Respuesta :
Answer:
Explanation:
Given
mass of can [tex]m_2=1.2\ kg[/tex]
mass of rock [tex]m_1=0.37\ kg[/tex]
upward of can [tex]v_2=6.8\ m/s[/tex]
velocity of rock [tex]v_1=8.6\ m/s(Horizontal)[/tex]
ricochets off angle [tex]\alpha =61^{\circ}[/tex]
Velocity of rock after collision [tex]v_3=5\ m/s[/tex]
Let velocity of can after collision is [tex]v_4[/tex]
suppose can makes an angle of [tex]\theta [/tex] with horizontal
as External motion is absent therefore we can conserve momentum
Conserving momentum in x direction
[tex]m_1v_1+0=m_1v_3\cos \alpha +m_2v_2\cos \theta[/tex]
[tex]0.37\times 8.6+0=0.37\times 5\times \cos (61)+1.2\times v_4\times \cos \theta [/tex]
[tex]v_4\cos \theta =1.904 -------1[/tex]
Conserving momentum in Y direction
[tex]0+m_2v_2=m_1v_3\sin \alpha +m_ 2v_4\sin \theta [/tex]
[tex]1.2\times 6.8=1.2\times 5\times \sin (61)+1.2\times v_4\sin \theta [/tex]
[tex]v_4\sin \theta =2.426 --------2[/tex]
squaring and adding 1 and 2 we get
[tex]v_4^2=9.511[/tex]
[tex]v_4=3.084\ m/s[/tex]
For angle [tex]\theta [/tex] divide 1 and 2
[tex]\tan \theta =\frac{2.426}{1.904}[/tex]
[tex]\tan \theta =1.274[/tex]
[tex]\theta =\tan ^{-1}(1.274)[/tex]
[tex]\theta =51.87^{\circ}[/tex]
