contestada

Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 kg ), who is initially at rest. After the collision, Jack is traveling at 5.00 m/s in a direction 34.0∘ north of east. Ignore friction.
A) What is the direction of the Jill's velocity after the collision?
B) What is the magnitude of the Jill's velocity after the collision?

Respuesta :

Answer:

Part(A): The magnitude of Jill's final velocity is [tex]\bf{6.59~m/s}[/tex].

Part(B): The direction is [tex]\bf{42.7^{0}}[/tex] south to east.

Explanation:

Given:

Mass of Jack, [tex]m_{1} = 59.0~Kg[/tex]

Mass of Jill, [tex]m_{2} = 47..0~Kg[/tex]

Initial velocity of Jack, [tex]v_{1i} = 8.00~m/s[/tex]

Initial velocity of Jill, [tex]v_{2i} = 0[/tex]

Final velocity of Jack, [tex]v_{1f} 5.00~m/s[/tex]

The final angle made by Jack after collision, [tex]\alpha = 34.0^{0}[/tex]

Consider that the final velocity of Jill be [tex]v_{2f}[/tex] and it makes an angle of [tex]\beta[/tex] with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

[tex]~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}[/tex]

where, [tex]v_{2f}^{x}[/tex] is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

[tex]v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s[/tex]

Conservation of momentum of the system along north direction is given by

[tex]~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s[/tex]

where, [tex]v_{2f}^{y}[/tex] is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

[tex]v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s[/tex]

Part(B):

The direction is given by

[tex]\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}[/tex]

Ver imagen WinterWhitney
okpalawalter8

Answer:

  • The direction of the Jill's velocity = [tex]5.98m/s[/tex]
  • The magnitude of the Jill's velocity = [tex]35.95^o[/tex] south of east

Explanation:

Traveling at [tex]5 m/s[/tex] in a direction [tex]34^o[/tex] north of east

Components [tex]5sin34 = 2.796[/tex]  north and [tex]5cos34 = 4.145[/tex] east

conservation of momentum(mv):

momentum before = momentum after

in the east direction:

[tex]59(8) + 47(0) = 59(4.145) + 47u[/tex]

Therefore,

[tex]U = 4.839m/s[/tex] east

In the north direction:

[tex]59(0) + 47(0) = 2.796(59) + 47v\\\\v = -3.509 m/s north \\\\v = 3.509 m/s south[/tex]

[tex]Velocity = \sqrt{(4.839)^2 + (3.509)^2}\\\\Velocity = 5.98m/s[/tex]

At angle,

[tex]tan^{-1}\frac{3.509}{4.839} = 35.95^o[/tex] south of east

For more information, visit

https://brainly.com/question/15700950

Otras preguntas