Respuesta :
We can solve this problem using only the law of conversion of energy. Because you don't need to calculate the angle [tex]\phi[/tex] you don't have to deal with momentum and this makes this problem a lot easier. A good portion of the information about the problem is redundant. We have masses of both bodies and velocity of body A before and after the collision. You can see that it's velocity decreased, that's because some of it's initial energy was transferred to body B.
Let us write down the law of conversion of energy.
[tex] \frac{m_av_a^2}{2} = \frac{m_av'_a^2}{2} + \frac{m_bv_b^2}{2} [/tex]
[tex]m_bv_b^2=m_a(v_a^2-v'_a^2)[/tex]
[tex]v_b= \sqrt{ \frac{m_a(v_a^2-v'_a^2)}{m_b} } [/tex]
The final answer is:
[tex]v_b=2.72 \frac{m}{s} [/tex]
Let us write down the law of conversion of energy.
[tex] \frac{m_av_a^2}{2} = \frac{m_av'_a^2}{2} + \frac{m_bv_b^2}{2} [/tex]
[tex]m_bv_b^2=m_a(v_a^2-v'_a^2)[/tex]
[tex]v_b= \sqrt{ \frac{m_a(v_a^2-v'_a^2)}{m_b} } [/tex]
The final answer is:
[tex]v_b=2.72 \frac{m}{s} [/tex]