CollisionsMedium

Elastic collision — two masses

Mass m₁ = 3 kg with velocity v₁ = 4 m/s collides elastically with m₂ = 1 kg at rest.\nCalculate the final velocities of both.

Given data

m₁ = 3 kgv₁ = 4 m/sm₂ = 1 kgv₂ = 0

Steps

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What is the velocity of m₁ after the collision?
Write your working — the numerical result must appear in the text m/s
What is the velocity of m₂ after the collision?
What is the total momentum of the system before the collision?
What is the total momentum of the system after the collision?

Full worked solution

What is the velocity of m₁ after the collision?
$v_1' = \dfrac{m_1 - m_2}{m_1 + m_2} v_1 = \dfrac{3-1}{3+1} \cdot 4$
$v_1' = \frac{m_1 - m_2}{m_1 + m_2}\,v_1 = \frac{3 - 1}{3 + 1} \times 4 = \frac{2}{4} \times 4 = \mathbf{2\,m/s}$ . After the elastic collision, $m_1$ continues in the same direction but at a lower speed.
What is the velocity of m₂ after the collision?
$v_2' = \dfrac{2m_1}{m_1 + m_2} v_1 = \dfrac{2 \cdot 3}{4} \cdot 4$
$v_2' = \frac{2m_1}{m_1 + m_2}\,v_1 = \frac{2 \times 3}{3 + 1} \times 4 = \frac{6}{4} \times 4 = \mathbf{6\,m/s}$ . The lighter mass $m_2$ is pushed forward at a higher speed than the incoming mass.
What is the total momentum of the system before the collision?
$p_i = m_1 v_1 = 3 \cdot 4$
$p_i = m_1 v_1 + m_2 v_2 = 3 \times 4 + 1 \times 0 = \mathbf{12\,kg\cdot m/s}$ . Only $m_1$ contributes since $m_2$ is initially at rest.
What is the total momentum of the system after the collision?
$p_f = m_1 v_1' + m_2 v_2' = 3 \cdot 2 + 1 \cdot 6$
$p_f = m_1 v_1' + m_2 v_2' = 3\times2 + 1\times6 = 6 + 6 = \mathbf{12\,kg\cdot m/s}$ ✓. Energy check: $KE_i = \frac{1}{2}\cdot3\cdot16 = 24\,\mathrm{J}$ ; $KE_f = \frac{1}{2}\cdot3\cdot4 + \frac{1}{2}\cdot1\cdot36 = 6 + 18 = 24\,\mathrm{J}$ ✓ — both momentum and kinetic energy are conserved.

Result:v₁' = 2 m/s, v₂' = 6 m/s. Both momentum and KE conservation verified.

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