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Q50PE

Expert-verifiedFound in: Page 291

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Two cars collide at an icy intersection and stick together afterward. The first car has a mass of **\(1200 kg\)**and is approaching at **\(8.00\;{\rm{m/s}}\)**due south. The second car has a mass of **\(850 kg\)**and is approaching at **\(17.0\;{\rm{m/s}}\)**due west. (a) Calculate the final velocity (magnitude and direction) of the cars. (b) How much kinetic energy is lost in the collision? (This energy goes into deformation of the cars.) Note that because both cars have an initial velocity, you cannot use the equations for conservation of momentum along the x -axis and y -axis; instead, you must look for other simplifying aspects.**

(a)The final velocity of the ball is\(8.5\;{\rm{m/s}}\) at an angle\(33.6^\circ \)with X-axis.

(b)The loss of energy is\(87819.5\;{\rm{J}}\).

**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 m_{1}=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

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