91Ó°ÊÓ

A final velocity is defined as the final speed of a moving object with an initial velocity and acceleration over some time.

The mass of the 1^st car is m₁=1200kg

The mass of the 2^nd car is\({m_2} = 850\;{\rm{kg}}\).

The velocity of the first car is\({v_1} = 8.00\;{\rm{m/s}}\).

The velocity of the 2nd car is\({v_2} = 17.0\;{\rm{m/s}}\).

(a)

Using the conservation of momentum along horizontal direction we get,

\(\begin{array}{c}{m_1}{v_1} + {m_2}{v_2} = \left( {{m_1} + {m_2}} \right)v\\v = \frac{{{m_1}{v_1} - {m_2}{v_2}}}{{\left( {{m_1} + {m_2}} \right)}}\end{array}\)

Substituting the values we get,

\(\begin{array}{c}v = \frac{{1200 \times \left( { - 8.00\;\hat j} \right) + 850 \times \left( { - 17.0\;\hat i} \right)}}{{1200 + 850}}\;{\rm{m/s}}\\ = \left( { - 7.05\;\hat i - 4.68\;\hat j} \right)\;{\rm{m/s}}\end{array}\)

The speed is,

\(\begin{array}{c}v = \sqrt {{{\left( {7.05} \right)}^2} + {{\left( {4.68} \right)}^2}} \;{\rm{m/s}}\\ = 8.5\;{\rm{m/s}}\end{array}\)

The angle with the X-axis is,

\(\begin{array}{c}\theta = {\tan ^{ - 1}}\frac{{ - 4.68}}{{ - 7.05}}\\ = 33.6^\circ \end{array}\)

(b)

The initial kinetic energy before collision is,

\(\begin{array}{c}{\left( {KE} \right)_i} = \left[ {\frac{1}{2} \times 1200 \times {8^2} + \frac{1}{2} \times 850 \times {{17}^2}} \right]\;{\rm{J}}\\ = 161225\;{\rm{J}}\end{array}\)

The final kinetic energy is,

\(\begin{array}{c}{\left( {KE} \right)_f} = \frac{1}{2} \times \left( {1200 + 850} \right) \times {8.5^2}\;{\rm{J}}\\ = 73405.5\;{\rm{J}}\end{array}\)

The loss of kinetic energy is,

\(\begin{array}{c}\Delta E = \left[ {161225 - 73405.5} \right]\;{\rm{J}}\\ = 87819.5\;J\end{array}\)

Therefore loss of kinetic energy is, 87819.5J