Collisions in two dimensions

In a nutshell

​When two or more objects collide, there is an exchange of momentum and kinetic energy between them. In elastic collisions kinetic energy is conserved, in inelastic collisions it is not. Momentum is always conserved in collisions. 

Equations

Variable definitions

Conservation of momentum

When two or more objects collide, there is an exchange of momentum and kinetic energy between them. The momentum is conserved in the collision, if there are no other forces acting in the system. This means the initial combined momentum of the two or more objects will equal the final combined momentum. The principle of conservation of momentum states:

for a system of interacting objects, the total momentum in a specified direction remains constant, as long as no external forces act on the system.

Types of collision

In a perfectly elastic collision, the kinetic energy of the system is conserved. In an inelastic collision however, the kinetic energy of the system is not conserved. This is because energy will be lost to other forms, such as heat and sound energy. In both systems, the momentum and total energy is conserved.

Collisions

When collisions occur in one direction, momentum in that specific direction is conserved. The principle of conservation of momentum allows you to derive the formula:

​$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$

lf objects are travelling in opposite directions, one direction must be taken as positive and the other negative to make this formula work.

Example

A bumper car of mass $$200 \ kg$$​ moving at $$3.5 \ ms^{−1}$$ collides head-on with another bumper car of mass $$250 \ kg$$ moving at $$2.0 \ ms^{−1}$$ in the opposite direction. The $$250 \ kg$$​ bumper car stops after the collision. Calculate the final velocity of the $$200 \ kg$$ bumper car.​

First, state the variables from the question:

$$m_1 = 200 \ kg, \ \ m_2 = 250 \ kg, \ \ u_1 = 3.5 \ ms^{-1}, \ \ u_2 = -2 \ ms^{-1}, \ \ v_2 = 0 \ ms^{-1}$$​​

Then state the equation needed:

​$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$​​

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Rearrange the equation:

​$$v_1 = \dfrac{m_1u_1 + m_2u_2 - m_2v_2 }{m_1}$$​​

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Sub in and solve:

​$$v= \dfrac{200 \times 3.5 + 250 \times -2 }{200}\newline \\[0.1in]v = 1 \ ms^{-1}$$

The final velocity of the bumper car is $$\underline{1 \ ms^{-1}}$$.​

Collisions in two dimensions

The conservation of momentum still applies in two dimensions. When solving collisions in two dimensions, resolve the momentums in either the horizontal or vertical direction. Just remember to keep it constant throughout. 

Considering momentum in the $$x$$​ direction:

​$$total \ initial \ momentum = total \ final \ momentum \\[0.1in]\newline p_{1ix} + p_{2ix} = p_{1fx} + p_{2fx}\\[0.1in]\newline m_1u_{1x} + m_2u_{2x} = m_1v_{1x} + m_2v_{2x}\\[0.1in]\newline m_1u_1cos\theta + m_2u_2cos\theta = m_1v_1cos\theta + m_2v_2cos\theta$$

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Example:  

Two identical balls of mass $$1 \ kg$$ collide with each other. Ball A is initially travelling horizontally at $$5 \ ms^{-1}$$​ and ball B is initially stationary. After the collision, ball A moves off at $$40 \degree$$ above the horizontal and ball B $$20 \degree$$ below the horizontal at $$2 \ ms^{-1}$$​. Calculate the velocity of ball A after the collision.

Resolving horizontally:

State variables:

​$$m = 1 \ kg, \ \ \theta_A = 40 \degree, \ \ \theta_B = 20 \degree, \ \ u_A = 5 \ ms^{-1}, \ \ u_B = 0 \ ms^{-1}, \, v_b=2\, ms^{-1}$$

State equation needed:

​$$m_1u_1cos\theta + m_2u_2cos\theta = m_1v_1cos\theta + m_2v_2cos\theta$$​​

Rearrange the equation:

​$$m_Au_Acos\theta_A + m_Bu_Bcos\theta_B = m_Av_Acos\theta_A + m_Bv_Bcos\theta_B \newline \\[0.1in] m_Av_{A}cos\theta_A = m_Au_{A}cos\theta_A - m_Bv_{B}cos\theta_B \newline \\[0.1in] v_{A}= \dfrac{m_A u_{A}cos\theta_A - m_Bv_{B}cos\theta_B}{m_Acos\theta_A }\newline \\[0.1in] v_{A}= \dfrac{u_{A}cos\theta_A - v_{B}cos\theta_B}{cos\theta_A }$$

Note: Be careful with the angles before and after. In this example Ball A is travelling horizontally so the angle $$\theta_A=0\degree$$ before the collision but $$\theta_A=40\degree$$ after the collision.

Sub in and solve:

$$v_A = \dfrac{ 5\cos(90) - 2 \cos(20) }{ \cos (40)} \newline \\[0.1in] v_A = 4.07 \ ms^{-1}\newline \\[0.1in] v_A = 4.1 \ ms^{-1} \ (2sf)$$​​

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Ball A moves away at a speed of $$\underline{4.1 \ ms^{-1}}$$.