The Motion of the Center of Mass

[Penguins66]

Homework Statement

A projectile is launched with speed v0 at an angle theta with respect to the horizontal. At the peak of its motion, it explodes into two pieces of equal mass, which continue to move in the original plane of motion. One piece strikes the ground a horizontal distance D further from the launch point than the point directly below the explosion at time (t < v0 sin(theta) / g) after the explosion. How high does the other piece go? Where does the other piece land? Answer in terms of v0, theta, D, and t.

Homework Equations

Conservation of momentum
$v = v_0 + a t$
$x = x_0 + v_0 t + (1/2) a t^2$
$v^2 = v_0^2 + 2 a \Delta x$
Others?

The Attempt at a Solution

I started out by drawing a picture of the particles. I assume since the first particle hits the ground at less than half of the time of flight, it must have been directed downward. I set up conservation of momentum formulae for both the x and y directions:
$2 v_0 cos\theta = v_1 cos\theta_1 + v_2 cos\theta_2$
$0 = v_1 sin\theta_1 + v_2 sin\theta_2$
Also:
$v_1 cos\theta_1 = \frac{D}{t}$
I'm not sure where to go from here. All attempts seem to lead to dead ends. Any help is appreciated. Thanks!

 

[PaintballerCA]

What do you know about the velocity in the x direction and in the y direction? If you have velocity and distance covered by the object, what else do you know about that object?

You know how fast the object is going in the x direction and how far it goes, so you can solve for time correct? Now, why was the ball only allowed to go as far as it did?

 

[Moetasim]

As a scientist, it is important to approach problems like these in a systematic and logical manner. Let's break down the problem and analyze each component separately.

First, we know that the projectile is launched with an initial velocity v0 at an angle theta with respect to the horizontal. This means that the initial velocity can be broken down into its x and y components as follows:

v0x = v0 cos(theta)
v0y = v0 sin(theta)

Next, we know that the projectile explodes at the peak of its motion, which means that the y component of its velocity is zero at this point. Therefore, the y components of the two pieces after the explosion will be equal and opposite, and can be represented as:

v1y = -v2y

Now, let's consider the x components of the velocities after the explosion. We know that the first piece will travel a horizontal distance D further than the point directly below the explosion. This means that its x component of velocity, v1x, will be equal to:

v1x = D/t

We can also use the conservation of momentum equation in the x direction to relate the initial x velocity, v0x, to the x velocities of the two pieces after the explosion:

2v0x = v1x + v2x

Substituting in the values we know for v0x and v1x, we get:

2v0 cos(theta) = D/t + v2x

Now, we can use this equation and the initial y velocity, v0y, to find the y velocities of the two pieces after the explosion. Using the conservation of momentum equation in the y direction, we get:

0 = v1y + v2y

Substituting in the value for v1y and the relationship between v1y and v2y, we get:

0 = -v2y + v2y

This means that the y component of the velocity for both pieces after the explosion is zero, which makes sense since we know that the y component of the velocity at the peak of the motion is also zero.

Now, we can use the equations for the motion of the center of mass to find the height of the second piece at any given time t after the explosion. The y position of the center of mass can be represented as:

y = (1/2)(v1y + v2y)t

Since we know that v2y