A shell is fired with velocity having horizontal and vertical components ux, uy. At the highest point of it's path the shell explodes into two fragments of mass m1, m2, which separate in a horizontal direction. The explosion generates kinetic energy E. Show that the distance between the Pieces when the hit the ground is:
(Uy/g)*(sqrt(2E(m1+m2)/m1m2))
To find the distance between the pieces when they hit the ground, we need to break down the problem into two parts: the horizontal motion and the vertical motion.
Let's consider the horizontal motion first. Since the fragments separate in a horizontal direction, the only force acting on them is due to the initial velocity component, ux. This force does not affect their horizontal motion. Therefore, both fragments will have the same horizontal velocity throughout their path. Let's call this velocity Vx.
Now, let's analyze the vertical motion. The initial vertical velocity of the shell when it was fired can be represented by uy. At the highest point of its path, the shell explodes and the fragments separate in a horizontal direction. After the explosion, there is no vertical force acting on the fragments except for gravity. Therefore, the only acceleration acting on the fragments is the acceleration due to gravity, g. Let's assume the time taken for the fragments to reach the ground is t.
Using the equations of motion, we can relate the initial vertical velocity (uy), the time taken (t), and the acceleration due to gravity (g):
s = ut + (1/2)gt^2
where s is the vertical distance traveled by the fragments. Since the shell explodes at the highest point of its path, the vertical distance traveled by the fragments will be equal to half of the maximum vertical distance of the shell. This maximum vertical distance is given by:
H = (uy^2) / (2g)
Therefore, the vertical distance traveled by the fragments is:
s = (1/2) H
Substituting the value of H in terms of uy and g, we have:
s = (1/2) [(uy^2) / (2g)]
simplifying, we get:
s = (uy^2) / (4g)
Now, let's find the time taken for the fragments to reach the ground. For the horizontal motion, the distance traveled by each fragment is the same, and it can be represented by X:
X = Vx * t
We need to relate X to the other variables in the problem. Since the horizontal velocity remains constant, we can use the equation:
X = Vx * t
To find Vx, we know that the initial velocity of the fragments, ux, is equal to Vx. Therefore:
X = ux * t
Now, substituting the value of X in terms of uy and g, we have:
X = (uy/g) * t
Simplifying further, we get:
t = (X * g) / uy
Now, we have expressions for both s and t. Let's substitute the values into our original equation:
s = (uy^2) / (4g)
t = (X * g) / uy
Finally, substituting s and t into the equation X = ux * t, we get:
(uy^2) / (4g) = (ux * X * g) / uy
Simplifying further, we have:
X = (uy^2) / 4*(ux * uy)
Since we are looking for the distance between the pieces when they hit the ground, X is equal to the distance between the pieces. Therefore, we have:
Distance = X = (uy^2) / 4*(ux * uy)
Now, let's simplify the expression:
Distance = (uy^2) / 4*(ux * uy)
= uy / (4 * ux)
= (1/4) * (uy / ux)
To express this in terms of the given variables, we need to substitute the expression for uy/ux:
Distance = (1/4) * (uy / ux)
= (1/4) * (uy / (E/m1)^0.5)
= (1/4) * (uy * (m1/E)^0.5)
= (1/4) * (uy * (m1/(m1+m2)*2E/m1m2)^0.5)
= (1/4) * (uy/g) * (sqrt(2E(m1+m2)/m1m2))
= (uy/g) * (sqrt(2E(m1+m2)/m1m2))
Hence, the distance between the pieces when they hit the ground is (uy/g) * (sqrt(2E(m1+m2)/m1m2)).