Elastic Collision of Two Masses – It Can Be Shown Exercise

An elastic collision is a collision where total momentum and total kinetic energy is conserved.

This illustration shows two objects A and B traveling towards each other. The mass of A is m_A and the moving with velocity V_Ai. The second object has a mass of m_B and velocity V_Bi. The two objects collide elastically. Mass A moves away at a velocity V_Af and mass B has a final velocity of V_Bf.

Given these conditions, textbooks give the following formulas for V_Af and V_Bf.

and

where
m_A is the mass of the first object
V_Ai is the initial velocity of the first object
V_Af is the final velocity of the first object
m_B is the mass of the second object
V_Bi is the initial velocity of the second object and
V_Bf is the final velocity of the second object.

These two equations are often just presented in this form in the textbook with little or no explanations. Very early in your science education, you will encounter the phrase “It can be shown …” between two steps of mathematics or “left as an exercise for the student”. This almost always translates into “homework problem”. This “It can be shown” example shows how to find the final velocities of two masses after an elastic collision.

This is a step by step derivation of these two equations.

First, we know total momentum is conserved in the collision.

total momentum before collision = total momentum after collision

m_AV_Ai + m_BV_Bi = m_AV_Af + m_BV_Bf

Rearrange this equation so the same masses are on the same side as each other

m_AV_Ai – m_AV_Af = m_BV_Bf – m_BV_Bi

Factor out the masses

m_A(V_Ai – V_Af) = m_B(V_Bf – V_Bi)

Let’s call this Equation 1 and come back to it in a minute.

Since we were told the collision was elastic, the total kinetic energy is conserved.

kinetic energy before collision = kinetic energy after collection

½m_AV_Ai² + ½m_BV_Bi² = ½m_AV_Af² + ½m_BV_Bf²

Multiply the entire equation by 2 to get rid of the ½ factors.

m_AV_Ai² + m_BV_Bi² = m_AV_Af² + m_BV_Bf²

Rearrange the equation so the like masses are together.

m_AV_Ai² – m_AV_Af² = m_BV_Bf² – m_BV_Bi²

Factor out the common masses

m_A(V_Ai² – V_Af²) = m_B(V_Bf² – V_Bi²)

Use the “difference between two squares” relationship (a² – b²) = (a + b)(a – b) to factor out the squared velocities on each side.

m_A(V_Ai + V_Af)(V_Ai – V_Af) = m_B(V_Bf + V_Bi)(V_Bf – V_Bi)

Now we have two equations and two unknowns, V_Af and V_Bf.

Divide this equation by equation 1 from before (the total momentum equation from above) to get

Now we can cancel out most of this

This leaves

V_Ai + V_Af = V_Bf + V_Bi

Solve for V_Af

V_Af = V_Bf + V_Bi – V_Ai

Now we have one of our unknowns in terms of the other unknown variable. Plug this into the original total momentum equation

m_AV_Ai + m_BV_Bi = m_AV_Af + m_BV_Bf

m_AV_Ai + m_BV_Bi = m_A(V_Bf + V_Bi – V_Ai) + m_BV_Bf

Now, solve this for the final unknown variable, V_Bf

m_AV_Ai + m_BV_Bi = m_AV_Bf + m_AV_Bi – m_AV_Ai + m_BV_Bf

subtract m_AV_Bi from both sides and add m_AV_Ai to both sides

m_AV_Ai + m_BV_Bi – m_AV_Bi + m_AV_Ai = m_AV_Bf + m_BV_Bf

2m_AV_Ai + m_BV_Bi – m_AV_Bi = m_AV_Bf + m_BV_Bf

factor out the masses

2 m_AV_Ai + (m_B – m_A)V_Bi = (m_A + m_B)V_Bf

Divide both sides by (m_A + m_B)

Now we know the value of one of the unknowns, V_Bf. Use this to find the other unknown variable, V_Af. Earlier, we found

V_Af = V_Bf + V_Bi – V_Ai

Plug in our V_Bf equation and solve for V_Af

Group the terms with the same velocities

The common denominator for both sides is (m_A + m_B)

Be careful of your signs in the first half of the expressions in this step

Now we’ve solved for both unknowns V_Af and V_Bf in terms of known values.

Note these match the equations we were supposed to find.

This was not a difficult problem, but there were a couple spots to trip you up.

First, all the subscripts can get tangled up if you aren’t careful or neat in your handwriting.

Second, sign errors. Subtracting a pair of variables inside parentheses will change the sign on BOTH variables. It is all too easy to carelessly turn – (a + b) into -a + b instead of -a – b.

Last, learn the difference between two squares factor. a² – b² = (a + b)(a – b) is an extremely useful factoring trick when trying to cancel something out of an equation.