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College Physics (Urone)
Found in: Page 290

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Short Answer

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\(266\,J\).

See the step by step solution

Step by Step Solution

Step 1: Definition of kinetic energy

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.

Step 2: Given data

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 \({M_x} = 60 + 75 = 135\;kg\)

Step 3: Velocity of the object after the collision

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 \(1.78\,m/s\)

Step 4: Kinetic energy lost:

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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