Answer to Question #161864 in Classical Mechanics for Jennifer Jace

Question #161864

Derive this formula for elastic collision

V1=(2m2 all over m1+m2)u1+(m2-m1 over m1+m2)u2

Where m1 is first mass

m2 us second mass


1
Expert's answer
2021-02-07T19:07:32-0500

Let’s derive the formula for the final velocities in case of elastic collision. From the law of conservation of momentum, we have:


"m_1u_1+m_2u_2=m_1v_1+m_2v_2. (1)"

Since collision is elastic, kinetic energy is conserved and we can write:


"\\dfrac{1}{2}m_1u_1^2+\\dfrac{1}{2}m_2u_2^2=\\dfrac{1}{2}m_1v_1^2+\\dfrac{1}{2}m_2v_2^2. (2)"

Let’s rearrange equations (1) and (2):


"m_1(u_1-v_1)=m_2(v_2-u_2), (3)""m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2). (4)"

Let’s divide equation (4) by equation (3):


"\\dfrac{(u_1-v_1)(u_1+v_1)}{u_1-v_1}=\\dfrac{(v_2-u_2)(v_2+u_2)}{v_2-u_2},""u_1+v_1=u_2+v_2. (5)"

Let's express "v_2" from the equation (5) in terms of "u_1", "u_2" and "v_1":


"v_2=u_1-u_2+v_1. (6)"

Let’s substitute equation (6) into equation (3). After simplification, we get:


"(m_1-m_2)u_1+2m_2u_2=(m_1+m_2)v_1."

From this equation we can find "v_1":


"v_1=\\dfrac{(m_1-m_2)}{(m_1+m_2)}u_1+\\dfrac{2m_2}{(m_1+m_2)}u_2."

Substituting "v_1"into the equation (6) we can find "v_2":


"v_2=\\dfrac{2m_1}{(m_1+m_2)}u_1+\\dfrac{(m_2-m_1)}{(m_1+m_2)}u_2."

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