Answered: During an ice show, a 100.0-kg skater…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

During an ice show, a 100.0-kg skater leaps with an initial of 4.00 m/s into the air and is caught by an initially stationary 75.0-kg skater. How much kinetic energy (KE) is lost (enter your energy in Joules)? Assume the friction force is negligible. (Hint: First find the final common velocity and subtract the final KE from the initial KE)

Expert Solution

Step 1

Given data:

Mass of first skater, $m_{1} = 100.0 kg$

Initial velocity of first skater, $u_{1} = 4.0 m/s$

Mass of the other skater, $m_{2} = 75.0 kg$

Initial velocity of another skater, $u_{2} = 0 m/s$

Collision is inelastic between two skaters.

Let the common velocity of the skaters be V.

Applying the conservation of momentum:

$\begin{array}{rcl} m_{1} u_{1} + m_{2} u_{2} & = & (m_{1} + m_{2}) V \\ ∴ \\ V & = & \frac{m_{1} u_{1} + m_{2} u_{2}}{(m_{1} + m_{2})} \end{array}$

Plugging in the given values:

$\begin{array}{rcl} V & = & \frac{100 \times 4.0 + 750 \times 0}{100 + 75} \\ = & 2.286 m/s \end{array}$

Lost in kinetic energy = initial kinetics energy - final kinetics energy

$\begin{array}{rcl} Δ K . E & = & \frac{1}{2} m_{1} {u_{1}}^{2} + \frac{1}{2} m_{2} {u_{2}}^{2} - \frac{1}{2} (m_{1} + m_{2}) V^{2} \\ = & \frac{1}{2} \times 100 \times {(4.0)}^{2} + 0 - \frac{1}{2} \times (100 + 75) \times {(2.286)}^{2} \\ = & 342.74 J \end{array}$

Answer:

Lost in kinetics energy: $342.74 J$

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