Question

A cannon of mass 5.71 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 73.5-kg shell horizontally with an initial velocity of 547 m/s. Suppose the cannon is then unbolted from the earth, and no external force hinders its recoil. What would be the velocity of a shell fired from this loose cannon

Answer 1

Answer:

541.14 m/s

Explanation:

We are given that

Mass of cannon=$m_1=5.71\times 10^3 kg$

Mass of shell,$m_2=73.5 kg$

Initial velocity of shell,v=547 m/s

We have to find the velocity of shell fired from this loose cannon.

According to law of conservation of momentum

$m_1v_1+m_2v_2=m_1u_1+m_2u_2$

Initial momentum of system=0

$m_1v_1=-m_2v_2$

$v_1=-\frac{m_2v_2}{m_1}$

When the cannon is bolted to the ground then only shell moves and kinetic energy of system equals to kinetic energy of shell

Kinetic energy of shell,$K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 73.5(547)^2}=1.09\times 10^7 J$

K.E of shell=$\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2$

K.E of shell=$\frac{1}{2}m_1(-\frac{m_2v_2}{m_1})^2+\frac{1}{2}m_2v^2_2$

K.E of shell=$\frac{1}{2}(\frac{m^2_2v^2_2}{m_1}+\frac{1}{2}m_2v^2_2)$

2K.E of shell=$m_2v^2_2(\frac{m_2}{m_1}+1)$

Velocity of shell fired from this loose cannon,v_2=$\sqrt{\frac{2k.E}{m_2(\frac{m_2}{m_1}+1)}$

$v_2=\sqrt{\frac{2\times 1.09\times 10^7}{73.5(\frac{73.5}{5.71\times 10^3}+1)}$

$v_2=541.14m/s$

Hence, the velocity of shell fired from this loose cannon would be 541.14 m/s