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Sagot :
Given,
The mass of puck A, m₁=0.0380 kg
The mass of puck B, m₂=0.0760 kg
The velocity of puck A before the collision, u=+6.29 m/s
The angle made by puck A with the x-axis, θ₁=65°
The angle made by puck B, θ₂=-37°
The momentum is conserved in both directions simultaneously and independently. That is, the sum x-components of the momentum before the collision and after the collision are equal. The same goes for the y-axis.
Considering the x-direction,
[tex]m_1u=m_1v_1\cos \theta_1+m_2v_2\cos \theta_2[/tex]Where v₁ is the velocity of puck A and v₂ is the velocity of puck B after the collision.
On substituting the known values,
[tex]\begin{gathered} 0.0380\times6.29=0.0380\times v_1\times\cos 65^{\circ}+0.0760\times v_2\times\cos (-37)^{\circ} \\ \Rightarrow0.24=0.016v_1+0.061v_2\text{ }\rightarrow\text{ (i)} \end{gathered}[/tex]Considering the y-direction,
[tex]0=m_1v_1\sin \theta_1+m_2v_2\sin \theta_2[/tex]On substituting the known values,
[tex]\begin{gathered} 0=0.0380\times v_1\times\sin 65^{\circ}+0.0760\times v_2\times\sin (-37)^{\circ} \\ 0=0.034v_1-0.046v_2\text{ }\rightarrow\text{ (ii)} \end{gathered}[/tex]On solving equations (i) and (ii),
[tex]\begin{gathered} v_1=3.93\text{ m/s} \\ v_2=2.90\text{ m/s} \end{gathered}[/tex]Thus the speed of pluck A is 3.93 m/s and the speed of pluck B is 2.90 m/s
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