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Physics · Ch 6 — Work, Energy and Power

Collisions in One Dimension

6.11.2

Collisions in One Dimension

Collisions in One Dimension

When two objects collide, the forces they exert on each other are internal to the two-body system. If no external force acts during the collision, the total momentum of the system is conserved. This is the bedrock principle. But what about kinetic energy? That depends entirely on the nature of the collision.

Elastic and Inelastic Collisions

A collision where the total kinetic energy of the system is conserved is called an elastic collision. A collision where some kinetic energy is transformed into other forms (heat, sound, deformation) is called an inelastic collision. In a perfectly inelastic collision, the two bodies stick together after impact and move with a common velocity.

Watch out

Momentum is always conserved in any collision, provided no external force acts. Energy is always conserved overall, but kinetic energy is not necessarily conserved — it can be converted into other forms.

Elastic Collision in One Dimension

Consider two bodies of masses m1m_1 and m2m_2 moving along a straight line with initial velocities u1u_1 and u2u_2 respectively. Let v1v_1 and v2v_2 be their velocities after the collision. We assume the collision is elastic and one-dimensional.

From conservation of momentum:

m1u1+m2u2=m1v1+m2v2(1)m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \qquad(1)

From conservation of kinetic energy:

12m1u12+12m2u22=12m1v12+12m2v22(2)\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \qquad(2)

We have two equations and two unknowns (v1v_1 and v2v_2). To solve them efficiently, we rearrange.

From (1):

m1(u1−v1)=m2(v2−u2)(3)m_1 (u_1 - v_1) = m_2 (v_2 - u_2) \qquad(3)

From (2):

m1(u12−v12)=m2(v22−u22)m_1 (u_1^2 - v_1^2) = m_2 (v_2^2 - u_2^2)

m1(u1−v1)(u1+v1)=m2(v2−u2)(v2+u2)(4)m_1 (u_1 - v_1)(u_1 + v_1) = m_2 (v_2 - u_2)(v_2 + u_2) \qquad(4)

Divide equation (4) by equation (3) — provided u1≠v1u_1 \neq v_1 and v2≠u2v_2 \neq u_2, which is true for a genuine collision:

u1+v1=v2+u2u_1 + v_1 = v_2 + u_2

u1−u2=v2−v1(5)u_1 - u_2 = v_2 - v_1 \qquad(5)

Important

Equation (5) is a key result: the relative velocity of approach before collision equals the relative velocity of separation after collision. This is a direct consequence of energy conservation in an elastic collision.

Now we can solve for v1v_1 and v2v_2. From (5), v2=u1+v1−u2v_2 = u_1 + v_1 - u_2. Substitute into (1):

m1u1+m2u2=m1v1+m2(u1+v1−u2)m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 (u_1 + v_1 - u_2)

m1u1+m2u2=m1v1+m2u1+m2v1−m2u2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 u_1 + m_2 v_1 - m_2 u_2

m1u1−m2u1+m2u2+m2u2=(m1+m2)v1m_1 u_1 - m_2 u_1 + m_2 u_2 + m_2 u_2 = (m_1 + m_2) v_1

u1(m1−m2)+2m2u2=(m1+m2)v1u_1 (m_1 - m_2) + 2 m_2 u_2 = (m_1 + m_2) v_1

v1=(m1−m2)u1+2m2u2m1+m2v_1 = \frac{(m_1 - m_2) u_1 + 2 m_2 u_2}{m_1 + m_2}

Similarly, substituting v1=v2+u2−u1v_1 = v_2 + u_2 - u_1 from (5) into (1) gives:

v2=(m2−m1)u2+2m1u1m1+m2v_2 = \frac{(m_2 - m_1) u_2 + 2 m_1 u_1}{m_1 + m_2}

These are the general formulas for velocities after a one-dimensional elastic collision.

Special Cases of Elastic Collision

Case 1: Equal masses (m1=m2m_1 = m_2)

Substitute m1=m2=mm_1 = m_2 = m into the formulas:

v1=(m−m)u1+2mu22m=2mu22m=u2v_1 = \frac{(m - m) u_1 + 2 m u_2}{2m} = \frac{2 m u_2}{2m} = u_2

v2=(m−m)u2+2mu12m=2mu12m=u1v_2 = \frac{(m - m) u_2 + 2 m u_1}{2m} = \frac{2 m u_1}{2m} = u_1

Important

When two bodies of equal mass undergo a one-dimensional elastic collision, they exchange their velocities. If one is initially at rest (u2=0u_2 = 0), then after collision v1=0v_1 = 0 and v2=u1v_2 = u_1 — the moving body stops and the stationary one moves with the original velocity.

Case 2: A very heavy body colliding with a very light body at rest

Let m1≫m2m_1 \gg m_2 and u2=0u_2 = 0. The formulas simplify. For v1v_1:

v1=(m1−m2)u1m1+m2≈m1u1m1=u1v_1 = \frac{(m_1 - m_2) u_1}{m_1 + m_2} \approx \frac{m_1 u_1}{m_1} = u_1

The heavy body continues with almost unchanged velocity.

For v2v_2:

v2=2m1u1m1+m2≈2m1u1m1=2u1v_2 = \frac{2 m_1 u_1}{m_1 + m_2} \approx \frac{2 m_1 u_1}{m_1} = 2 u_1

Note

A massive object (like a wall) barely slows down when hit by a light object. The light object rebounds with approximately twice the speed of the massive one (if the massive one was moving toward it). If the massive object is stationary, the light object bounces back with nearly its original speed.

Case 3: A very light body colliding with a very heavy body at rest

Let m1≪m2m_1 \ll m_2 and u2=0u_2 = 0. Then:

v1=(m1−m2)u1m1+m2≈−m2u1m2=−u1v_1 = \frac{(m_1 - m_2) u_1}{m_1 + m_2} \approx \frac{- m_2 u_1}{m_2} = -u_1

v2=2m1u1m1+m2≈0v_2 = \frac{2 m_1 u_1}{m_1 + m_2} \approx 0

Tip

A light ball hitting a massive stationary wall rebounds with almost the same speed in the opposite direction (v1≈−u1v_1 \approx -u_1), while the wall barely moves. This is why a ball bounces back from a wall.

Inelastic Collision in One Dimension

In an inelastic collision, kinetic energy is not conserved. Momentum is still conserved. For a perfectly inelastic collision, the two bodies stick together and move with a common velocity vv after collision.

From momentum conservation:

m1u1+m2u2=(m1+m2)vm_1 u_1 + m_2 u_2 = (m_1 + m_2) v

v=m1u1+m2u2m1+m2v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}

The loss in kinetic energy is:

ΔK=12m1u12+12m2u22−12(m1+m2)v2\Delta K = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 - \frac{1}{2} (m_1 + m_2) v^2

Substituting vv and simplifying gives:

ΔK=12m1m2m1+m2(u1−u2)2\Delta K = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (u_1 - u_2)^2 …