Momentum is defined as the product of mass and velocity.
A rapidly moving truck has a large momentum. It takes a large or prolonged force to get the truck up to this speed, or to bring it to a stop afterwards.
A slow moving snail has a small momentum. It would be easy to slow down or speed up a little snail.
$$p = mv$$
\(p\) = momentum [kg m/s] vector\(m\) = mass [kg]
\(v\) = velocity[m/s] vector
Google a typical mass and velocity.
solution
$$\text{convert to m/s}$$ $$ 18 \left( \mathrm{ \frac{\color{red}{mile}}{\color{blue}{hour}}} \right)\left(\frac{1609\,\mathrm{ m}}{1 \,\color{red}{\mathrm{mile} }}\right)\left(\frac{1\, \color{blue}{ \mathrm{hour} }}{3600\, \mathrm{s} }\right) = 8.0 \mathrm{\tfrac{m}{s}} $$$$p=mv$$ $$p=(5 \, \mathrm{kg})(7.6 \, \mathrm{\tfrac{m}{s}})$$ $$p=38 \,\mathrm{kg \tfrac{m}{s}}$$
Impulse
An impulse is defined as both a force multiplied by time (FΔt) and a change in momentum (Δp). The longer a force is applied, the more the momentum will change.
Examples of impulse:cars crashing
rockets accelerating
punching
jumping
derivation of impulse
$$F = ma$$ $$a = \frac{\Delta v}{\Delta t}$$ $$\text{substitute acceleration with Δv/Δt}$$ $$F = m \frac{\Delta v}{\Delta t}$$ $$F \Delta t = m \Delta v$$$$F \Delta t = m(v_f - v_i)$$ $$F \Delta t = mv_f - mv_i$$ $$F \Delta t = p_f - p_i$$ $$F \Delta t = \Delta p$$
$$J = F \Delta t = \Delta p = m\Delta v$$
\(J\) = impulse [Ns, kg m/s] vector\(F\) = force [N, kg m/s²] vector
\(\Delta t\) = time period [s]
\(\Delta p\) = change in momentum [kg m/s] vector
\(m\) = mass [kg]
\(v\) = velocity[m/s] vector
solution
$$F \Delta t = m\Delta v$$ $$\frac{F \Delta t}{\Delta v} = m$$ $$\frac{(-100)(0.1)}{14-15} = m$$ $$\frac{(-100)(0.1)}{-1} = m$$ $$10 \, \mathrm{kg} = m$$solution
$$ 60 \left( \mathrm{ \frac{\color{red}{mile}}{\color{blue}{hour}}} \right)\left(\frac{1609\,\mathrm{ m}}{1 \,\color{red}{\mathrm{mile} }}\right)\left(\frac{1\, \color{blue}{ \mathrm{hour} }}{3600\, \mathrm{s} }\right) = 26.8 \mathrm{\tfrac{m}{s}} $$$$F \Delta t = m\Delta v$$ $$F = \frac{m\Delta v}{\Delta t}$$ $$F = \frac{(2250)(26.8)}{2.8}$$ $$F = \frac{60300}{2.8}$$ $$F = 21536 \, \mathrm{N}$$
How does the 0 to 60 miles/hour acceleration compare to the acceleration of gravity?
solution
$$F = 21536 N$$ $$F = ma$$ $$\frac{F}{m}=a$$ $$\frac{21536}{2250}=a$$ $$a = 9.57 \,\mathrm{ \tfrac{m}{s^2} }$$ $$ g = 9.81 \, \mathrm{\tfrac{m}{s^2}} $$It's just a bit under the acceleration of gravity. How would that feel?
Conservation of Momentum
Momentum is useful because the total momentum for a system of objects is conserved. This means the sum of all momentums won't change even if the objects collide.
Click the simulation above a few times and note that the total momentum is the same before and after a collision.
Conservation of momentum isn't true if one of the objects in the system interacts with an object outside the system. For example: conservation doesn't hold if an object hits the ground, and the ground isn't part of our system.
Conservation of momentum means we can set the total momentums at two points in time equal to each other. (as long as there are no outside forces)
$$ \sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$m_1 u_1 + m_2u_2 + \cdots = m_1v_1 + m_2v_2 + \cdots$$
\(p\) = momentum [kg m/s] vector\(u\) = initial velocity [m/s] vector
\(v\) = final velocity [m/s] vector
\(m\) = mass [kg]
solution
$$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$\text{green + white = green + white}$$ $$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(2)(3)+(10)(-1)=(2)v_{1}+(1)(10)$$ $$6-10=(2)v_{1}$$ $$-4=(2)v_{1}$$ $$-2 \mathrm{\tfrac{m}{s}} = v_{1}$$green box | blue box | orange box |
---|---|---|
m = 1.60kg | m = 4.90kg | m = 3.60kg |
u = 3.0m/s | u = 2.0m/s | u = -1.0m/s |
v = ??? | v = 1.18m/s | v = 1.58m/s |