Answer:
Step-by-step explanation:
Given
mass of cannon and cart is
![m_1=6475.91 kg](https://img.qammunity.org/2021/formulas/physics/college/8a9k7vuj47tfd9bm4z3x3nixjfgsxx39f9.png)
Spring constant
![k=20000 N/m](https://img.qammunity.org/2021/formulas/physics/college/nnmn092y7tz8tjyfom79cc4vb9ervh0kp5.png)
mass of Projectile
![m_2=234 kg](https://img.qammunity.org/2021/formulas/physics/college/tszc7up5sirubzasa5pjnfssz30v8ncday.png)
Launch Velocity of cannon
![v_2=170 m/s](https://img.qammunity.org/2021/formulas/physics/college/dabkvrw3pivk5vjbky510r6t34qcie01lx.png)
Launch angle
![\theta =31.3^(\circ)](https://img.qammunity.org/2021/formulas/physics/college/wqihrgbiefa4w1pi2ejqsoexds64vydr2t.png)
As the External Force is zero therefore we can conserve momentum
Initially both Projectile and cannon is at rest
Conserving momentum in horizontal direction
![m_1u_1+m_2u_2=m_1v_1+m_2v_2\cos 31.3](https://img.qammunity.org/2021/formulas/physics/college/w95wkuq7zg7igmeayl3wi3tn344wao97x0.png)
![0+0=6475.91* v_1+234* 170\cdot \cos (31.3)](https://img.qammunity.org/2021/formulas/physics/college/38kjuw9px9z0w90ghfvbqv3qo4k7q5q43v.png)
![v_1=-(234* 170\cdot \cos (31.3))/(6475.91)](https://img.qammunity.org/2021/formulas/physics/college/mlp14bnt4vqqmqwu8gztx79z399sziu3b2.png)
![v_1=-5.24 m/s](https://img.qammunity.org/2021/formulas/physics/college/fbillxbho5q3gacbeabc95ga59vjhjcz81.png)
i.e. cannon is moving in opposite direction of Projectile