Answer:
Explanation:
Given
mass of cannon and cart is [tex]m_1=6475.91 kg[/tex]
Spring constant [tex]k=20000 N/m[/tex]
mass of Projectile [tex]m_2=234 kg[/tex]
Launch Velocity of cannon [tex]v_2=170 m/s[/tex]
Launch angle [tex]\theta =31.3^{\circ}[/tex]
As the External Force is zero therefore we can conserve momentum
Initially both Projectile and cannon is at rest
Conserving momentum in horizontal direction
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\cos 31.3[/tex]
[tex]0+0=6475.91\times v_1+234\times 170\cdot \cos (31.3)[/tex]
[tex]v_1=-\frac{234\times 170\cdot \cos (31.3)}{6475.91}[/tex]
[tex]v_1=-5.24 m/s[/tex]
i.e. cannon is moving in opposite direction of Projectile