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A cannon of mass${m_1}$ = 10000 Kg located on a smooth horizontal platform fires a shell of mass${m_2}$ = 200 kg in horizontal direction with a velocity${v_2}$ = 300$m{s^{ - 1}}$ . Find the velocity of the cannon after it is shot

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Last updated date: 20th Jun 2024
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Answer
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Hint:- According to the conservation of linear momentum, the total momentum before and after the collision will be the same if there is no external force acting on the colliding objects. This implies the momentum when the cannon is at rest is equal to the momentum of the shell fired and then the momentum after firing.

Complete step-by-step solution
Step 1:
We are given that a cannon of mass${m_1}$ = 10000 Kg located on a smooth horizontal platform.
Cannon fires a shell of mass${m_2}$ = 200 kg in horizontal direction.
The velocity with which it has been shot is ${v_2}$ = 300$m{s^{ - 1}}$
We are going to find the velocity of the cannon after it is shot, say V.
There is no external force to the system. Therefore we can use linear momentum conservation theorem.
The linear momentum conservation theorem states that if two objects collide, then the total momentum before and after the collision will be the same if there is no external force acting on the colliding objects.
Step 2:
From the definition of the conservation of the linear momentum we can state that:
Momentum of the system before collision = momentum of the system before collision.
Momentum is the product of mass of the object with its respective velocity or P=MV, where P is denoted as momentum, m is the mass and V is the velocity.
Applying conservation of linear momentum we get, ${m_1}{v_1}$ = ${m_2}{v_2}$ + ${m_1}V$ ……. (1)
Where, ${m_1}$ = mass of the cannon = 10000 kg
${v_1}$= initial velocity of the cannon = 0
${m_2}$= mass of shell = 200 kg
${v_1}$= velocity of shell = 300$m{s^{ - 1}}$
V = final velocity of cannon
Substituting the value in equation 1 we get, 0=$200 \times 300 + 10000 \times V$
On solving it we will get 10000V = −60000
From here we will get the final velocity of canon is −6$m{s^{ - 1}}$ (recoil velocity)
Hence, the velocity of the cannon after it is 6$m{s^{ - 1}}$

Note:- Total momentum after firing equals to the total momentum before firing. The recoil velocity is the velocity of a body after the ejection of the object from the body. This is the reason that we have got our velocity in negative sign. As velocity can never be negative it is said as recoil velocity.