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Ivenika [448]
3 years ago
9

Jack (mass 52.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. he collides with jill (mass 49.0 k

g ), 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. part a what is the direction of the jill's velocity after the collision?

Physics
1 answer:
dybincka [34]3 years ago
6 0
We must write down laws of conservation of momenta and energy. 
For the law of conservation of momenta will we will use two axes. One will be x-axis that will correspond to the east, and the other one will be y-axis corresponding to the north. Jack will be marked as 1 and Jill will be marked as 2.
Law of conservation of energy:
\frac{m_1v_1^2}{2}=\frac{m_1v'_1^2}{2}+ \frac{m_2v_2^2}{2}\\
m_2v_2^2=m_1v_1^2-m_1v_1'^2\\
v_2=\sqrt{\frac{m_1v_1^2-m_1v_1'^2}{m_2}
This will give us Jill's velocity after the colision.
v_2=6.43\frac{m}{s}
Law of conservation of momenta:
x: m_1v_1=m_1v_1'cos(34)+m_2v_2cos(\theta)\\
y: 0=m_1v_1'sin(34)-m_2v_2sin(\theta)\\
We will use the second equation to get the angle at which the Jill is traveling:
0=m_1v_1'sin(34)-m_2v_2sin(\theta)\\
m_2v_2sin(\theta)=m_1v_1'sin(34)\\
sin(\theta)=\frac{m_1v_1'sin(34)}{m_2v_2}\\
\theta=sin^{-1}(\frac{m_1v_1'sin(34)}{m_2v_2})
When we plug all the number we get:
\theta=27.45^\circ
Please note that this is the angle below the x-axis.

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