# Elastic Collision Formula

An elastic collision is a collision where both kinetic energy, KE, and momentum, p, are conserved. This means that KE_{0} = KE_{f} and p_{o} = p_{f}. Recalling that KE = 1/2 mv^{2}, we write
1/2 m_{1}(v_{1i})^{2} + 1/2 m_{2}(v_{i})^{2} = 1/2 m_{1}(v_{1f})^{2} + 1/2 m_{2} (v_{2f})^{2}, the final total KE of the two bodies is the same as the initial total KE of the two bodies. And, since p = linear momentum = mv, then we write m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}.

[A] m_{1}v_{1i} + m_{2v2i = m1v1f + m2v2f}

[B] 1/2 m_{1}(v_{1i})^{2} + 1/2 m_{2}(v_{i})^{2} = 1/2 m_{1}(v_{1f})^{2} + 1/2 m_{2} (v_{2f})^{2}

KE = kinetic energy

p = momentum

m = mass, kg

m_{i} = mass of 1st object

m_{2}= mass of 2nd object

v = velocity, m/s

v_{1} = velocity of 1st object

v_{2} = velocity of 2nd object

v_{i} = initial velocity

v_{f} = final velocity

Elastic Collision Formula Questions:

1) A red ball of mass 0.2 kg hits a blue ball of mass 0.25 kg, in an elastic collision, and the red ball comes to a stop. The red ball has a velocity of 5 m/s, and the blue ball was at rest. What is the final velocity of the blue ball?

Answer: The mass of the 1st ball, m_{1} = 0.2 kg; the mass of the 2nd ball, m_{2} = 0.25kg. The initial velocity of the 1st ball, v_{1i} = 5 m/s; the initial velocity of the 2nd ball, v_{2i} = 0; the final velocity of the 1st ball, (v_{1f}) = 0.

m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}

(0.2 kg)(5 m/s) + (0.25 kg)(0 m/s) = (0.2 kg)(0) + (0.25 kg)(v_{2f})

1.0 kg.m/s + 0 = 0 + (0.25 kg)(v_{2f})

1.0 kg.m/s = (0.25 kg)(v_{2f})

(1.0 kg.m/s) / 0.25 kg = (v_{2f})

4 m/s = (v_{2f})

2) Use the equation for conservation of kinetic energy in an elastic collision to determine the final velocity for the blue ball.

Answer: The mass of the 1st ball, m_{1} = 0.2 kg; the mass of the 2nd ball, m_{2} = 0.20kg. The initial velocity of the 1st ball, v_{1i} = 5 m/s; the initial velocity of the 2nd ball, v_{2i} = 0; the final velocity of the 1st ball, (v_{1f}) = 0.

1/2 m_{1}(v_{1i})^{2} + 1/2 m_{2}(v_{i})^{2} = 1/2 m_{1}(v_{1f})^{2} + 1/2 m_{2} (v_{2f})^{2}

1/2 (0.2 kg)(5m/s)^{2} + 1/2 (0.2 kg)(0) = 1/2 (0.2 kg)(0) + 1/2 (0.2 kg)(v_{2f})^{2}

1/2 (0.2 kg)(5m/s)^{2} = 1/2 (0.2 kg)(v_{2f})^{2}

(5m/s)^{2} = (v_{2f})^{2}

25 m^{2}/s^{2} = (v_{2f})^{2}

v_{2f} = √25 m^{2}/s^{2}

v_{2f} = 5 m/s

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