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Q32PE

Expert-verifiedFound in: Page 290

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**During an ice show, a **\(60kg\)**skater leaps into the air and is caught by an initially stationary **\(75kg\)**skater. (a) What is their final velocityassuming negligible friction and that the **\(60kg\)**skater’s originalhorizontal velocity is **\(4\;.00m/s\)**? (b) How much kinetic energy is lost?**

(a) Velocity of the system after the collision is\(1.78\,m/s\).

(b) The kinetic energy lost is\(266\,J\).

**Kinetic energy is the energy that an item or particle has as a result of its movement. When work is done on an object by exerting a net force, the object accelerates and gains kinetic energy as a result.**

Here the collision is happening, where one skater is catching other. The collision in this case is inelastic.

The two object before the collision are separated, but after the collision they are joint.

The mass of the first object before collision\({m_a} = 60\;kg\)

The mass of the second object before collision=\({m_b} = 75\;kg\)

The velocity of the first object before the collision= \({v_a} = 4\;m/s\)

The velocity of the second object before the collision= \({V_b} = 0\;m/s\)

Mass of the objects when they are together after the collision \({M_x} = 60 + 75 = 135\;kg\)

By putting all the value into the equation we get

\(\begin{aligned}{M_a}{V_a} + {M_b}{V_b} &= \left( {{M_a} + {M_b}} \right){V_x}\\{V_x} &= \dfrac{{{M_a}{V_a} + {M_b}{V_b}}}{{\left( {{M_a} + {M_b}} \right)}}\\{V_x} &= \dfrac{{\left( {60} \right)\left( {40} \right) + \left( {75} \right)\left( 0 \right)}}{{\left( {135} \right)}}\\{V_x} &= 1.78\,m/s\end{aligned}\)

Velocity of the system after the collision is \(1.78\,m/s\)

The kinetic energy lost will be the difference of the initial kinetic energy to the final kinetic energy.

\(\begin{aligned}K{E_{lost}} &= K{E_i} - K{E_f}\\K{E_{lost}} &= \dfrac{1}{2}{m_a}{v_a}^2 - \dfrac{1}{2}{m_x}{v_x}^2\\K{E_{lost}} &= \dfrac{1}{2}\left( {{m_a}{v_a}^2 - {m_x}{v_x}^2} \right)\\K{E_{lost}} &= \dfrac{1}{2}\left( {\left( {60} \right){{\left( 4 \right)}^2} - \left( {135} \right){{\left( {1.78} \right)}^2}} \right)\end{aligned}\)

\(K{E_{lost}} = 266\,J\)

The kinetic energy lost is \(266\,J\).

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