energy of the explosion is 30 J.

Question 17

A futuristic design for a car is to have a large flywheel within the car to store kinetic energy. The flywheel is a solid uniform disk of mass 370 kg with a radius of 0.50 m, and it can rotate up to $200 \mathrm { rev } / \mathrm { s }$ Assuming all of this stored kinetic energy could be transferred to the linear speed of the $1500 - \mathrm { kg }$ car, find the maximum attainable speed of the car.

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Question 18

A solid uniform sphere of mass 120 kg and radius 1.7 m starts from rest and rolls without slipping down an inclined plane of vertical height 5.3 m. What is the angular speed of the sphere at the bottom of the inclined plane?

Accepted Answer

The angular speed of the sphere at the bottom can be found using energy conservation principles. The potential energy at the top is converted into translational and rotational kinetic energy at the bottom. The potential energy at the top is given by $PE = mgh$, where $m = 120\,kg$, $g = 9.8\,m/s^2$, and $h = 5.3\,m$. The kinetic energy at the bottom is the sum of translational ($\frac{1}{2}mv^2$) and rotational ($\frac{1}{2}I\omega^2$) kinetic energies. For a solid sphere, $I = \frac{2}{5}mr^2$, and the condition for rolling without slipping is $v = r\omega$.Setting the potential energy equal to the total kinetic energy gives:$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$Substituting $I = \frac{2}{5}mr^2$ and $v = r\omega$, we get:$$mgh = \frac{1}{2}m(r\omega)^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\omega^2$$Simplifying and solving for $\omega$, we find:$$gh = \frac{7}{10}r^2\omega^2$$$$ \omega = \sqrt{\frac{10gh}{7r^2}}$$Plugging in the given values:$$ \omega = \sqrt{\frac{10 \times 9.8 \times 5.3}{7 \times (1.7)^2}}$$$$ \omega \approx 5.1\,rad/s$$

Question 19

A baseball pitcher is employing a ballistic pendulum to determine the speed of his fastball. A $3.3 - \mathrm { kg }$ lump of clay is suspended from a cord 2.0 m long. When the pitcher throws his fastball aimed directly at the clay, the ball suddenly becomes embedded in the clay and the two swing up to a maximum height of 0.080 m. If the mass of the baseball is 0.21 kg, find the speed of the pitched ball.

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Question 20

In a certain nuclear reactor, neutrons suddenly collide with carbon nuclei, which have 12 times the mass of neutrons. In a head-on elastic collision with a stationary carbon nucleus, what fraction of its initial speed does the neutron have after the collision?

Accepted Answer

In an elastic collision between a neutron of mass $m_1$ and a carbon nucleus of mass $m_2$, the velocity of the two objects after the collision can be related to their velocities before the collision through the following equation: $$v_{	ext{after}}^{(1)}=\frac{(m_1-m_2)v_{	ext{before}}^{(1)}}{m_1+m_2}+\frac{2m_2v_{	ext{before}}^{(2)}}{m_1+m_2},$$ where $v_{	ext{before}}^{(1)}$ and $v_{	ext{before}}^{(2)}$ are the initial velocities of the neutron and carbon nucleus, respectively.
Because the carbon nucleus is initially stationary, $v_{	ext{before}}^{(2)}=0$. Additionally, because the neutron has much less mass than the carbon nucleus ($m_1\ll m_2$), we can simplify the above equation to $$v_{	ext{after}}^{(1)}\approx\frac{2v_{	ext{before}}^{(1)}}{13}.$$ Thus, the fraction of the neutron's initial speed that it retains after the collision is $\boxed{	extbf{(C) }11/13}$.

Quiz 10: Energy and Work

A wheel having a moment of inertia of 5.00 kg ∙ m2 starts from rest and accelerates under a constant torque of 3.00 N ∙ m for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s?