91Ó°ÊÓ

Momentum is a fundamental concept in physics. It is defined as the product of an object's mass and velocity, represented as \( p = mv \). Momentum is a vector quantity, meaning it has both magnitude and direction. In a closed system, the total momentum is conserved. This means that before and after a collision, the total momentum of the system remains unchanged.
In our exercise, a 3 kg object is moving with a velocity of 8 m/s and collides with an object of unknown mass \( M \), initially at rest. After the collision, the second object moves at 6 m/s. Using the conservation of momentum principle, we set up the equation:
\[m_1v_1 + Mv_2 = m_1v_1' + Mv_2'\]
Substituting the known values gives us:

• The initial momentum of the 3 kg object: \(3 imes 8 = 24\)
• The initial momentum of mass \( M = 0 \) because it is stationary
• The final momentum of \( M = M imes 6 \)
• The final momentum of the 3 kg object is unknown, so we have \(3v_1'\)

Bringing it all together yields the equation: \(24 = 3v_1' + 6M\), from which we solve for \( v_1' \) and later derive \( M \) using additional equations.

Kinetic energy is the energy that an object possesses due to its motion, described by the formula \( KE = \frac{1}{2}mv^2 \). During elastic collisions, not only is momentum conserved, but kinetic energy is as well. This means the total kinetic energy before collision is equal to the total after the collision.
For our problem, we write the energy conservation equation as:
\[\frac{1}{2}m_1v_1^2 + \frac{1}{2}Mv_2^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}Mv_2'^2\]
Since the second object was initially at rest, we simplify to:

• The initial kinetic energy of the 3 kg object: \(\frac{1}{2} \times 3 imes 8^2 = 96\)
• The final kinetic energy of \( M \): \(\frac{1}{2}M \times 6^2 = 18M\)
• The final kinetic energy of 3 kg object: \(\frac{1}{2} \times 3v_1'^2 \)

The energy conservation equation thus becomes: \(96 = 1.5v_1'^2 + 18M\). This is used alongside the momentum conservation equation to solve for \( M \).

One-dimensional collisions involve objects moving along a single straight line. This simplifies calculations of momentum and energy because all vector directions point along the same axis, removing the need to consider separate components.
In our scenario, we have an elastic one-dimensional collision. We assume no external forces, so the conservation laws apply cleanly. Given that the objects in the problem travel only along the x-axis, we can set up equations that reflect this linear movement. This means:

• All velocities are either positive or negative along this single axis.
• Calculations can be straightforwardly performed using scalar values for mass, velocity, momentum, and energy.
• The conservation laws enable us to deduce the unknown values, exploiting the symmetry and principles inherent in these linear interactions.

By combining both the conservation of momentum and kinetic energy, which are inherently built into this one-dimensional collision setup, we successfully determined the mass \( M \) to be 5 kg.