Hi everyone, hopefully someone will be able to point me in the right direction with this problem, I get as far as combining the two equations together and no matter how I rearrange them they don't seem to cancel nicely and leave me with an awful quadratic, I can find v(adsbygoogle = window.adsbygoogle || []).push({}); _{x}, but the question specificity states to findmfirst.

Question:

Consider two railway trucks that can travel along a straight railway track which deﬁnes the x-axis. A laden truck of massM =6 .6×103 kgand an empty truck of unknown massmapproach each other at constant velocity. The laden truck has an initial velocity component ofuwhich is half that of the initial velocity component of the empty truck (but in the opposite direction). After an elastic collision, the velocity component of the laden truck is halved, but it continues in the same direction. The unknown post-collision velocity component of the empty truck is_{x}= +2 .0ms−1v._{x}

(b) Rearrange the momentum equation into the form:v...,and rearrange the energy equation into the form_{x}=m =....Then combine these simultaneous equations to determine values, ﬁrst for m and then forv, using the known values of_{x}Mandu._{x}

The attempt at a solution

Rearranging equations:

Momentum:

$$ m_1u_1+m_2u_2=m_1v_1+m_2v_2 $$

$$ m_1u_1+m_2u_2-m_1v_1=m_2v_2 $$

$$ v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2} $$

$$ v_2=\frac{m_1u_1-m_1v_1}{m_2}+u_2 $$

$$ v_2=\frac{m_1\left(u_1-v_1\right)}{m_2}+u_2 $$

Kinetic energy:

$$ \frac{1}{2}m_1u_1^2\ +\ \frac{1}{2}m_2u_2^2\ =\ \frac{1}{2}m_1v_1^2\ +\ \frac{1}{2}m_2v_2^2 $$

$$ m_1u_1^2+m_2u_2^2=m_1v_1^2+m_2v_2^2 $$

$$ m_1u_1^2=m_1v_1^2+m_2v_2^2-m_2u_2^2 $$

$$ m_1u_1^2-m_1v_1^2=m_2v_2^2-m_2u_2^2 $$

$$ m_1\left(u_1^2-v_1^2\right)=m_2\left(v_2^2-u_2^2\right) $$

$$ m_2=\frac{m_1\left(u_1^2-v_1^2\right)}{\left(v_2^2-u_2^2\right)} $$

This is where It starts getting messy, I cant seem to find a way to get the square terms to vanish which makes this problem a mess, can anyone point me in the right direction?

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# Homework Help: Combining conservation of momentum and kinetic energy

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