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.
(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.
To solve this problem, we can apply the principle of conservation of linear momentum.
(a) To find the speed with which the barrel begins to recoil, we need to consider the initial momentum and the final momentum of the system.
The initial momentum of the system is zero since the gun is at rest. The final momentum of the system is the sum of the momentum of the gun and the momentum of the shell after it is fired.
The momentum of the gun is given by:
M * v_gun
The momentum of the shell is given by:
m * v_shell
According to the conservation of linear momentum, the initial momentum is equal to the final momentum:
0 = M * v_gun + m * v_shell
Since the barrel and the shell move in opposite directions horizontally, the final momentum is negative.
Therefore,
-M * v_gun = m * v_shell
Now, divide both sides of the equation by M:
-v_gun = (m/M) * v_shell
Let's call the ratio (m/M) as x:
-v_gun = x * v_shell
Now, we can solve for v_gun:
v_gun = -x * v_shell
So, the speed with which the barrel begins to recoil is -x times the speed of the shell.
(b) To find the constant force required to bring the gun to rest in 3 seconds, we can use Newton's second law, which states that the force applied to an object is equal to the rate of change of its momentum.
The rate of change of momentum is given by:
Force = (change in momentum) / (time)
The change in momentum of the gun is equal to the initial momentum of the gun, which we found to be M * v_gun.
The time is given as 3 seconds.
Therefore, the constant force required to bring the gun to rest is:
Force = (M * v_gun) / (3)
Substituting the value of v_gun:
Force = (M * (-x * v_shell)) / (3)
Simplifying, we get:
Force = -(M * x * v_shell) / (3)
So, the constant force required to bring the gun to rest in 3 seconds is -(M * x * v_shell) / 3.