Problem 64 - College Physics

Question

Two objects of masses m and 3m are moving toward each other along the x-axis with the same initial speed v0. The object with mass m is traveling to the left, and the object with mass 3m is traveling to the right. They undergo an elastic glancing collision such that m is moving downward after the collision at right angles from its initial direction. (a) Find the final speeds of the two objects. (b) What is the angle u at which the object with mass 3m is scattered?

Step-by-Step Solution

Answer

The final speeds are given by \(v1 = 4*v0/3\) for mass \(3m\), and \(v2 = sqrt(2)*v0\) for mass \(m\). The scattering angle is \(u = arctan(3sqrt(2)/4)\).

Step 1: Conservation of momentum and kinetic energy for calculating the final speeds

Set up the conservation of momentum on the x-axis and the y-axis. For the x-axis, since the masses were moving towards each other initially, we have \(m*v0 + 3m*v0 = 3m*v1 + m*v2*cos(90)\) where \(m*v0\) is the momentum of the mass \(m\) and \(3m*v0\) is the momentum of the mass \(3m\). \(v1\) and \(v2\) are the final velocities of mass \(3m\) and \(m\) respectively. Since \(cos(90) = 0\), the equation reduces to \(4m*v0 = 3m*v1\). Thus, the final velocity of object with mass \(3m\) is \(v1 = 4*v0/3\). Apply the conservation of energy, \(1/2*m*v0^2 + 1/2*3m*v0^2 = 1/2*3m*v1^2 + 1/2*m*v2^2\) which simplifies to \(2v0^2 = 2v1^2 + v2^2\). Solving the equation, we get the final speed of mass \(m, v2 = sqrt(2)*v0\)

Step 2: Calculate the scattering angle

Use the phrase 'glancing collision' to infer that the angle of deviation is 90 degrees. Therefore, we can draw a right angled triangle with the y-axis as the line of deviation, the vector component in the x direction as the adjacent side and the resultant velocity as the hypotenuse. In this right triangle, we can calculate the scattering angle \(u\) using trigonometry - specifically tangent which is ration of the opposite to the adjacent in a right triangle. Therefore, \(tanu = v2sin(90)/v1\). By substituting the values of \(v1\) and \(v2\), we can solve for \(u\).

Step 3: Plugging the values

We can get \(u = arctan(v2/v1)\). Substituting the obtained values of \(v1\) and \(v2\), we get \(u = arctan(sqrt(2)*v0/(4*v0/3)) = arctan(3sqrt(2)/4)\)

Key Concepts

Conservation of Momentum

Understanding the conservation of momentum is crucial in physics, particularly in the study of collisions. Momentum is a measure of an object's motion, which is the product of its mass and velocity. The principle of conservation of momentum states that, in a closed system where no external forces are acting, the total momentum before a collision is equal to the total momentum after the collision.

In the context of the problem, when the two objects with masses m and 3m collide elastically along the x-axis, the conservation of momentum can be applied separately along the x-axis and y-axis. For the x-axis, the initial momentum of the two-object system can be summed up as \(m*v0 - 3m*v0\). Since the collision is elastic and no external forces are involved, this total momentum must be the same after the collision. By setting up the equations accordingly, the final velocities can be determined.

Conservation of Kinetic Energy

In an elastic collision, not only is momentum conserved, but kinetic energy is also conserved. Kinetic energy is the energy that an object possesses due to its motion and is given by the formula \(\frac{1}{2}mv^2\). For elastic collisions, this means that the total kinetic energy before and after the collision remains constant.

In our problem, we leverage the conservation of kinetic energy to solve for the final speeds of the objects. Before the collision, the total kinetic energy is the sum of the kinetic energies of each object, expressed as \(\frac{1}{2}m*v0^2 + \frac{1}{2}3m*v0^2\). After the collision, the total kinetic energy should be equal to the sum of the individual kinetic energies of the objects. Setting these two expressions equal and solving for the unknowns provides the final velocities without losing any energy to heat or deformation.

Scattering Angle

The scattering angle is a term used to describe the angle at which an object is deflected or scattered after a collision. It is a critical factor in collision problems as it helps to understand the directional change of objects. In the stated problem, the object with mass m moves downwards at right angles after the collision which implies a 90-degree change from its initial direction.

The scattering angle for the object with mass 3m can be found by drawing a right triangle with velocity vectors and using trigonometric relationships. The use of tangent, which relates the opposite side to the adjacent side in a right triangle, aids in calculating the angle. This step not only helps in visualizing the physics involved but also provides a more profound understanding of the post-collision behavior of the objects.

Trigonometry in Physics

Trigonometry, the branch of mathematics that deals with the relationships between the sides and angles of triangles, finds extensive applications in physics, especially when analyzing vectors and motion. In collision problems, trigonometry is used to resolve vector components and to calculate angles of deflection, such as the scattering angle mentioned previously.

Using trigonometric functions such as sine, cosine, and tangent can simplify the complex motions into more manageable calculations. These functions are particularly handy when objects move in two dimensions, as in the case of our elastic glancing collision where they allow us to express the movement of the objects post-collision in terms of their components along the x and y axes. Thus, a firm grasp of trigonometry is essential for problem-solving in physics.