I Don't Know How to Solve This Physics Problem

Question

A cannon elevated at some angle 𝜃 (unknown) fires a shell with some initial speed v₀ (unknown) that has a range R when fired over level ground. During a test fire, a shell explodes at the top of the trajectory into two unequal mass fragments. The shell explodes in such a manner that neither fragment experiences a change in momentum in the y direction. However, 18% of the shell (m₂) continues onward and the other fragment (m₁) falls straight down. Determine the distance D (in terms of R) that the forward moving fragment lands from the cannon.

Answer 1

HI Elle ,

To determine the distance D in terms of R that the forward-moving fragment lands from the cannon, you can use the principle of conservation of horizontal momentum. Since neither fragment experiences a change in momentum in the y-direction, we'll focus on the horizontal motion.

Let's denote the mass of the entire shell as M and the initial horizontal velocity of the shell as v0x. After the explosion, the mass m1 falls straight down, and the mass m2 continues forward. We can write the following conservation of momentum equation in the x-direction:

Initial momentum = Final momentum

M * v0x = m2 * v2f

Here:

- M is the mass of the entire shell.

- v0x is the initial horizontal velocity of the shell.

- m2 is the mass of the forward-moving fragment.

- v2f is the final horizontal velocity of the forward-moving fragment.

Now, we also know that the forward-moving fragment has lost some speed due to the explosion, and it still has to travel the same range R. The time it takes to reach R is the same for both the forward-moving fragment and the shell's initial motion.

So, we can use the range formula to relate the final velocity v2f to R:

R = (v2f * t)

Now, we need to find an expression for v2f. We know that the forward-moving fragment (m2) contains 18% of the shell's mass:

m2 = 0.18 * M

Now, we can express the initial velocity v0x in terms of m2 and m1:

v0x = (m2 * v2f + m1 * 0) / M

Since neither fragment experiences a change in momentum in the y-direction, we can write this as:

v0x = m2 * v2f / M

Now, we can solve for v2f:

v2f = (v0x * M) / m2

Now, substitute this expression for v2f into the range formula:

R = ((v0x * M) / m2) * t

Now, we have an expression for R. We can express M in terms of m1 and m2, and we need to determine the time t.

Since m1 falls straight down, its horizontal velocity remains zero, and it falls under gravity. You can use the following equation to find the time t for m1 to fall:

0.5 * g * t^2 = H

Where:

- g is the acceleration due to gravity.

- H is the height from which m1 falls.

Once you find the time t, you can substitute it into the expression for R to find D in terms of R.

Keep in mind that the specific values of v0, m1, m2, and H would be needed to find the numerical value of D in terms of R.

I hope this will help. I am happy to tutor you on any other questions you may have; please feel free to send me a message!