Question 1

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:The total mechanical energy of the system is conserved. The initial energy is all potential energy stored in the first bob, and the final energy is all potential energy stored in the second bob. Therefore, we have:$$m_1gy_1 = m_2gy_2$$where $m_1$ and $y_1$ are the mass and height of the first bob, and $m_2$ and $y_2$ are the mass and height of the second bob. Solving for $y_2$, we get:$$y_2 = \frac{m_1y_1}{m_2} = \frac{(0.25 \mathrm{~kg})(0.45 \mathrm{~m})}{0.54 \mathrm{~kg}} = 0.21 \mathrm{~m} = 21 \mathrm{~cm}$$Therefore, the second bob will rise 21 cm above its initial position.

Question 2

As shown in the figure, a bullet of mass 0.010 kg moving horizontally suddenly strikes a block of wood of mass 1.5 kg that is suspended as a pendulum. The bullet lodges in the wood, and together they swing upward a vertical distance of 0.40 m. The length of the string is 2.0 m. What was the speed of the bullet just before it struck the wooden block?

Accepted Answer

The answer of As shown in the figure, a bullet...

Question 3

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?

Accepted Answer

The answer of A uniform solid cylinder with a radius...

Question 4

A baseball pitcher is employing a ballistic pendulum to determine the speed of his fastball. A $\text { 3.3-kg }$ 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.

Accepted Answer

The answer of A baseball pitcher is employing a ballistic...

Question 5

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.

Accepted Answer

The answer of A futuristic design for a car is...

Question 6

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, the velocity of a particle after collision can be determined using the formula: $v' = \frac{(m_1 - m_2)}{(m_1 + m_2)}v$, where $v'$ is the velocity of the particle after collision, $m_1$ is the mass of the moving particle (neutron), $m_2$ is the mass of the stationary particle (carbon nucleus), and $v$ is the initial velocity of the moving particle. Given that the carbon nucleus has 12 times the mass of the neutron, we substitute $m_1 = 1m$ and $m_2 = 12m$ into the formula, resulting in $v' = \frac{(1m - 12m)}{(1m + 12m)}v = \frac{-11m}{13m}v$. The negative sign indicates direction change, but the question asks for the fraction of the initial speed, which is the absolute value, hence $|-\frac{11}{13}|$ or $\frac{11}{13}$.

Question 7

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?

Accepted Answer

The answer of In a certain nuclear reactor, neutrons suddenly...

Question 8

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 of a solid cylinder is given by:

$I = \frac{1}{2}mr^2$

Plugging in the given values, we get:

$I = \frac{1}{2}(400	ext{ kg})(0.6	ext{ m})^2 = 72	ext{ kg}\cdot	ext{m}^2$

The rotational kinetic energy of the flywheel is given by:

$K = \frac{1}{2}I\omega^2$

where $\omega$ is the angular speed of the flywheel.

We want to find $\omega$ such that $K = 3.2	imes 10^{17}$ J. Plugging in the values for $I$ and $K$, we get:

$3.2	imes 10^{17} = \frac{1}{2}(72	ext{ kg}\cdot	ext{m}^2)\omega^2$

Solving for $\omega$, we get:

$\omega = \sqrt{\frac{2K}{I}} \approx 940	ext{ rad/s}$

Therefore, the best choice is (C).

Question 9

An 8.0-g bullet is suddenly shot into a 4.0-kg block that is at rest on a frictionless horizontal surface, as shown in the figure. The bullet remains lodged in the block. The block then moves against a spring and compresses it by 3.7 cm. The force constant (spring constant) of the spring is $2500 \mathrm {~N} / \mathrm { m }$ What was the initial speed v of the bullet?

Accepted Answer

The answer of An 8.0-g bullet is suddenly shot into...

Question 10

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 11

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 work done by the flywheel is equal to its change in kinetic energy. The kinetic energy of a rotating disk is given by:$$K = \frac{1}{2}I\omega^2$$where I is the moment of inertia and $\omega$ is the angular velocity. The moment of inertia of a solid uniform disk is given by:$$I = \frac{1}{2}MR^2$$where M is the mass and R is the radius. Substituting these expressions into the equation for kinetic energy, we get:$$K = \frac{1}{4}MR^2\omega^2$$The change in kinetic energy is:$$\Delta K = K_f - K_i = \frac{1}{4}MR^2(\omega_f^2 - \omega_i^2)$$where $\omega_i$ is the initial angular velocity and $\omega_f$ is the final angular velocity. Substituting the given values, we get:$$\Delta K = \frac{1}{4}M(1.2 \mathrm{~m})^2[(0 \mathrm{~rad/s})^2 - (500 \mathrm{~rpm})(2\pi \mathrm{~rad/min}/60 \mathrm{~s/min})^2]$$$$\Delta K = \frac{1}{4}M(1.44 \mathrm{~m}^2)(1.05 \times 10^4 \mathrm{~rad^2/s^2})$$$$\Delta K = 4.0M \mathrm{~J}$$Setting this equal to the work done by the flywheel, we get:$$4.0M \mathrm{~J} = 3.9 \mathrm{~kJ}$$$$M = \frac{3.9 \mathrm{~kJ}}{4.0 \mathrm{~J/kg}} = 0.975 \mathrm{~kg}$$Therefore, the mass of the flywheel is 0.975 kg, which is closest to choice A) 4.0 kg.

Question 12

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?

Accepted Answer

The answer of The L-shaped object shown in the figure...

Question 13

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?

Accepted Answer

The answer of A 1200-kg pick-up truck traveling south at...

Question 14

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 15

In a police rifle test, a 15-g bullet traveling 213 m/s in a vertical direction suddenly buries itself in a 2.4-kg block of wood at rest directly above it. As a result, the bullet-block combination moves vertically upward.
(a) Determine the velocity of the bullet-block combination just after the impact.
(b) Determine the maximum height reached by the bullet/block combination.
(c) Is kinetic energy conserved in this collision?

Accepted Answer

The answer of In a police rifle test, a 15-g...

Question 16

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?

Accepted Answer

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

Question 17

An 8.0-g bullet is suddenly shot into a 4.0-kg block, at rest on a frictionless horizontal surface, as shown in the figure. The bullet remains lodged in the block. The block then moves against a spring and compresses it by 8.9 cm. The force constant (spring constant) of the spring is $1400 \mathrm {~N} / \mathrm { m }$ What is the magnitude of the impulse on the block (including the bullet inside) due to the spring during the entire time interval during which the block compresses the spring?

Accepted Answer

The answer of An 8.0-g bullet is suddenly shot into...

Question 18

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:$$K_1 = \frac{1}{2}mv_1^2 = \frac{1}{2}(2.6 \mathrm{~kg})(3.6 \mathrm{~m/s})^2 = 18.48 \mathrm{~J}$$The kinetic energy of the second fragment is:$$K_2 = \frac{1}{2}mv_2^2 = \frac{1}{2}(3.3 \mathrm{~kg})v_2^2$$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 equal to the total kinetic energy, so:$$U = K = 18.48 \mathrm{~J} + \frac{1}{2}(3.3 \mathrm{~kg})v_2^2$$We don't know the value of $v_2$, but we can use the conservation of momentum to find it. The total momentum of the system before the explosion is zero, so the total momentum after the explosion must also be zero:$$0 = (2.6 \mathrm{~kg})(3.6 \mathrm{~m/s}) + (3.3 \mathrm{~kg})v_2$$Solving for $v_2$, we get:$$v_2 = -2.7 \mathrm{~m/s}$$Substituting this value 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 19

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?

Accepted Answer

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

Question 20

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?

Accepted Answer