|A small device of mass 3.00 kg is sliding at 4.00 m/s along thex-axis on a horizontal frictionless surface. When it reaches the origin, it explodes intotwo pieces. One, of mass 1.00 kg, slides off at a velocity of 5.55 m/s at an angle of 50.0°above the x-axis. The other piece, of mass 2.00 kg slides off with an unknown velocity02f. What is the kinetic energy of that second piece?

Name: |A small device of mass 3.00 kg is sliding at 4.00 m/s along thex-axi
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Description: |A small device of mass 3.00 kg is sliding at 4.00 m/s along thex-axis on a horizontal frictionless surface. When it reaches the origin, it explodes intotwo pieces. One, of mass 1.00 kg, slides off at a velocity of 5.55 m/s at an angle of 50.0°above

Answered: |A small device of mass 3.00 kg is…

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter9: Momentum And Its Conservation

Section: Chapter Questions

Problem 96A

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|A small device of mass 3.00 kg is sliding at 4.00 m/s along the
x-axis on a horizontal frictionless surface. When it reaches the origin, it explodes into
two pieces. One, of mass 1.00 kg, slides off at a velocity of 5.55 m/s at an angle of 50.0°
above the x-axis. The other piece, of mass 2.00 kg slides off with an unknown velocity
02f. What is the kinetic energy of that second piece?

Expert Solution

Step 1

Use conservation of momentum to calculate the kinetic energy of the second piece.

Equate the horizontal components of the momenta of the system.

$m v = m_{1} v_{1} \cos θ_{1} + m_{2} v_{2} \cos θ_{2} (1)$

Equate the vertical components of the momenta of the system.

$m_{1} v_{1} \sin θ_{1} = m_{2} v_{2} \sin θ_{2} (2)$

Step 2

substitute the value of $v_{2}$ in equation (1) using equation (2).

$m v = m_{1} v_{1} \cos θ_{1} + m_{2} [\frac{m_{1} v_{1} \sin θ_{1}}{m_{2} \sin θ_{2}}] \cos θ_{2} m v - m_{1} v_{1} \cos θ_{1} = m_{1} v_{1} \sin θ_{1} c o t θ_{2} c o t θ_{2} = \frac{m v - m_{1} v_{1} \cos θ_{1}}{m_{1} v_{1} \sin θ_{1}} θ_{2} = c o t^{- 1} (\frac{m v - m_{1} v_{1} \cos θ_{1}}{m_{1} v_{1} \sin θ_{1}}) (3)$

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