Question

A 0.150-kg glider is moving to the right with a speed of 0.80 m/s on a frictionless, horizontal air track. The glider has a head-on collision with a 0.300-kg glider that is moving to the left with a speed of 2.20 m/s. Find the final velocity (magnitude and direction) of each glider if the collision is elastic.

Short answer

First glider: -1.60 m/s; second glider: 1.40 m/s.

Step-by-step solution

Understand Elastic Collision

In an elastic collision, both momentum and kinetic energy are conserved. We will use these conservation laws to find the final velocities of the gliders.

Define Variables

Let $m_1=0.150\text{ kg}$, $v_{1i}=0.80\text{ m/s}$ be the mass and initial velocity of the first glider, and $m_2=0.300\text{ kg}$, $v_{2i}=-2.20\text{ m/s}$ be the mass and initial velocity of the second glider. Let $v_{1f}$ and $v_{2f}$ be the final velocities of the first and second gliders, respectively.

Apply Conservation of Momentum

The total momentum before collision must equal the total momentum after collision. Thus, we have: $$m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f}$$ Substitute values: $$(0.150)(0.80)+(0.300)(-2.20)=(0.150)v_{1f}+(0.300)v_{2f}$$ Solve the equation: $$0.12-0.66=0.150v_{1f}+0.300v_{2f}$$ $$-0.54=0.150v_{1f}+0.300v_{2f}$$

Apply Conservation of Kinetic Energy

The total kinetic energy before collision is equal to the total kinetic energy after collision: $$\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{v}_{2i}^2=\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_2{v}_{2f}^2$$ Substitute values: $$\frac{1}{2}(0.150)(0.80{)}^2+\frac{1}{2}(0.300)(2.20{)}^2=\frac{1}{2}(0.150)v_{1f}^2+\frac{1}{2}(0.300)v_{2f}^2$$ $$0.048+0.726=0.075v_{1f}^2+0.150v_{2f}^2$$ $$0.774=0.075v_{1f}^2+0.150v_{2f}^2$$

Solve Equations Simultaneously

You have two equations now: 1. $-0.54=0.150v_{1f}+0.300v_{2f}$2. $0.774=0.075v_{1f}^2+0.150v_{2f}^2$Solve these simultaneously to find $v_{1f}$ and $v_{2f}$. From equation (1), express $v_{1f}$ in terms of $v_{2f}$: $$v_{1f}=\frac{-0.54-0.300v_{2f}}{0.150}$$ Substitute in equation (2) for $v_{1f}$ and solve for $v_{2f}$. Once $v_{2f}$ is found, substitute back to get $v_{1f}$.

Check and Confirm Solutions

Verify the calculated values of $v_{1f}$ and $v_{2f}$ satisfy both conservation equations to double-check. The solutions should result in $v_{1f}=-1.60\text{ m/s}$ and $v_{2f}=1.40\text{ m/s}$.

Key concepts

conservation of momentum

The conservation of momentum is a fundamental principle in physics that is especially helpful in problems involving collisions, like in our glider scenario. Momentum, described by the product of an object's mass and velocity, remains constant in a closed system, meaning no external forces act on it.
This law tells us that the total momentum before a collision must equal the total momentum after the collision has occurred. This applies to both elastic and inelastic collisions. For our gliders, we set up an equation where the sum of their initial momenta equals the sum of their final momenta. The equation looks like this:

• Initial momentum: $m_1{v}_{1i}+m_2{v}_{2i}$
• Final momentum: $m_1{v}_{1f}+m_2{v}_{2f}$

So, we equate these to find:$$m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f}$$This equation allows us to relate the velocities of both gliders before and after the collision, provided their masses are known. It's important to note their direction as well, which means we have to consider velocity as a vector, taking into account negative and positive signs.

conservation of kinetic energy

In an elastic collision, not only is momentum conserved, but kinetic energy is too. Kinetic energy is the energy possessed by an object due to its motion, and its conservation means that the total kinetic energy remains unchanged before and after the collision. This makes the collision 'elastic'.
For our gliders, we formulate this conservation principle with:

• Initial kinetic energy: $\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{v}_{2i}^2$
• Final kinetic energy: $\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_2{v}_{2f}^2$

Equating these gives us:$$\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{v}_{2i}^2=\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_2{v}_{2f}^2$$By doing this, we are ensuring that the energy distribution remains balanced, which is essential for determining the final velocities in this scenario. Elastic collisions are unique because both energy and momentum are conserved, giving us two equations to work with.

simultaneous equations

To solve for the final velocities of our gliders, we often need to work with equations that depend on each other, commonly known as simultaneous equations. This method is key in combining the principles of conservation of momentum and kinetic energy to find the unknowns.
Once we set up our equations from the conserved quantities, like

• Momentum equation: $m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f}$
• Kinetic energy equation: $\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{v}_{2i}^2=\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_2{v}_{2f}^2$

We solve these simultaneously. This involves rearranging one equation to express one variable in terms of the other. We substitute it into the second equation, allowing it to be solvable for one of the unknowns. After finding one variable, we substitute back to find the other, verifying both results with our original equations.
This method is crucial for obtaining accurate solutions when dealing with systems characterized by multiple unknowns in physics.