PHYS 170

Lecture 08 - Momentum and Collision

Introduction

Momentum is defined by: $$ p = mv $$ In a collsion the total momentum of the system is conserved. The SI unit of momentum is \(kgm/s\) or \(Ns\).

Below is a typical collision, with \(u_1, u_2\) denoting the initial velocity before the collision and \(v_1, v_2\) being the final velocity after the collision:
two simple vectors

Applied to the picture above, the law of conservation of momentum is: $$ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 $$ which basically means "initial total momentum = final total momentum".

Types of Collision

We will study two types of collisions:

Name Definition After collision Equation
Elastic collision \(KE_i = KE_f\) objects separate cleanly \(\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \)
Inelastic collision \(KE_i \neq KE_f\) objects stick together \(v_1 = v_2\)

In other words, in an elastic collision, the system retains all its kinetic energy from before the collision. While in an inelastic collision, some of the kinetic energy is lost and is converted into heat and other forms of energy that are not kinetic. In the macroscopic world, there is no perfect elastic collision because some kinetic energy is always lost. Similarly, the condition \(v_1 = v_2\) actually only applies to what is known as "totally inelastic collision" when the objects completely stick together. Most collisions are in-between cases, but we will not worry about such partially elastic collisions in this course.

No matter what type of collision it is, conservation of momentum (\(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\)) still applies.

Do not use conservation of energy to solve collision problems because generally some energy of the colliders is lost to the surroundings during the collision.

Simulation - Collision and Conservation of Momemtum (click to hide)

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For completely inelastic collision, slide the Elasticity slider to 0%.

Video - Elastic vs Inelastic Collision (click to show)

Try It Yourself (click to show)

Note that the negative \(u_2\) means it is moving toward the negative x-direction. $$ \begin{eqnarray} p_{total} &=& p_1 + p_2 = m_1 u_1 + m_2 u_2 \\ &=& (1)(2) + (2)(-4) = -6kgm/s \\ KE_{total} &=& \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \\ &=& \frac{1}{2} (1)( 2^2) + \frac{1}{2} (2)(-4)^2 \\ &=& 2J + 16J = 18J \end{eqnarray} $$

Examples

Since the two objects stick together in the end, they will have the same final velocity, which we will call \(v_{12} = v_1=v_2\). $$ \begin{eqnarray} m_1 u_1 + m_2 u_2 &=& m_1 v_1 + m_2 v_2 \\ \Rightarrow m_1 u_1 + m_2 u_2 &=& (m_1 + m_2) v_{12} \\ \Rightarrow v_{12} &=& \frac{ m_1 u_1 + m_2 u_2}{m_1 + m_2} \\ &=& \frac{ (2)(3) + (4)(0)}{2+4} = 1m/s \end{eqnarray} $$
$$ \begin{eqnarray} KE_i &=& \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \\ &=& \frac{1}{2} (2) (3^2) + \frac{1}{2} (4) (0^2) \\ &=& 9J \\ KE_f &=& \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \\ &=& \frac{1}{2} m_1 v_{12}^2 + \frac{1}{2} m_2 v_{12}^2 \\ &=& \frac{1}{2} (m_1 + m_2 )v_{12}^2 \\ &=& \frac{1}{2} (2+4) (1^2)\\ &=& 3J \\ \Delta KE &=& KE_f - KE_i = 3 - 9 \\ &=& -6J \end{eqnarray} $$ It is very important that you have the minus sign in the front of \(\Delta KE\). It means that some kinetic energy is lost in the collision (as one expects for an inelastic collision). If you mistakenly write \(\Delta KE = +9J\), it means the kinetic energy actually went up after the collision, an impossibility for this type of collision.

Try It Yourself (click to show)

Since \(m_2\) is moving left, its initial velocity is negative: $$ u_1 = +2m/s \\ u_2 = -2m/s $$ If you get the sign right, computing the momentum is straight forward: $$ \begin{eqnarray} p_{1i}&=& m_1 u_1 \\ &=& (4)(2) = 8kgm/s \\ p_{2i}&=& m_2 u_2 \\ &=& (6)(-2) = -12kgm/s \\ p_{total}&=& p_{1i} + p_{2i} \\ &=& 8-12 = -4kgm/s \end{eqnarray} $$ The negative momentum is due to the fact that the more massive object is moving left, therefore giving the system an overall negative momentum.
$$ \begin{eqnarray} m_1 u_1 + m_2 u_2 &=& m_1 v_1 + m_2 v_2 \\ \Rightarrow m_1 u_1 + m_2 u_2 &=& (m_1 + m_2) v_{12} \\ \Rightarrow v_{12} &=& \frac{ m_1 u_1 + m_2 u_2}{m_1 + m_2} \\ &=& \frac{ (4)(2) + (6)(-2)}{4+6} \\ &=& \frac{-4}{10}= -0.4m/s \end{eqnarray} $$
$$ \begin{eqnarray} KE_i &=& \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \\ &=& \frac{1}{2} (4) (2^2) + \frac{1}{2} (6) (2^2) \\ &=& 8J + 12J = 20J \\ KE_f &=& \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \\ &=& \frac{1}{2} m_1 v_{12}^2 + \frac{1}{2} m_2 v_{12}^2 \\ &=& \frac{1}{2} (m_1 + m_2 )v_{12}^2 \\ &=& \frac{1}{2} (4+6) (-0.4)^2\\ &=& 0.8J \\ \Delta KE &=& KE_f - KE_i = 0.8 - 20 \\ &=& -19.2J \end{eqnarray} $$ The calculation shows that the objects lost most of their kinetic energy.

We start with two equations: $$ \begin{eqnarray} m_1 u_1 + m_2 u_2 &=& m_1 v_1 + m_2 v_2 \tag{conservation of momentum}\\ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 &=& \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \tag{definition of elastic collision} \end{eqnarray} $$ Now we have two equations, and only two unknows (\(m_2, v_2\)).

Put the numbers in the first equation: $$ \begin{eqnarray} m_1 u_1 + m_2 u_2 &=& m_1 v_1 + m_2 v_2\\ \Rightarrow (4)(2) &=& (4)(-1) + m_2 v_2 \\ \Rightarrow m_2 v_2 &=& 12 \end{eqnarray} $$ The second equation gives: $$ \begin{eqnarray} \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 &=& \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \\ \Rightarrow \frac{1}{2} (4) (2)^2 &=& \frac{1}{2} (4) (-1)^2 + \frac{1}{2} m_2 v_2^2 \\ \Rightarrow 8 &=& 2 + \frac{1}{2} m_2 v_2^2 \\ \Rightarrow m_2 v_2^2 &=& 12 \\ \end{eqnarray} $$ This gives the following equations: $$ \begin{eqnarray} m_2 v_2 &=& 12 \\ m_2 v_2^2 &=& 12 \end{eqnarray} $$ The first one gives \(m_2 = \frac{12}{v_2}\), putting it into the second one we have: $$ \begin{eqnarray} (\frac{12}{v_2})v_2^2&=& 12\\ v_2 &=& 1 m/s \\ \Rightarrow m_2 &=& \frac{12}{v_2} = 12kg \end{eqnarray} $$

Impulse

"Impulse" measures an amount of force applied over a period of time. If the time duration (\(\Delta t\)) of the force is short (which is typical of collisions), we can write impulse as: $$ I \approx F \Delta t $$ The unit of impulse is \(Ns\), which is actually identical to \(kgm/s\), the unit of momentum. (The general equation of impulse is actually \(I = \int F dt\), but the approximation above will be good enough for the study of collisions in this course.)

The main result concerning impulse is the following: $$ I = \Delta p $$ In other words, the impulse applied to an object will determine its change of momentum. Intuitively this makes sense because the harder you push and the longer you push (i.e. larger \(F\Delta t\)), the more you could change the motion of an object. This could be proven mathematically using Newton's Second Law: $$ \begin{eqnarray} F&=&ma \\ &=& m \frac{dv}{dt} = \frac{d (mv)}{dt} \\ &=& \frac{dp}{dt} \\ \Rightarrow F &\approx& \frac{\Delta p}{\Delta t} \\ \Rightarrow F \Delta t &\approx& \Delta p \\ \Rightarrow I &\approx& \Delta p \end{eqnarray} $$

The impulse of the force over \(10s\) is: $$ I = F \Delta t = (5N)(10s) = 50Ns $$ This gives the change in momentum: $$ \Delta p = I = 50Ns = 50kgm/s $$ The change in velocity could be found by noting \(\Delta p = \Delta (mv) = m \Delta v\): $$ m \Delta v = 50kgm/s \Rightarrow \Delta v = \frac{50kgm/s}{2kg} = 25 m/s $$

Try It Yourself (click to show)

$$ \begin{eqnarray} p_i &=& m u = (2)(3) = 6kgm/s\\ p_f &=& m v = (2) (-5) = -10kgm/s\\ \Delta p &=& p_f - p_i = -16kgm/s\\ \end{eqnarray} $$
$$ \begin{eqnarray} I &=& F \Delta t \\ \Rightarrow F &=& \frac{I}{\Delta t} = \frac{-16Ns}{0.1s} = -160N \end{eqnarray} $$ The negative sign means the force is pushing the ball to the left.

Center of Mass

Imagine you throw out a stick in a spinning motion (in the absence of gravity):

The motion looks really complicated. However, with the help of a special point, the center of mass, represented by a red dot below, the motion above could be separated into 2 simple motion:

Center of mass enables us to split complicated motion into linear motion and rotation, which we can then calculate separately.

Here is another example, in the presence of gravity:

You already learned in an earlier chapter how to compute the motion of a point mass in projectile motion, once you learn (in the next few chapters) how to compute the rotation of the hammer, you will be able to describe the complicated motion of the hammer and any other big (non-point-like) objects completely.

The basic formula for the center of mass is: $$ x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} $$

The formula could be easily generalized to three or more masses, and in higher dimensions. For example, the center of mass of three objects along the y axis is: $$ y_{CM} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} $$

$$ \begin{eqnarray} x_{CM} &=& \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \\ &=& \frac{(4)(2)+ (6)(5)}{4+6} \\ &=& 3.8m \end{eqnarray} $$
The coordinates of the three mass can be read off from the figure: $$ \begin{eqnarray} m_1: (0, 0) \\ m_2: (0, 3) \\ m_3: (4, 3) \end{eqnarray} $$ $$ \begin{eqnarray} x_{CM} &=& \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2+ m_3} \\ &=& \frac{(1)(0)+ (2)(0)+ (3)(4)}{1+2+3} \\ &=& 2m \\ y_{CM} &=& \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2+ m_3} \\ &=& \frac{(1)(0)+ (2)(3)+ (3)(3)}{1+2+3} \\ &=& 2.5m \end{eqnarray} $$

In summary the center of mass is at: $$ (2, 2.5)m $$

Simulation - Center of Mass (click to hide)

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Drag more masses out from the box on the top right hand corner. Drag back in to remove a mass.
The mass of each object can be changed by selecting it and then dragging on the slider.

Try It Yourself (click to show)

The coordinates of the two masses can be read off from the figure: $$ \begin{eqnarray} x_1= 2m \\ x_2=-2m \\ \end{eqnarray} $$ $$ \begin{eqnarray} x_{CM} &=& \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \\ &=& \frac{(4)(2)+ (6)(-2)}{4+6} \\ &=& -0.4m \end{eqnarray} $$ The negative answer means the CM is left of the origin.

Video - CM explanation (click to show)

Video - more examples on CM (click to show)

The mass of each square is not given, so we will assume each to have mass \(M\) (which will be canceled out in the calculation below). The main trick here is to group a few squares together and replace the group by their center of mass, as in the figure on the left.

This turns the object into these two point masses at positions: $$ \begin{eqnarray} 3M: (15, 15) \\ 2M: (10, 5) \end{eqnarray} $$ From this it is now straight forward to find the center of mass: $$ \begin{eqnarray} x_{CM}&=& \frac{(3M)(15)+(2M)(10)}{3M+2M} \\ &=& \frac{65M}{5M} = 13cm\\ y_{CM}&=& \frac{(3M)(15)+(2M)(5)}{3M+2M} \\ &=& \frac{55M}{5M} = 11cm \end{eqnarray} $$

Try It Yourself (click to show)

The mass of each square is not given, so we will assume each to have mass \(M\). The main trick here is to group a few squares together and replace the group by their center of mass, as in the figure on the left.

This turns the object into these two point masses at positions: $$ \begin{eqnarray} 1M: (25, 15) \\ 3M: (15, 5) \end{eqnarray} $$ From this it is now straight forward to find the center of mass: $$ \begin{eqnarray} x_{CM}&=& \frac{(M)(25)+(3M)(15)}{M+3M} \\ &=& \frac{70M}{4M} = 17.5cm\\ y_{CM}&=& \frac{(M)(15)+(3M)(5)}{M+3M} \\ &=& \frac{30M}{4M} = 7.5cm \end{eqnarray} $$

Center of Mass and Conservation of Momentum

Center of mass can be used to prove conservation of momentum. While the arguement could be made mathematically rigorous, we will simply point out the rough idea here.

Suppose two masses \(m_1\) and \(m_2\) are about to collide, their combined center of mass will not be affected by interactions between the two masses. If for the duration of their collision there is negligible external force, then the center of mass will essentially behave like a point object described in Newton's First Law, traveling at constant velocty (this can be seen in the simulation Collision and Conservation of Momemtum above, click on the "Show CM" button to see): $$ v_{CM} = constant $$ But the center of mass velocity \(v_{CM}\) is just the time derivative of the position of the center of mass: $$ \begin{eqnarray} constant&=& v_{CM} = \frac{dx_{CM}}{dt} \\ &=& \frac{d}{dt} (\frac{m_1 x_1+m_2 x_2}{m_1+m_2}) \\ &=&\frac{m_1 \dot{x_1}+m_2 \dot{x_2}}{m_1+m_2} \\ &=& \frac{m_1 v_1+m_2 v_2}{m_1+m_2} \\ \Rightarrow m_1 v_1+m_2 v_2 &=& constant \end{eqnarray} $$ The last line is the law of conservation of momemtum.

Notations

Name Symbol Unit Meaning
Momentum \(p\) \( kgm/s = Ns\) a quantity conserved in collisions
Impulse \(I\) \( Ns = kgm/s\) a measure of how hard you push and for how long, leads to change in momentum