Answer to Question #116085 in Mechanics | Relativity for Jackson

Question #116085
Two objects are involved in a perfectly elastic head-on collision on a frictionless air hockey table. The objects have masses of 6.21kg and 9.4kg respectively. If the first object has a final velocity of 1.9m/s [right]and the second object has a final velocity of 0.80 m/s [left], calculate the initial velocities of each object
Expert's answer

Let "m_1=6.21\\,\\mathrm{kg}" and "m_2= 9.4\\,\\mathrm{kg}." Let "v_{1} \\; \\mathrm{and}\\; v_2" be the initial velocities of bodies and "u_{1} = 1.9\\,\\mathrm{m\/s}\\; \\mathrm{and}\\; u_2 = 0.80\\,\\mathrm{m\/s}" be the velocities after the collision.

We may write the law of conservation of impulses (see

"m_1\\vec{v_1} + m_2\\vec{v_2} = m_1\\vec{u_1} + m_2\\vec{u_2}."

"\\vec{v_1}" is directed left because the final velocity of the first body is directed right. In projection

"-m_1v_1+ m_2v_2 = m_1u_1 - m_2u_2." Here "v_1 \\, \\mathrm{and}\\; v_2" are positive values.

Next, we substitute the values of mass and velocity and get

"-m_1v_1 + m_2v_2 = 4.279\\,\\mathrm{kg\\,m\/s}."

Let us write the law of conservation of energy:

"\\dfrac{m_1v_1^2}{2} + \\dfrac{m_2v_2^2}{2} =\\dfrac{m_1u_1^2}{2} + \\dfrac{m_2u_2^2}{2}\\,."

We may multiply the expression on 2 and calculate the right hand.

"{m_1v_1^2} + {m_2v_2^2} =28.4341\\,\\mathrm{J}."

So we have a system of equations, namely

"\\begin{cases}\n-m_1v_1 + m_2v_2 = 4.279\\,\\mathrm{kg\\,m\/s}, \\\\\n{m_1v_1^2} + {m_2v_2^2} =28.4341\\,\\mathrm{J}.\n\\end{cases}"


"\\begin{cases}\n-6.21v_1 + 9.4v_2 = 4.279, \\\\\n{6.21v_1^2} + {9.4v_2^2} =28.4341.\n\\end{cases}" "\\begin{cases}\n6.21v_1 = 9.4v_2-4.279, \\\\\n{6.21^{-1}\\cdot( 9.4v_2-4.279)^2} + {9.4v_2^2} =28.4341.\n\\end{cases}"

Solving the last equation, we get "v_{2,1} = -0.8\\,\\mathrm{m\/s}\\; (u_{1,1} =-1.9\\,\\mathrm{m\/s} )" and "v_{2,2} = 1.35\\,\\mathrm{m\/s}\\; (u_{1,2} =1.35\\,\\mathrm{m\/s} )." The first pair of values is negative, so we should choose the second pair.

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