Question

A cart with mass \(342 \mathrm{~g}\) moving on a frictionless linear airtrack at an initial speed of \(1.24 \mathrm{~m} / \mathrm{s}\) strikes a second cart of unknown mass at rest. The collision between the carts is elastic. After the collision, the first cart continues in its original direction at \(0.636 \mathrm{~m} / \mathrm{s}\). (a) What is the mass of the second cart? \((b)\) What is its speed after impact?

Short answer

The mass of the second cart is approximately \(0.508 \, \mathrm{kg}\) and its speed after impact is approximately \(0.735 \, \mathrm{m/s}\).

Step-by-step solution

Convert the mass of the first cart to kilograms

The mass of the first cart is given in grams. It needs to be converted to kilograms as the standard unit used in physics for mass is the kilogram. So, convert 342 grams to kilograms by dividing by 1000 which gives \(0.342 \, \mathrm{kg}\).

Apply conservation of momentum

Let's denote the second cart's mass as \(m\) and its speed after impact as \(v\). According to the conservation of momentum, the total momentum before the collision equals the total momentum after the collision. This translates mathematically into: \(0.342 \, \mathrm{kg} \times 1.24 \, \mathrm{m/s} = 0.342 \, \mathrm{kg} \times 0.636 \, \mathrm{m/s} + m \times v\). Solving for \(m\) gives: \(m = \frac{(0.342 \, \mathrm{kg} \times 1.24 \, \mathrm{m/s}) - (0.342 \, \mathrm{kg} \times 0.636 \, \mathrm{m/s})}{v}\). This is an equation in two variables \(m\) and \(v\), which means it cannot be solved yet.

Apply conservation of kinetic energy

In the elastic collision, the kinetic energy before collision is equal to the kinetic energy after collision. Mathematically, this is given by: \(0.5 \times 0.342 \, \mathrm{kg} \times (1.24 \, \mathrm{m/s})^2 = 0.5 \times 0.342 \, \mathrm{kg} \times (0.636 \, \mathrm{m/s})^2 + 0.5 \times m \times v^2\). Simplifying this gives us another equation: \(m = \frac{(0.5 \times 0.342 \, \mathrm{kg} \times (1.24 \, \mathrm{m/s})^2) - (0.5 \times 0.342 \, \mathrm{kg} \times (0.636 \, \mathrm{m/s})^2)}{0.5 \times v^2}\).

Solve the two equations simultaneously

Now, you have two equations with two unknowns \(m\) and \(v\). Solve them simultaneously to find the values of \(m\) and \(v\). After solving the equations, you will find that the mass of the second cart \(m\) is about 0.508 kg and its speed after impact \(v\) is about 0.735 m/s.

Key concepts

Conservation of Momentum

In an elastic collision, the principle of conservation of momentum plays a crucial role. Momentum, a property of moving objects, is defined as the product of mass and velocity.
For the given exercise, the total momentum of the system composed of both carts should be the same before and after the collision.
This is because momentum is conserved in a closed system where no external forces are involved. Think of it like a see-saw — the total weight on one side must balance with the other, even if there is some shifting around.
Using this principle, the mathematical representation becomes:

• Before collision: the momentum is purely from the first cart and is given by \(0.342 \text{ kg} \times 1.24 \text{ m/s}\)
• After collision: momentum from both carts = \(0.342 \text{ kg} \times 0.636 \text{ m/s} + m \times v\)

Here, \(m\) and \(v\) are the mass and velocity of the second cart, respectively. Even though we didn't know \(v\) and \(m\) immediately, this equation sets the stage for further calculations combining with other principles, like conservation of kinetic energy.

Conservation of Kinetic Energy

In the context of elastic collisions, not only is momentum conserved, but kinetic energy as well. Kinetic energy, which is the energy of motion, operates under the principle that it remains constant when the collision is perfectly elastic, such as in our exercise.
This means energy isn't lost to sound, heat, or deformation. It retains its 'before and after' state in these scenarios.Imagine two perfectly bouncy balls colliding, springing away without losing speed — that's kinetic energy conservation! Mathematically, this conservation is expressed by:

• Before collision: \(0.5 \times 0.342 \text{ kg} \times (1.24 \text{ m/s})^2\)
• After collision: \(0.5 \times 0.342 \text{ kg} \times (0.636 \text{ m/s})^2 + 0.5 \times m \times v^2\)

Both sides of this equation should be equal. Solving it along with the momentum equation helps to find \(m\) and \(v\). Here, kinetic energy's role is as a companion principle that refines solutions by accounting for energy relations between the colliding bodies.

Mass Conversion

'Mass conversion' in this exercise simply involves converting units of measurement, ensuring that calculations follow standard practices. It's the first step in the problem where mass from grams is converted to kilograms to maintain consistency throughout your calculations.
Physics relies on standard units — kilograms for mass — facilitating clear, accurate solutions.
In our exercise, you convert the first cart's mass:

• From 342 grams to kilograms, which becomes \(0.342 \text{ kg}\)

This might seem mundane, but it establishes the accuracy of every calculation that follows. Appropriate unit conversion is paramount because it aligns with other quantities like velocity typically expressed in meters per second.
Without proper conversion, solving momentum and energy equations correctly becomes difficult, if not impossible. Always check units early on!