# Two rocks launched up/collison (Kinematics)

• Moe777
In summary, the question is asking for the velocity of the second rock launched at time T from the same initial position as the first rock, in order for the two rocks to collide with the first rock having a velocity of -V0/2. The solution involves setting up equations using the basic kinematics formulas and assuming conditions for the collision. The times at which the rocks are thrown, t1 and t2, must be related to the time interval between launches, T, and the time of collision.
Moe777

## Homework Statement

This is a quiz question that was given in my AP physics C class. "A rock is launched upward with a speed V0. With what velocity must a second rock be launched from the same initial position at time T later so that it hits the first rock when it has a velocity of -V0/2."

## Homework Equations

The only formulas we needed for this question were the basic velocity, displacement, and acceleration formulas (as functions of time) and their variations. (1D kinematics).

## The Attempt at a Solution

1) I drew a diagram on an axis with a vertical line (y axis) and the origin being zero (the initial position of the balls).

2) I listed all of the knowns and unknowns given from the question. We know that:
• Rock1 (first rock) has an initial velocity of V0.
• Time= T
• Acceleration= -g= -9.81 m/s2
• When the second rock hits the first rock, the velocity of the first rock will be -V0/2.
We don't know the velocity that is necessary for rock2 (second rock), which allows the given condition ( When the second rock hits the first rock, the velocity of the first rock will be -V0/2.) to happen. Basically, we don't know the initial velocity of rock2.

I then assumed the conditions for the collision:
When the rocks collide, they will be the same distance from the ground, so d1=d2
and I also assumed the time of travel for both rocks at the collision, which I put as T=T, which is where I'm sure that I made a mistake.

Since this is from a collected quiz, I don't remember exactly what I did. However, I remember doing something similar to this:

(-V0rock1 /2)T- (1/2)gT2=( V0rock2)T- (1/2)gT2
I forgot the process, but my solution was something like V0/2T, which is incorrect because the units come out to be m/s2. I feel that I did something completely off track, and I really need help as to how I should approach these problems and how to think more effectively. I felt as though this question should have been easier for me, which kind of discouraged me after that quiz was collected. This isn't the first time we get a question like this, and help for the question and questions like these would be much appreciated. Thank you in advance.

Last edited:
You seem to be confused about T. The given T is the interval between launch times. If the first rock is in flight for time t1 before the collision, how long is the second rock in flight until then?

Moe777 said:
Time= T
That is not a good idea if more than one time is involved. Labels like t1 and so on make it easier to see what happens when.

Oh. I see what's going on. So I drew a graph just to help me visualize, where I assumed arbitrary times, where the first rock is thrown at 1s and the second is thrown at 4s. The time interval between them is 3s. This helped me confirm the equation that T=t2- t1, which means that t2=t1+T

Moe777 said:
Oh. I see what's going on. So I drew a graph just to help me visualize, where I assumed arbitrary times, where the first rock is thrown at 1s and the second is thrown at 4s. The time interval between them is 3s. This helped me confirm the equation that T=t2- t1, which means that t2=t1+T
Maybe. It depends how you are defining t1 and t2. Are these times at which the rocks are thrown, or the times they spend in the air?

I defined them as the times that the rocks were thrown.

Ok. I still assumed that d1=d2.

V01t1+(1/2)gt12= V02(T+t1)+(1/2)g(T+t1)2

Now you gave the second object more time of flight than the first one.

Also, your use of t1 doesn't match the definition above, as it seems to be the total flight time of the first object here.

Moe777 said:
I defined them as the times that the rocks were thrown.
I'm still unclear what you mean. The times at which they were thrown or for which they were aloft?
If they are the times at which the rocks are thrown, you might as well choose t1=0, making t2=T. But then you need another variable to represent the time of the collision.
If they are the times the rocks spend in the air, which one should be the greater?

Ok I think it's becoming clearer. I was going off of your first reply that T is the time interval between the launching of the rocks, and may have gotten confused from that. I now understand that T is the time in which they collide. Clearly, Rock1 is in the air longer because it was thrown first. Therefore, t1 is greater than t2. Therefore, the equation, T-t1=t2 must be true.

Moe777 said:
Ok I think it's becoming clearer. I was going off of your first reply that T is the time interval between the launching of the rocks, and may have gotten confused from that. I now understand that T is the time in which they collide.
No, T is the time interval between launches. Reread the question.
What relationship does that give you between the times they spend in the air, t1 and t2?

## 1. How do you calculate the velocity of two rocks launched up?

The velocity of an object is calculated by dividing the distance traveled by the time it takes to cover that distance. In the case of two rocks launched up, the velocity can be calculated using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and t is the time taken. The initial velocity can be determined by measuring the speed at which the rocks are launched, and the acceleration due to gravity is a constant value of 9.8 m/s^2.

## 2. What factors affect the collision of two rocks in flight?

The collision of two rocks in flight can be affected by several factors, such as the initial velocity and angle of launch, the mass and size of the rocks, air resistance, and external forces like wind. These factors can affect the trajectory and speed of the rocks, ultimately determining the point and manner of their collision.

## 3. How does the angle of launch affect the collision of two rocks?

The angle of launch plays a crucial role in determining the collision of two rocks. If the rocks are launched at the same angle, they will collide at the same height and distance from the launch point. However, if the angles of launch are different, the rocks will have different trajectories and may collide at a different point or not collide at all.

## 4. Can the collision of two rocks be predicted using kinematics equations?

Yes, the collision of two rocks can be predicted using kinematics equations. By analyzing the initial conditions, such as the velocity and angle of launch, the mass and size of the rocks, and the effects of external forces, kinematics equations can be used to calculate the time and location of collision. However, these predictions may not be entirely accurate as they do not take into account factors like air resistance and potential changes in the initial conditions.

## 5. What are the different types of collisions that can occur between two rocks?

There are two main types of collisions that can occur between two rocks - elastic and inelastic collisions. In an elastic collision, the total kinetic energy of the system before and after the collision remains the same. In contrast, in an inelastic collision, some kinetic energy is lost, usually in the form of heat or sound. The type of collision between two rocks can be determined by analyzing the conservation of momentum and energy equations.

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