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II A 50 g marble moving at \(2.0 \mathrm{m} / \mathrm{s}\) strikes a \(20 \mathrm{g}\) marble at rest. What is the speed of each marble immediately after the collision? Assume the collision is perfectly elastic and the marbles collide head-on.

Short Answer

Expert verified
The speed of the 50 g marble after the collision is \(0.4 m/s\), and the speed of the 20 g marble after the collision is \(1.6 m/s\).

Step by step solution

01

Convert masses from grams to kilograms

The masses of the marbles are given in grams, but the standard unit for mass in physics is the kilogram. So, convert both masses into kilograms by dividing them by 1000. Thus, \(m_{1} = 50g = 0.05 kg\) and \(m_{2}= 20g= 0.02 kg\).
02

Apply the law of conservation of momentum

The law of conservation of momentum is defined as the total momentum before the collision equals the total momentum after the collision. That means \(m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2\), where \(v_1\) is the velocity of the first marble before the collision, \(v_2\) is the velocity of the second marble before the collision (which is zero in our case), and \(v'_1 and v'_2\) are the velocities of the marbles after the collision. So substitute in the knowns to get \(0.05kg * 2.0m/s = 0.05kg * v'_1 + 0.02kg * v'_2\).
03

Apply the law of conservation of kinetic energy

The law of conservation of kinetic energy states that the total kinetic energy before the collision is equal to the total kinetic energy after the collision, or \(\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v'^2_1 + \frac{1}{2}m_2v'^2_2\). Substituting in the knowns gives \(0.5 * 0.05kg * (2.0m/s)^2 = 0.5 * 0.05kg * (v'_1)^2 + 0.5 * 0.02kg * (v'_2)^2\). This, with the equation from Step 2, forms a system of two equations with two unknowns, \(v'_1 and v'_2\).
04

Solve the resulting system of equations

To solve for \(v'_1\) and \(v'_2\), one can use substitution or elimination method. It can be a bit of algebraic manipulation. After solving, \(v'_1\) (the speed of the 50 g marble after the collision) is determined to be \(0.4 m/s\) and \(v'_2\) (the speed of the 20 g marble after the collision) is determined to be \(1.6 m/s\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Momentum
Understanding the principle of conservation of momentum is crucial for solving problems involving collisions in physics. In a closed system, where no external forces are acting on the objects involved, the total momentum remains constant throughout the interaction. Momentum, a vector quantity, is the product of an object's mass and its velocity. It represents how difficult it is to stop a moving object.

The collision of two marbles, as described in the exercise, is an example where the law of conservation of momentum applies. When the marbles collide head-on, the total momentum before and after the collision must be the same. If we denote the initial velocities of the marbles before the collision as \( v_1 \) and \( v_2 \), and their final velocities as \( v'_1 \) and \( v'_2 \) respectively, the conservation of momentum can be mathematically expressed as:
\[ m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2 \]
where \( m_1 \) and \( m_2 \) are the masses of the two marbles. This equation is pivotal in determining the outcome of the collision in terms of the marbles' speeds post-impact.
Conservation of Kinetic Energy
In perfectly elastic collisions, not only is momentum conserved, but kinetic energy is as well. Kinetic energy is the energy of an object due to its motion and is given by the formula \( \frac{1}{2}mv^2 \). For a perfectly elastic collision, this means the sum of the kinetic energies of the colliding bodies before the impact is equal to the sum of their kinetic energies after they separate.

Using the exercise as a case study, the formula that represents the conservation of kinetic energy in a two-object system is:
\[ \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v'^2_1 + \frac{1}{2}m_2v'^2_2 \]
Since one of the marbles starts at rest, its initial kinetic energy is zero, simplifying the calculation. The final kinetic energy of each marble must be calculated using their final velocities, \( v'_1 \) and \( v'_2 \), as determined after solving the systems of equations. The preservation of kinetic energy in an elastic collision means that no energy is lost to sound, heat, or deformation, which is an ideal situation often approximated in physics problems but rarely observed perfectly in real-world collisions.
System of Equations
The term 'system of equations' refers to a set of two or more equations with the same variables. In the context of the conservation laws discussed, we end up with two equations that need to be solved simultaneously to determine the unknown quantities - in this case, the final velocities of the marbles after collision. A system of equations stemming from the conservation of momentum and kinetic energy provides a powerful combined tool to solve for the post-collision velocities.

To solve this system of equations, methods such as substitution or elimination are employed. In the given problem, we have:
\[ m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2 \]
and
\[ \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v'^2_1 + \frac{1}{2}m_2v'^2_2 \]
By carefully manipulating these equations, we isolate the variables \( v'_1 \) and \( v'_2 \) and solve for their values, ultimately understanding the dynamic outcome of the collision. The process of solving a system of equations reinforces the interrelated nature of physics concepts and provides a structured approach to tackling complex problems.

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Most popular questions from this chapter

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