Finding the value of the coefficient of elasticity for a 1D collision.

[Lemansky]

Hi all first post. Myself and my friends are very much stumped by this question. Any help would be much appreciated!

Homework Statement

A body with mass m collides directly with a body of mass 3m. If two thirds of the original kinetic energy is lost, find the coefficient of elasticity for this collision.

Homework Equations

Conservation of momentum: m1u1+ m2u2=m1v1+m2v2.

Newton's empirical law: v2-v1= -e(u2-u1)

Loss in kinetic energy: (1/2)m1(u1)^2+(1/2)m2(u2)^2=(1/2)m1(v1)^2+(1/2)m2)v2)^2.

The Attempt at a Solution

Using all the info given and giving the sphere of mass m initial velocity u and final velocity v1, and the sphere of mass 3m initial velocity of 0 and a final velocity of v2 we derived three equations, as follows,

1) u=v1+3v2 from the conservation of momentum.

2)u^2=3(v1)^2+9(v2)^2 from the fact that the final kinetic energy equals a third of the initial kinetic energy,

3)(v2-v1)/e =u from rearranging Newton's empirical law.


By squaring out equation 1 and setting it equal to equation 2 we found that v1=3v2.

No matter which way we approached the problem the same equations always fell out and by juggling them around we were able to get values for e with one problem, they were always negative (but had they been positive they would have been realistic). We are probably making some simple mistake with signs or we're cancelling out somehting which can't be canceled but we really are stuck.


Any help much appreciated!

 

[Nate810]

From the equations you derived, you can solve for the coefficient of elasticity. From equation 1, you have v1 = u-3v2, and from equation 3, you have u = (v2-v1)/e. Substituting the expression for v1 into equation 3 and rearranging, you get e = (v2-u+3v2)/u = (4v2-u)/u. Using the expression for u from equation 2, you can substitute it into the equation for e and solve for v2. Then, substitute the expression for v2 into the equation for e to get your answer.

 

[Moetasim]

Hello, thank you for reaching out for help with this problem. I can understand your frustration and confusion. Let's take a closer look at the equations that you have derived and see if we can find the mistake or missing piece of information.

First, let's define our variables for clarity:

m1 = mass of the first body
m2 = mass of the second body
u1 = initial velocity of the first body
u2 = initial velocity of the second body
v1 = final velocity of the first body
v2 = final velocity of the second body
e = coefficient of elasticity

Now, let's examine the equations:

1) Conservation of momentum: m1u1 + m2u2 = m1v1 + m2v2

This equation is correct and represents the conservation of momentum in a 1D collision. However, we need to be careful with the signs of our velocities. In this equation, the positive direction is chosen to be in the direction of the second body's initial velocity (u2). This means that the initial velocity of the first body (u1) should be negative, as it is moving in the opposite direction. So the equation should be: m1(-u1) + m2u2 = m1v1 + m2v2.

2) Newton's empirical law: v2-v1= -e(u2-u1)

This equation is also correct, but we need to be careful with the signs of our velocities again. In this equation, the positive direction is chosen to be in the direction of the first body's initial velocity (u1). This means that the initial velocity of the second body (u2) should be negative, as it is moving in the opposite direction. So the equation should be: v2-(-u2) = -e(u2-(-u1)).

3) Loss in kinetic energy: (1/2)m1(u1)^2 + (1/2)m2(u2)^2 = (1/2)m1(v1)^2 + (1/2)m2(v2)^2

This equation is also correct, but again we need to be careful with the signs. The initial kinetic energy is given as (1/2)m1(u1)^2 + (1/2)m2(u2)^2, which is correct. However, the final kinetic energy should be (