Physics, 20.03.2020 04:59 edjiejwi

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

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Answer from: rjsimpson73

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

Answer from: Quest

explanation:

as the height from which the ruler was dropped increased, the speed of the ruler at ryan's hand increased.

Answer from: Quest

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hope this