How Can Physics Homework Challenges Enhance Problem-Solving Skills?

[wayfarer]

I was wondering is someone could help me with the following problems:

1. Two children are trying to shoot a marble of mass m into a small box using a spring-loaded gun that is fixed on a table and shoots horizontally from the edge of the table. View Figure The edge of the table is a height H above the top of the box (the height of which is negligibly small), and the center of the box is a distance d from the edge of the table. The spring has a spring constant k. The first child compresses the spring a distance x_1 and finds that the marble falls short of its target by a horizontal distance d_12.
By what distance, x_2, should the second child compress the spring so that the marble lands in the middle of the box? Express answer in terms of m, k, g, H and d.

2. In a ballistic pendulum an object of mass m is fired with an initial speed v_0 at a pendulum bob. The bob has a mass M, which is suspended by a rod of length L and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement theta. Find an expression for v_0, the initial speed of the fired object.

3. At time t=0 a grinding wheel has an angular velocity of 21.0 rad/s. It has a constant angular acceleration of 30.0 rad/s^2 until a circuit breaker trips at time t = 2.30 s. From then on, the wheel turns through an angle of 437 rad as it coasts to a stop at constant angular deceleration.
Through what total angle did the wheel turn between t=0 and the time it stopped?
At what time does the wheel stop?
What was the wheel's angular acceleration as it slowed down?

4. Consider a turntable to be a circular disk of moment of inertia I_t rotating at a constant angular velocity omega_i around an axis through the center and perpendicular to the plane of the disk (the disk's "primary axis of symmetry"). The axis of the disk is vertical and the disk is supported by frictionless bearings. The motor of the turntable is off, so there is no external torque being applied to the axis.
Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is I_r. The initial angular velocity of the second disk is zero.
There is friction between the two disks.
After this "rotational collision," the disks will eventually rotate with the same angular velocity.
What is the final angular velocity of the two disks?
Because of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final rotational kinetic energy, K_f, of the two spinning disks?
Assume that the turntable deccelerated during time Deltat before reaching the final angular velocity ( Deltat is the time interval between the moment when the top disk is dropped and the time that the disks begin to spin at the same angular velocity). What was the average torque, \avg{\tau}, acting on the bottom disk due to friction with the record?

 

[cristo]

Welcome to the forum, wayfarer. Please note that we ask you to show some work before we can give you any help. Also, in future, please post your questions in the relevant homework forum. (There is no need to re-post this, since it will be moved to the correct homework forum).

So.. what thoughts have you had for any of the problems?

 

[Moetasim]

1. To find the required distance x_2 for the second child to compress the spring, we can use the principle of conservation of energy. Since the marble is initially at rest and there is no external force acting on it, the total energy at the start of the experiment is equal to the total energy at the end. The initial energy is potential energy stored in the compressed spring, which is given by 1/2*k*x_1^2. The final energy is the sum of potential energy and kinetic energy of the marble, which is given by 1/2*m*v^2 + m*g*d_12, where v is the horizontal velocity of the marble. Equating the two energies and solving for v, we get v = sqrt((k*x_1^2 + 2*m*g*d_12)/m). Now, using the equation for horizontal displacement, x = v*t, we can find the required distance x_2 as x_2 = (d - d_12) + v^2/(2*g). Substituting the value of v, we get x_2 = (d - d_12) + (k*x_1^2 + 2*m*g*d_12)/(2*m*g). This can be simplified as x_2 = (k*x_1^2 + 4*m*g*d_12)/(4*m*g) or x_2 = (k*x_1^2 + 4*m*g*H)/(4*m*g) if we assume that the height of the table is equal to H.

2. In a ballistic pendulum, the initial kinetic energy of the object is converted into potential energy when it collides with the pendulum bob. This potential energy is then converted back into kinetic energy as the pendulum swings up to its maximum displacement. Using the conservation of energy, we can equate the initial kinetic energy, 1/2*m*v_0^2, with the potential energy at maximum displacement, M*g*L*(1 - cos(theta)). Solving for v_0, we get v_0 = sqrt(2*M*g*L*(1 - cos(theta))/m).

3. The total angle turned by the wheel can be found by integrating the angular velocity with respect to time. Since the acceleration is constant, we can use the equation theta = theta_0 + omega_0*t + 1/2*alpha*t^2, where theta