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**(KE=1/2*m*V^2)**. I assumed that the collision is 100% elastic. I also assumed that the total Kinetic Energy created by the acceleration of the two masses by gravity was transferred to both masses equally to create the following formula:

**1/2*(M1+M2)*Vt^2=(1/2*M1*VM1^2)+(1/2*M2*VM2^2)**

Where,

M1= Mass one.

M2= Mass two.

Vt= Velocity of both masses at the instant before impact with surface of earth.

VM1= Velocity of M1 at the instant it breaks contact with M2.

VM2= Velocity of M2 at the instant it breaks contact with M1.

This formula is correct but it complicates the process by needing the velocity of the more massive object in order to find the velocity of the less massive object. After discovering this i decided to go about this with a new thought process. I assumed, as i did above, that the Kinetic Energy of both masses at the instant before impact was equal to the Kinetic energy transferred to the larger mass AND the lesser mass. Therefore:

**KEM1=KEM2=KEt**

Where:

KEM1= The kinetic energy transferred to mass one.

KEM2= The kinetic energy transferred to mass two.

KEt= The total kinetic energy created by the displacement of the two masses in earths gravitational field.

From this i determined that:

KEt=1/2*(M1+M2)*Vt^2

Because of the above statement i will set this equal to the kinetic energy of the lesser mass:

**1/2*(M1+M2)*Vt^2=1/2*M1*VM1^2**

In order to find Vt i used a simplification of:

Vf^2=Vi^2+2*acceleration*change in distance

Which is:

**Vf=square root(2*acceleration*change in distance)**

**Vf=Vt**

Therefore:

1/2*(M1+M2)*(2*acceleration*change in distance)=1/2*M1*VM1^2

I then solved for VM1:

**VM1=square root(((M1+M2)*(19.6*x))/M1)**

Where:

2*acceleration=19.6 meters/second^2

x= change in the distance in the direction of the acceleration

Is this correct? I have never taken calculus so the math that is used may be beyond my current comprehension.