1. 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.

2. Originally Posted by jarny
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.

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

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

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

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

And we know that $m_s = 2m_b$

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

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

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

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

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.

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

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

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

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

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

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

$v_b = 6.27696 \, m/s$

-Dan