Elastic & Inelastic Collisions (College Board AP® Physics 1: Algebra-Based): Study Guide

A mass $m_1$ with initial velocity $v$ collides elastically with a stationary mass $m_2$.

(A) Derive expressions for $v_1$ and $v_2$, the final velocities of $m_1$ and $m_2$ respectively, in terms of $m_1$, $m_2$ and $v$.

(B) For the final velocities of the two masses to be in the same direction, indicate whether $m_1$ must be greater than, less than, or equal to $m_2$. Justify your reasoning.

(C) For the final velocities of the two masses to be in opposite directions, indicate whether $m_1$ must be greater than, less than, or equal to $m_2$. Justify your reasoning.

Answer:

Part (A)

Step 1: Analyze the scenario

• Before the collision, only mass $m_1$ is moving, so it will provide the total momentum of the system

• After the collision, mass $m_2$ will be given a velocity $v_2$ and the velocity of mass $m_1$ will decrease from $v$ to $v_1$

• Without knowing the relative sizes of the masses, the direction of $v_1$ is unknown

• In an elastic collision, both momentum and kinetic energy are conserved

• Therefore, we can write expressions for

  • total momentum before = total momentum after

  • total kinetic energy before = total kinetic energy after

Step 2: Write expressions for the momentum and kinetic energy of the system

• The momentum of the system is

$\sum {\overrightarrow{p}}_i=\sum {\overrightarrow{p}}_f$

$m_{1}v=m_1{v}_1+m_2{v}_2$

• The kinetic energy of the system is

$\sum K_i=\sum K_f$

$\frac{1}{2}{m}_1{v}^2=\frac{1}{2}{m}_1{v_1}^2+\frac{1}{2}{m}_2{v_2}^2$

$m_1{v}^2=m_1{v_1}^2+m_2{v_2}^2$

Step 3: Rearrange each expression to collect the mass terms and make $v_1$ or $v_2$ the subject

• The momentum expression can be arranged to give

$m_2{v}_2=m_{1}v-m_1{v}_1$

equation 1: $m_2{v}_2=m_1\left(v-v_1\right)$

• The kinetic energy expression can be arranged to give

$m_2{v_2}^2=m_1{v}^2-m_1{v_1}^2$

$m_2{v_2}^2=m_1\left(v^2-{v_1}^2\right)$

equation 2: $m_2{v_2}^2=m_1\left(v+v_1\right)\left(v-v_1\right)$

Step 4: Determine the relationship between the velocities

• Divide equation 2 by equation 1, and simplify:

$\frac{m_2{v_2}^2}{m_2{v}_2}=\frac{m_1\left(v+v_1\right)\left(v-v_1\right)}{m_1\left(v-v_1\right)}$

$\frac{\cancel{m_2}{v_2}^{\cancel{2}}}{\cancel{m_2}\cancel{v_2}}=\frac{\cancel{m_1}\left(v+v_1\right)\cancel{\left(v-v_1\right)}}{\cancel{m_1}\cancel{\left(v-v_1\right)}}$

$v_2=v+v_1$

• Now we have an expression relating the velocities, we can use this along with equation 1 to

  • eliminate $v_2$, giving an expression for $v_1$ in terms of $m_1$, $m_2$ and $v$

  • eliminate $v_1$, giving an expression for $v_2$ in terms of $m_1$, $m_2$ and $v$

Step 5: Write an expression for the final velocity ${\mathit{v}}_{\mathbf{1}}$

• Substitute the expression for $v_2$ back into equation 1:

$m_2\left(v+v_1\right)=m_1\left(v-v_1\right)$

• Expand and collect the velocity terms:

$m_{2}v+m_2{v}_1=m_{1}v-m_1{v}_1$

$v_1\left(m_1+m_2\right)=m_{1}v-m_{2}v$

$v_1=\frac{m_{1}v-m_{2}v}{m_1+m_2}$

Step 6: Write an expression for the final velocity ${\mathit{v}}_{\mathbf{2}}$

• Substitute the expression for $v_1$ into the expression for $v_2$:

$v_2=v+v_1=v+\frac{m_{1}v-m_{2}v}{m_1+m_2}$

• Combine the fractions, then expand and simplify:

$v_2=\frac{v\left(m_1+m_2\right)}{m_1+m_2}+\frac{m_{1}v-m_{2}v}{m_1+m_2}$

$v_2=\frac{m_{1}v+m_{2}v+m_{1}v-m_{2}v}{m_1+m_2}$

$v_2=\frac{2m_{1}v}{m_1+m_2}$

Part (B)

Step 1: Analyze the scenario

• The expression for $v_2$ indicates that it will always have a positive value, as it does not contain any minus signs

• The expression for $v_1$ indicates it can have a positive or negative value as it does contain a minus sign

• Therefore, the final velocities of the two masses will be in the same direction for positive values of $v_1$

Step 2: Deduce and justify the relationship

• For $v_1$ to be in the same direction as $v_2$, $m_1$ must be greater than $m_2$

• Justification:

  • From the equation, $v_1=\frac{\left(m_1-m_2\right)v}{m_1+m_2}$, positive values of $v_1$ are possible when $\left(m_1-m_2\right)v>0$

  • This is only possible when $m_1>m_2$

Part (C)

Step 1: Analyze the scenario

• The final velocities of the two masses will be in opposite directions for negative values of $v_1$

Step 2: Deduce and justify the relationship

• For $v_1$ to be in the opposite direction to $v_2$, $m_1$ must be less than $m_2$

• Justification:

  • From the equation, $v_1=\frac{\left(m_1-m_2\right)v}{m_1+m_2}$, negative values of $v_1$ are possible when $\left(m_1-m_2\right)v<0$

  • This is only possible when $m_1<m_2$

Every AP Physics exam you take will include a question asking you to "derive an expression". Derivations are a key problem-solving skill in physics and serve as a powerful tool for determining relationships and making reliable predictions for numerous scenarios.

If you are finding it tricky to master derivations, try these steps:

1. Analyze the scenario to identify exactly what is being asked in the derivation and come up with a rough plan for where you want to end up

• In the example above, we needed two expressions, one for $v_1$ and one for $v_2$in terms of the three quantities given in the question $\left(m_1,m_2,v\right)$, so it was clear we had to relate the two velocities to each other, and then use algebra to write separate equations for each one.

2. Begin your solution by identifying one or more fundamental equations from the AP Physics Exam equation sheet

• In the example above, we started by applying the fundamental equations for momentum $\left(p=mv\right)$ and kinetic energy $\left(K=\frac{1}{2}m{v}^2\right)$ to each object which gave us two equations as a basis for the rest of the derivation.

3. Obtain a final expression using the algebraic manipulation skills you know, such as factorizing, expanding, and simplifying variables

• In the example above, once we set up the equations containing the required terms, all we had left to do was to rearrange, manipulate, and simplify the expressions algebraically to get them into the required forms

This process should help you to apply a structure to derivations that exam markers can easily follow to ensure you receive maximum credit, but more importantly, you must practice as many of these types of questions as you can to improve your confidence with them.

A car of mass $m$, traveling with a velocity $v_0$, strikes a stationary vehicle of mass $2m$ in a head-on inelastic collision and the bumpers lock together.

What fraction of the initial kinetic energy is lost in this collision?

A      $\frac{1}{2}$

B      $\frac{1}{3}$

C      $\frac{1}{4}$

D      $\frac{2}{3}$

The correct answer is D

Answer:

Step 1: Analyze the scenario

• Before the collision, only the car of mass $m$ is moving, so it will provide the total momentum of the system

• After the collision, the cars become one object of total mass $3m$ and velocity $v$

• This is an example of a perfectly inelastic collision

Step 2: Use conservation of momentum to relate the velocities

• Conservation of momentum: the total momentum before the collision is equal to the total momentum after

$\sum {\overrightarrow{p}}_i=\sum {\overrightarrow{p}}_f$

$m{v}_0=3mv$

$v_0=3v$

Step 3: Determine the initial and final kinetic energies

• The initial kinetic energy of the system is:

$K_i=\frac{1}{2}m{v_0}^2=\frac{1}{2}m{\left(3v\right)}^2=\frac{9}{2}m{v}^2$

• The final kinetic energy of the system is:

$K_f=\frac{3}{2}m{v}^2$

Step 4: Determine the fraction of kinetic energy lost

• The change in kinetic energy is equal to:

$∆K=K_i-K_f$

$∆K=\frac{9}{2}m{v}^2-\frac{3}{2}m{v}^2$

$∆K=3m{v}^2$

• The fraction of kinetic energy lost is therefore:

$\frac{∆K}{K_i}=\frac{3m{v}^2}{{\displaystyle \frac{9}{2}}m{v}^2}=\frac{2}{3}$

• Therefore, the correct option is D

It can be helpful to think about collisions and explosions as if there are four types rather than two:

• elastic - kinetic energy conserved

• perfectly elastic - kinetic energy conserved and no energy transferred between objects

• inelastic - kinetic energy not conserved

• perfectly inelastic - kinetic energy not conserved and maximum energy transferred to surroundings

However, in AP Physics 1, be aware that the 'perfectly elastic' collision type will not be tested.