91Ó°ÊÓ

Momentum is a fundamental concept in physics. When two objects collide, the total momentum of the system before collision remains the same after collision, given no external forces interfere. An elastic collision is a special type of interaction in which total momentum is preserved.

In our problem, we have ball 1 with mass \(2m\) moving towards a stationary ball 2 with mass \(m\). Let's denote the initial velocity of ball 1 as \(v\), and the initial velocity of ball 2 is \(0\) since it's at rest. Thus, the total initial momentum can be expressed as:

• \(p_{initial} = (2m)v + (m)(0) = 2mv\).

After the collision, the momentum for each ball changes, but the total must equal the initial. Let the final velocities of ball 1 and ball 2 be \(v_1'\) and \(v_2'\) respectively. Therefore, the equation for the conservation of momentum is:

• \(2mv = 2mv_1' + mv_2'\).

This relationship helps us to determine the velocity changes for each object involved.

In an elastic collision like the one described, not only is momentum conserved but so is kinetic energy. This means the total kinetic energy before the collision equals the total kinetic energy after the collision. Initially, only ball 1 is moving, so its initial kinetic energy is given by:

• \(KE_{initial} = \frac{1}{2}(2m)v^2 = mv^2\).

After the collision, both balls have some velocity, so we write their kinetic energies in terms of their final velocities:

• \(KE_{final} = \frac{1}{2}(2m)(v_1'^2) + \frac{1}{2}(m)(v_2'^2)\).

By setting \(KE_{initial} = KE_{final}\), we arrive at the equation:

• \(mv^2 = mv_1'^2 + \frac{1}{2}mv_2'^2\).

This second equation gives us the second set of values needed to solve for the final velocities of both balls.

After establishing our equations from conservation laws, the next phase is calculating the final velocities using algebra. We have two equations derived from momentum and kinetic energy:

• From Momentum: \(2v = 2v_1' + v_2'\)
• From Kinetic Energy: \(v^2 = v_1'^2 + \frac{1}{2}v_2'^2\)

We can solve these equations simultaneously. Let's manipulate and substitute. First, express \(v_2'\) in terms of \(v_1'\) using momentum:

• \(v_2' = 2(v - v_1')\)

By substituting this into the kinetic energy equation, we next solve:

• \(v^2 = v_1'^2 + 2(v - v_1')^2\)

After simplifying, we find:

• \(v_1' = -\frac{v}{3}\)
• \(v_2' = \frac{4v}{3}\)

These values tell us the post-collision speeds and directions of the two masses.

A head-on collision means the centers of the two colliding objects are aligned along the same line of motion, leading to a straightforward interaction. This setup simplifies analysis because all motion is along a single axis.

In such collisions, understanding momentum conservation becomes straightforward since there isn't any angular momentum involved. Only linear velocities need consideration. Using our momentum and energy equations, we found that:

• The heavier ball (mass \(2m\)) reverses direction after the collision. It ends up moving with a velocity of \(-\frac{v}{3}\).
• The lighter ball (mass \(m\)) goes in the positive direction with a speed of \(\frac{4v}{3}\).

These outcomes demonstrate how momentum and energy are distributed in a perfectly elastic collision, allowing us to predict motion accurately.