This is a simulation of a "Newton's cradle". A real Newton's cradle is made of metal balls suspended by two strings. Click and drag a ball to fling it.
When two bodies collide, momentum is transferred. Momentum is the velocity of a body multiplied by its mass. A small force can quickly stop an object with low momentum, but a large or prolonged force is required to stop an object with high momentum.
$$p = mv$$
\(p\) = momentum [kg m/s] vector\(m\) = mass [kg]
\(v\) = velocity [m/s] vector
Google a typical mass and velocity.
solution
$$ 18 \left( \mathrm{ \frac{\color{red}{mile}}{\color{Teal}{hour}}} \right)\left(\frac{1609\,\mathrm{m}}{1 \,\color{red}{\mathrm{mile} }}\right)\left(\frac{1\, \color{Teal}{ \mathrm{hour} }}{3600\,\mathrm{s}}\right) = 8.0 \mathrm{\tfrac{m}{s}} $$$$p=mv$$ $$p=(5.0 \, \mathrm{kg})(8.0 \, \mathrm{\tfrac{m}{s}})$$ $$p=40 \,\mathrm{kg\tfrac{m}{s}}$$
Conservation of Momentum
The law of conservation of momentum states that total momentum doesn't change over time, even if two objects collide. When a property doesn't change over time we say that the property is conserved.
$$m_1v_1 + m_2v_2 = \text{total momentum}$$ Click the simulation below a few times and watch the sum of the total horizontal momentum before and after the collision.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)
derivation of conservation of momentum
First we are going to rewrite force in terms of a change in velocity.
$$F = ma $$ $$a = \frac{\Delta v}{\Delta t}$$ $$F = m \frac{\Delta v}{\Delta t}$$ $$F = m \frac{(v-u)}{\Delta t}$$When two bodies collide they each experience an equal but opposite force for an equal period of time as explained by Newton's third law.
$$ F_1 \Delta t = -F_2 \Delta t $$ $$m_1 \frac{(v_1-u_1)}{\Delta t} \Delta t= -m_2 \frac{(v_2-u_2)}{\Delta t}\Delta t$$ $$m_1 (v_1-u_1) = -m_2 (v_2-u_2)$$ $$m_1 v_1- m_1 u_1 = -m_2 v_2 + m_2 u_2$$ $$ m_1 u_1 + m_2 u_2= m_2 v_2 + m_1 v_1$$\(u\) = initial velocity [m/s] vector
\(v\) = final velocity [m/s] vector
\(m\) = mass [kg]
solution
$$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$\text{green + white = green + white}$$ $$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(2)(3)+(10)(-1)=(2)v_{1}+(10)(0)$$ $$6-10=(2)v_{1}$$ $$-4=(2)v_{1}$$ $$-2 \mathrm{\tfrac{m}{s}} =v_{1}$$answer
The total momentum changes when one of the objects collides with the walls.
Question: Why does the total momentum change? Is this a violation of conservation of momentum?
answer
The total momentum is conserved, but we aren't including the walls in our calculations. It can be very difficult to measure the change in velocity for very massive objects, like the ground or a wall.
green box | blue box | orange box |
---|---|---|
m = 1.60 kg | m = 4.90 kg | m = 3.60 kg |
u = 3.0 m/s | u = 2.0 m/s | u = -1.0 m/s |
v = ??? | v = 1.18 m/s | v = 1.58 m/s |
solution
$$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$ {\color{Lime} \blacksquare} \quad +\quad {\color{Blue} \blacksquare} \quad+\quad {\color{orange} \blacksquare} \quad = \quad {\color{Lime} \blacksquare} \quad+\quad {\color{Blue} \blacksquare} \quad+\quad {\color{orange} \blacksquare}$$ $$m_{1}u_{1} + m_{2}u_{2} + m_{3}u_{3} = m_{1}v_{1}+m_{2}v_{2}+m_{3}v_{3}$$ $$(1.6)(3)+(4.9)(2)+(3.6)(-1)=(1.6)v_{1}+(4.9)(1.18)+(3.6)(1.58)$$ $$11=1.6v_{1}+11.47$$ $$-0.29 \mathrm{\tfrac{m}{s}} = v_{1}$$Perfectly Inelastic Collisions
When two objects collide and stick together we say the collision is perfectly inelastic. In those situations the equations for conservation of momentum simplify a bit.
$$m_1u_1 + m_2u_2 = m_1v_1+m_2v_2$$ $$m_1u_1 + m_2u_2 = \left(m_1+m_2\right)v$$solution
Objects can also start as one mass and separate, like in an explosion.
$$\left(m_1+m_2\right)u = m_1v_1 + m_2v_2$$solution
solution
$$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$\text{goalie holding ball = goalie + ball}$$ $$(m_{1}+m_{2})u=m_{1}v_{1}+m_{2}v_{2}$$ $$(100.0+1.1)(3.2)=(100.0)v_{1}+(1.1)(10.0)$$ $$323.52=(100.0)v_{1}+11$$ $$3.13 \, \mathrm{\tfrac{m}{s}} = v_{1}$$2-D Conservation of Momentum
Momentum is a vector. It has a direction and a magnitude. We can solve for the horizontal and vertical components separately just like how we solved 2-D Motion problems. Although in this case there isn't a time variable to link up the vertical and horizontal equations.
solution
$$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$\text{hexagon + triangle = hexagon + triangle}$$ $$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(5.20)(2)+(1.17)(-1)=(5.20)v_{1}+(1.17)(2.72)$$ $$10.40-1.17=(5.20)v_{1}+3.18$$ $$6.05=(5.20)v_{1}$$ $$1.16 \, \mathrm{\tfrac{m}{s}}= v_{1}$$Vertical: Lets take a look at the vertical aspect of the pink hexagon and cyan triangle collision. Before the collision the hexagon has a mass of 5.20 kg and is not moving up or down. The final vertical velocity of the hexagon is 0.23 m/s down. The final vertical velocity of the triangle is 0.67 m/s up. What is the triangle's initial vertical velocity?