Please Solve The Following Problems. Show All Work
Please Solve The Following Problems You Must Show All Work
Please solve the following problems. You must show all work. 1.A rail car of mass 4500 kg is moving with a velocity of 12.4 m/s collides with a stationary rail car of mass 6700 kg. What is the velocity of the two cars if they stick after they collide? 2.
A tennis player hits a 1.45 kg tennis ball with a racket of mass 2.5 kg. If he hits the ball with a velocity of 7.5 m/s and then stops, what impulse did he imply on the ball? What is the ball’s velocity? 3. A 35 g bullet is fired in a ballistic pendulum device (see page 178 In text) and increases the height of the center of mass of the device and the bullet by 2.1 cm.
If the level arm has a mass of 3.4 kg, what was the initial velocity of the bullet? 4. A puck slides on a surface for 5.6 m before it stops. If it was moving at a velocity of 15 m/s and has a mass of 1.2 kg, what is the force of friction exerted on the puck? There is a non-conservative work done on the puck equal to the change in kinetic energy, or: 5.
If a tire on a semi-truck is 1.1 m. How many FULL rotations will the tire roll if the truck moves 24.5 meters? 6. A large circus ride has a centripetal acceleration at 6.0 m/s 2 . If the circular ride has a radius of 11.0 m, what is its velocity at the edge? 7. If a 56 diameter circular disk is uniformly angularly accelerating at a rate of 5.0 rad/s 2 , what will the final angular_velocity be after 4.0 seconds if it starts at a rate of 10.0 rad/s? What angle is subtended over that time? 8. A 40.0 N force is applied to a 46.0 cm long torque wrench. What is the net torque applied to the nut? Also if this force is applied for 3.0 seconds and the bolt starts from rest and ends with an angular velocity of 0.5 rad/s, what is the bolt’s moment of inertia? Notes on homework: On number 2, it means the racket stops after it strikes the ball. On number 3, it refers to page 178, but it really means example 7-9 on page 181. For the second part of number 4, it sounds confusing, but it's asking about the momentum of the puck. Number 5 is referring to the DIAMETER of the tire. If it meant the radius, the tire would be over 7 feet tall. Answer the second part of number 7 in radians or revolutions. Don't use degrees. With these rotational problems, you should get used to putting your calculator in Radian mode. On number 8, you can assume the force is being applied perpendicular to the wrench.
Paper For Above instruction
Introduction
Physics problems involving linear and rotational motion encompass fundamental principles such as conservation of momentum, energy, and kinematics. These concepts are essential for analyzing real-world systems, from collisions to rotational dynamics. This paper provides detailed solutions to eight diverse physics problems, illustrating the application of these core principles with step-by-step calculations and explanations.
Problem 1: Colliding Rail Cars
A rail car of mass 4500 kg moving at 12.4 m/s collides with a stationary rail car of mass 6700 kg and they stick together. To find their combined velocity after collision, we use the conservation of momentum:
\[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \]
Given:
\[ m_1 = 4500\,\text{kg} \]
\[ v_1 = 12.4\,\text{m/s} \]
\[ m_2 = 6700\,\text{kg} \]
\[ v_2 = 0\,\text{m/s} \]
Calculating:
\[ (4500)(12.4) + (6700)(0) = (4500 + 6700) v_f \]
\[ 55800 = 11200 v_f \]
\[ v_f = \frac{55800}{11200} \approx 4.987\,\text{m/s} \]
The combined cars move at approximately 4.99 m/s after the collision.
Problem 2: Impulse and Ball Velocity Post-Impact
A tennis player hits a 1.45 kg tennis ball with a 2.5 kg racket at 7.5 m/s, then stops, implying a change in momentum (impulse). The impulse \( J \) is:
\[ J = \Delta p = m_{ball} v_{f} - m_{ball} v_{initial} \]
Since the racket stops after hitting:
\[ \text{Impulse} = m_{ball} v_{f} \]
and the impulse exerted on the ball is equal in magnitude to the change in momentum:
\[ J = m_{ball} (v_{f} - 0) \]
The impulse:
\[ J = 1.45 \times 7.5 = 10.875\,\text{kg·m/s} \]
The force applied over the contact time relates to impulse. Assuming the force acts over a very short contact interval \( \Delta t \), the average force:
\[ F_{avg} = \frac{J}{\Delta t} \]
Without \( \Delta t \), we focus on the impulse value, which is about 10.88 kg·m/s.
The ball's velocity immediately after impact is 7.5 m/s, as given.
Problem 3: Bullet in Ballistic Pendulum
A 35 g (0.035 kg) bullet strikes a ballistic pendulum, raising the combined mass by 2.1 cm (\( 0.021\,\text{m} \)). The level arm mass is 3.4 kg, and the total increased height relates to the initial velocity \( v \) of the bullet via conservation of momentum:
\[ m_{bullet} v = (m_{level\,arm} + m_{bullet}) v_{pendulum} \]
The potential energy gained:
\[ PE = (m_{total}) g h \]
Expressing the initial velocity:
\[ v_{pendulum} = \sqrt{2 g h} \]
Total mass:
\[ m_{total} = 3.4 + 0.035 = 3.435\,\text{kg} \]
Calculations:
\[ v_{pendulum} = \sqrt{2 \times 9.81 \times 0.021} \approx \sqrt{0.412} \approx 0.64\,\text{m/s} \]
Initial bullet velocity:
\[ v = \frac{m_{total} v_{pendulum}}{m_{bullet}} = \frac{3.435 \times 0.64}{0.035} \approx \frac{2.198}{0.035} \approx 62.8\,\text{m/s} \]
Thus, the initial velocity of the bullet was approximately 62.8 m/s.
Problem 4: Frictional Force on Puck
A puck with mass 1.2 kg slides 5.6 m at an initial velocity of 15 m/s before stopping. The work done by friction equals the change in kinetic energy:
\[ W_{friction} = \Delta KE = \frac{1}{2} m v_{final}^2 - \frac{1}{2} m v_{initial}^2 \]
As:
\[ v_{final} = 0 \]
\[ \Delta KE = - \frac{1}{2} \times 1.2 \times (15)^2 = -135\,\text{J} \]
The force of friction:
\[ F_{friction} = \frac{W_{friction}}{d} = \frac{-135}{5.6} \approx -24.11\,\text{N} \]
The negative sign indicates direction opposite to motion.
In magnitude, the force exerted on the puck is approximately 24.11 N.
Problem 5: Tire Rotations
A tire with diameter 1.1 m rolls a distance of 24.5 m:
\[ \text{Number of rotations} = \frac{\text{Distance}}{\text{Circumference}} \]
Circumference:
\[ C = \pi D = \pi \times 1.1 \approx 3.455\,\text{m} \]
Rotations:
\[ N = \frac{24.5}{3.455} \approx 7.09 \]
The tire completes approximately 7.09 full rotations.
Problem 6: Circular Ride Velocity
Centripetal acceleration \( a_c = \frac{v^2}{r} \)
Given:
\[ a_c = 6.0\,\text{m/s}^2 \]
\[ r = 11.0\,\text{m} \]
Solve for \( v \):
\[ v = \sqrt{a_c \times r} = \sqrt{6.0 \times 11.0} \approx \sqrt{66} \approx 8.12\,\text{m/s} \]
The velocity at the edge of the ride is approximately 8.12 m/s.
Problem 7: Rotational Kinetics
A disk with diameter 56 cm (0.56 m) and radius \( r = 0.28\,\text{m} \) is angularly accelerating at \( 5.0\,\text{rad/s}^2 \), starting from \( \omega_i = 10.0\,\text{rad/s} \):
Final angular velocity:
\[ \omega_f = \omega_i + \alpha t = 10.0 + 5.0 \times 4.0 = 30.0\,\text{rad/s} \]
Subtended angle \( \theta \):
\[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 = 10.0 \times 4.0 + 0.5 \times 5.0 \times 16 = 40 + 40 = 80\,\text{rad} \]
Revolutions:
\[ \text{Revolutions} = \frac{\theta}{2\pi} \approx \frac{80}{6.283} \approx 12.74 \]
The disk rotates through about 12.74 revolutions.
Problem 8: Torque and Moment of Inertia
Force \( F = 40.0\,\text{N} \) applied perpendicular to a wrench of length \( L = 46\,\text{cm} = 0.46\,\text{m} \):
Net torque:
\[ \tau = F \times L = 40.0 \times 0.46 = 18.4\,\text{Nm} \]
Angular acceleration \( \alpha \):
\[ \alpha = \frac{\omega_{final} - \omega_{initial}}{t} = \frac{0.5\,\text{rad/s} - 0}{3.0\,\text{s}} \approx 0.167\,\text{rad/s}^2 \]
Moment of inertia \( I \):
\[ I = \frac{\tau}{\alpha} = \frac{18.4}{0.167} \approx 110.2\,\text{kg·m}^2 \]
This approximates the moment of inertia of the bolt.
Conclusion
These problems span a broad spectrum of classical mechanics concepts, demonstrating the application of conservation laws, kinematic equations, and rotational dynamics calculations. Understanding these principles allows for the analysis of complex physical systems relevant to engineering, sports, transportation, and amusement rides. Accurate application of formulas, careful attention to units, and correct interpretation of problem statements are essential for effective problem-solving in physics.
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