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
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
The equation for impulse can be produced by replacing acceleration in F = ma with Δv/Δt.
$$F = ma$$ $$a = \frac{\Delta v}{\Delta t}$$ $$F = m \frac{\Delta v}{\Delta t}$$ $$F \Delta t = m \Delta v$$$$F \Delta t = m(v_f - v_i)$$ $$F \Delta t = mv_f - mv_i$$ $$F \Delta t = p_f - p_i$$ $$F \Delta t = \Delta p$$
$$J = F \Delta t = \Delta p = m\Delta v$$
\(J\) = 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
$$F \Delta t = m\Delta v$$ $$\frac{F \Delta t}{\Delta v} = m$$ $$\frac{(-100)(0.1)}{14-15} = m$$ $$\frac{(-100)(0.1)}{-1} = m$$ $$10kg = m$$solution
$$ 60 \left( \mathrm{ \frac{\color{red}{mile}}{\color{blue}{hour}}} \right)\left(\frac{1609\,\mathrm{ m}}{1 \,\color{red}{\mathrm{mile} }}\right)\left(\frac{1\, \color{blue}{ \mathrm{hour} }}{3600\, \mathrm{s} }\right) = 26.8 \mathrm{\tfrac{m}{s}} $$$$F \Delta t = m\Delta v$$ $$F = \frac{m\Delta v}{\Delta t}$$ $$F = \frac{(2250)(26.8)}{2.8}$$ $$F = \frac{60300}{2.8}$$ $$F = 21536 N$$
solution
$$F = 21536 N$$ $$F = ma$$ $$\frac{F}{m}=a$$ $$\frac{21536}{2250}=a$$ $$a = 9.57 \small \frac{m}{s^2}$$Is that a high acceleration? Compare the car's acceleration to gravity. How would the extra acceleration feel?$$ g = 9.81 \small \frac{m}{s^2} $$
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)
$$p_{initial} = p_{final}$$ $$p_{1i}+p_{2i}+...=p_{1f}+p_{2f}+...$$
$$m_{1}u_{1}+m_{2}u_{2}+...=m_{1}v_{1}+m_{2}v_{2}+...$$
\(u\) = initial velocity [m/s] vector
\(v\) = final velocity [m/s] vector
\(m\) = mass [kg]
solution
initial momentum = final momentum
green + white = green + white
$$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(2)(3)+(10)(-1)=(2)v_{1}+(1)(10)$$ $$6-10=(2)v_{1}$$ $$-4=(2)v_{1}$$ $$-2 \tfrac{m}{s}= v_{1}$$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 |