Horizontal Speed of Fragment

[shimizua]

Homework Statement

A 14.0-kg shell is fired from a gun with a muzzle velocity 160.0 m/s at 27.0o above the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. What is the horizontal speed of the other fragment?

Homework Equations

The velocity is 160/s at an angle of 27deg - velocity must always have a speed and direction.
That's equivalent to 260 sin(27) vertically and 160 cos(27) horizontally. You can see this if you draw a triangle.

A nice feature about physics is that you can treat forces at right angles independantly so we can ignore the horizontall bit fro now and just look at the vertical.
It starts off going up at 125sin(33) and slows down at 9.8m/s^2
v^2 = u^2 + 2 a s At the top when u=0 and a = -9.8
160sin() ^2 = 2 * 9.8 * s where s (the height) = 164.31

The kinetic energy is = 1/2 m v^2 and potential energy is = m g h
At the top of the curve you have used up some of the initial KE as PE - so we now work out how much KE is left.
KE = 1/2m v^2 - mgh = 1/2* M * 160^2 - M * 9.8 * 164.31 = M ( 1/2*160^2 - 9.8*164.31).

Now the shell is only moving horizontally at the top of the curve so all this kinetic energy is going to go into the horizontal velocity.
But the mass has just halved - so:

M * those numbers = 1/2 1/2 M V^2 where V is the new velocity.

The Attempt at a Solution

so i tried doing this and it came up wrong. i saw that someone had already asked this question before but i don't know what i am doing wrong. i also just tried doing m1v1=m2v2 and that was wrong also

 

[tiny-tim]

Hi CzarValvador!

i] You didn't need to caclulate h … your KE + PE equation gives you V without it.

Now the shell is only moving horizontally at the top of the curve so all this kinetic energy is going to go into the horizontal velocity …

ii] An explosion is a collision in reverse.

Collisions are not elastic unless the question says so.

(and explosions go BANG!, so they can't conserve energy, can they?)

So forget energy, and use something else.

 

[thomate1]

I would approach this problem by first analyzing the given information and identifying the relevant equations and principles that can be applied.

From the given information, we know that the shell has a mass of 14.0 kg and is fired from a gun with a muzzle velocity of 160.0 m/s at an angle of 27.0o above the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass, with one fragment falling vertically and the other fragment having an unknown horizontal speed.

The relevant equations that can be applied to this problem are the equations of motion for projectiles, which describe the motion of an object in a gravitational field. These equations are:

1. v = u + at (where v is final velocity, u is initial velocity, a is acceleration, and t is time)
2. s = ut + 1/2 at^2 (where s is displacement)
3. v^2 = u^2 + 2as
4. v = u + gt (where g is acceleration due to gravity)

From the given information, we can determine the vertical velocity of the shell at the top of the trajectory using the equation v = u + gt. Plugging in the values, we get:

v = 0 + 9.8 * t
0 = 9.8 * t
t = 0

This means that at the top of the trajectory, the shell has a vertical velocity of 0 m/s.

Next, we can use the equation v^2 = u^2 + 2as to determine the height of the shell at the top of the trajectory. Plugging in the values, we get:

0 = 160^2 + 2 * (-9.8) * s
0 = 25600 - 19.6s
19.6s = 25600
s = 1306.12 m

Now, we can use the equation KE = 1/2mv^2 to determine the kinetic energy of the shell at the top of the trajectory. Plugging in the values, we get:

KE = 1/2 * 14.0 * 160^2
KE = 179200 J

Since the shell explodes into two fragments of equal mass, we can assume that each fragment has a mass of 7.0 kg. We can also assume that the total kinetic energy of the fragments after the