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anonymous

  • one year ago

Sphere A has mass m and is moving with velocity v. It makes a head-on elastic collision with a stationary sphere B of mass 2m. After the collision their speeds (vA and vB) are:

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  1. anonymous
    • one year ago
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    @michele_laino

  2. anonymous
    • one year ago
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    @irishboy123

  3. anonymous
    • one year ago
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    Conserve momentum and also energy. Initial total momentum and energy would be mv and 0.5mv^2 Final total momentum and energy would be mvA + 2mvB and 0.5mvA^2+ 0.5(2m)vB^2

  4. Abhisar
    • one year ago
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    Hint: \(\sf V_B = \Large \frac{2M_1V}{M_1+M_2}\)

  5. anonymous
    • one year ago
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    @mashy and @abhisar it would be very kind of you to lead me to the conclusion and get me the final answer....

  6. IrishBoy123
    • one year ago
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    step 1 formally write out the equation for conservation of momentum as it applies here do that and we can go to step 2

  7. Michele_Laino
    • one year ago
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    since we have an elastic collision, and such collision happen along a horizontal line, then we have to apply the conservation of momentum (our mechanical system is isolated) and the conservation of kinetic energy ( there is no potential energy of interaction), so we can write this: \[\large \left\{ \begin{gathered} mv = m{u_1} + \left( {2m} \right){u_2}\quad \left( {{\text{conservation of momentum}}} \right) \hfill \\ \hfill \\ \frac{{m{v^2}}}{2} = \frac{{mu_1^2}}{2} + \frac{{\left( {2m} \right)u_2^2}}{2}\quad \left( {{\text{conservation of kinetic energy}}} \right) \hfill \\ \end{gathered} \right.\] where \( \large u_1, \; u_2 \) are the speed after collision, and \( \large v \) is the speed of the sphere \( \large A \) before collision. Please solve that system for \( \large u_1, \; u_2\)

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