(a) To find the speed with which the barrel begins to recoil, we can use the principle of conservation of momentum. Initially, the total momentum of the gun and the shell is zero since they are at rest. However, after the shell is fired, the momentum of the system should still be zero.
Let the speed with which the barrel begins to recoil be denoted by V. The momentum of the gun and the shell after firing can be expressed as M * (-V) + m * v = 0.
Simplifying the equation, we get M * V = m * v.
Therefore, the speed with which the barrel begins to recoil is V = (m * v) / M.
(b) To find the constant force required to bring the gun to rest in 3 seconds, we can use the equation F = m * a, where F is the force, m is the mass, and a is the acceleration.
In this case, the force required to bring the gun to rest is the force that counteracts the momentum of the gun and the shell.
From part (a), we know that the initial momentum of the gun and the shell is M * V, and the final momentum is 0. The change in momentum is therefore -M * V.
Using the formula F = m * a, we can determine the force required by substituting m with M and a with -V/t, where t is the time taken.
Therefore, the force required is F = M * (-V) / t.
Given that the time taken is 3 seconds, the force required to bring the gun to rest in 3 seconds is F = -(3M * V) / t.
A gun of mass M fires a shell of mass m and recoils horizontally. If the barrel of the gun is horizontal and the shell travels along it with speed v, find the: (a) speed with which the barrel begins to recoil.
(b) constant force required to bring the gun to rest in 3 seconds.
1 answer
1 answer