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    urgent help!!!! physics problem

    Can someone please help me? i need this problem by 9:00 pm tonight eastern time: A running student has HALF the kinetic energy that his younger brother has. The student speeds up by 1.3m/s at which point he has the same kinetic energy as his brother. I that student's mass is twice as large as his brother's mass, what were the original speeds of both the student and his brother? I would really appreciate if someone could help me, THANKS! The formula for this problem is KE= (1/2)m(v^2) where KE is the kinetic energy in joules, m is mass, and v is the velocity.
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    Quote Originally Posted by jarny View Post
    Can someone please help me? i need this problem by 9:00 pm tonight eastern time: A running student has HALF the kinetic energy that his younger brother has. The student speeds up by 1.3m/s at which point he has the same kinetic energy as his brother. I that student's mass is twice as large as his brother's mass, what were the original speeds of both the student and his brother? I would really appreciate if someone could help me, THANKS! The formula for this problem is KE= (1/2)m(v^2) where KE is the kinetic energy in joules, m is mass, and v is the velocity.
    Any variable with a subscript "s" is the student and with the subscript "b" is the brother.

    $\displaystyle K_s = \frac{1}{2}K_b$

    So
    $\displaystyle \frac{1}{2}m_sv_s^2 = \frac{1}{4}m_bv_b^2$

    (1) $\displaystyle m_sv_s^2 = \frac{1}{2}m_bv_b^2$

    Now, the student speeds up.
    (2) $\displaystyle \frac{1}{2}m_s(v_s + 1.3)^2 = \frac{1}{2}m_bv_b^2$

    And we know that $\displaystyle m_s = 2m_b$

    Start by putting the mass relation in equations 1 and 2.
    (1) $\displaystyle (2m_b)v_s^2 = \frac{1}{2}m_bv_b^2$ <-- Cancel the $\displaystyle m_b$

    (1) $\displaystyle 2v_s^2 = \frac{1}{2}v_b^2$

    (2) $\displaystyle \frac{1}{2}(2m_b)(v_s + 1.3)^2 = \frac{1}{2}m_bv_b^2$ <-- Cancel the $\displaystyle m_b$

    (2) $\displaystyle (v_s + 1.3)^2 = \frac{1}{2}v_b^2$

    Note that the RHS of both equation 1 and equation 2 are the same. Thus the LHS of equation 1 and equation 2 are also the same.

    $\displaystyle 2v_s^2 = (v_s + 1.3)^2$

    $\displaystyle 2v_s^2 = v_s^2 + 2.6v_s + 1.69$

    $\displaystyle v_s^2 - 2.6v_2 - 1.69 = 0$

    You can solve this using the quadratic formula. I get that
    $\displaystyle v_s = 3.13848 \, m/s$ or $\displaystyle v_s = -0.538478 \, m/s$. Now the v quantities in these equations are speeds, which are always positive or zero. So we discard the negative solution.

    Thus
    (1) $\displaystyle 2v_s^2 = \frac{1}{2}v_b^2$

    $\displaystyle 2(3.13848)^2 = \frac{1}{2}v_b^2$

    $\displaystyle v_b = 6.27696 \, m/s$

    -Dan
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