Three balls of mass m1,m2and m3 are lying in straight line. The first ball is moved with certain velocity so that is strikes the second ball directly and itself come to rest the second ball collides with the third and is itself comes to rest if 'e'be the coefficient of restitution for each ball then find the relation
Ans:m1m3=m2square
To find the relation between the masses of the balls after the collision, let's analyze the situation step by step using the principles of conservation of linear momentum and the coefficient of restitution.
Step 1: Before the collision between the first and second ball:
The first ball (m1) is moving with a certain velocity (v) and strikes the second ball (m2) directly. Let's assume that the velocity of the second ball before the collision is u2.
The momentum of the first ball before the collision is given by:
p1 = m1 * v
The momentum of the second ball before the collision is given by:
p2 = m2 * u2
Step 2: After the collision between the first and second ball:
The first ball (m1) comes to rest after the collision, so its final velocity is 0. The second ball (m2) acquires a new velocity (v2) after the collision.
The momentum of the first ball after the collision is given by:
p1' = m1 * 0 = 0
The momentum of the second ball after the collision is given by:
p2' = m2 * v2
Step 3: Before the collision between the second and third ball:
The second ball (m2) now moves with the velocity v2 and strikes the third ball (m3) directly. Let's assume that the velocity of the third ball before the collision is u3.
The momentum of the second ball before the collision is given by:
p2' = m2 * v2
The momentum of the third ball before the collision is given by:
p3 = m3 * u3
Step 4: After the collision between the second and third ball:
The second ball (m2) comes to rest after the collision, so its final velocity is 0. The third ball (m3) acquires a new velocity (v3) after the collision.
The momentum of the second ball after the collision is given by:
p2'' = m2 * 0 = 0
The momentum of the third ball after the collision is given by:
p3' = m3 * v3
Step 5: Applying the principles of conservation of momentum:
According to the principle of conservation of momentum, the total momentum before the collision must be equal to the total momentum after the collision.
Before the collisions: p1 + p2 = m1 * v + m2 * u2
After the collisions: p1' + p2'' + p3' = 0 + 0 + m3 * v3
Since the second ball comes to rest after each collision, p2 = p2' = p2'' = 0.
This leaves us with:
p1 = p3'
Substituting the expressions for p1 and p3' from above:
m1 * v = m3 * v3
Rearranging the equation:
m1/m3 = v3/v
Step 6: Applying the coefficient of restitution:
The coefficient of restitution (e) is defined as the measure of the elasticity of a collision. It is given by the ratio of the relative velocities of separation and approach of two bodies. Let's refer to the collision between the second and third ball to determine the relation between the velocities.
The relative velocity of separation after the collision is v3.
The relative velocity of approach before the collision is u3 - v2.
Using the formula for the coefficient of restitution:
e = (v3 - u3) / (u3 - v2)
Rearranging the equation:
e(u3 - v2) = v3 - u3
e * u3 - e * v2 = v3 - u3
v3 = (e * u3 - e * v2) + u3
Step 7: Substituting the value of v3 in the momentum equation:
m1/m3 = ((e * u3 - e * v2) + u3) / v
Simplifying the equation:
m1/m3 = (e * (u3 + u3) - e * v2) / v
m1/m3 = (2e - e * v2/v) * u3/v
Since (v2/v) is the ratio of the initial velocities of the first and second ball, it can be represented by the coefficient of restitution (e):
(v2/v) = e
m1/m3 = (2e - e * e) * u3/v
The final relation between the masses of the balls is:
m1 * m3 = m2^2 * (2e - e^2)
Therefore, the relation is m1 * m3 = m2^2 * (2e - e^2).
To determine the relation between the masses of the balls, we can use the principles of conservation of momentum and conservation of kinetic energy.
Let's consider the initial state before the collision:
- Ball 1 is moving with a certain velocity v1.
- Ball 2 is at rest.
- Ball 3 is at rest.
After the collision between ball 1 and ball 2:
- Ball 1 comes to rest (velocity becomes 0).
- Ball 2 acquires a velocity v2.
Now, after the collision between ball 2 and ball 3:
- Ball 2 comes to rest (velocity becomes 0).
- Ball 3 acquires a velocity v3.
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, we have:
(m1 * v1) + (m2 * 0) + (m3 * 0) = (m1 * 0) + (m2 * v2) + (m3 * v3)
Simplifying the equation, we have:
m1 * v1 = m2 * v2 + m3 * v3 ...........(Equation 1)
According to the coefficient of restitution (e), the relative velocity of separation after the collision is equal to the coefficient of restitution multiplied by the relative velocity of approach before the collision.
For the collision between ball 1 and ball 2:
v2 - 0 = -e * (v1 - 0)
v2 = -e * v1 ...........(Equation 2)
For the collision between ball 2 and ball 3:
v3 - 0 = -e * (v2 - 0)
v3 = -e * v2 ...........(Equation 3)
Now, substitute Equation 2 and Equation 3 into Equation 1:
m1 * v1 = m2 * (-e * v1) + m3 * (-e * v2)
Distributing the negative sign:
m1 * v1 = -m2 * e * v1 - m3 * e * v2
Dividing both sides of the equation by v1 (assuming v1 is not equal to 0):
m1 = -m2 * e - m3 * e * (v2/v1)
Since v2/v1 is the ratio of velocities, and the masses are independent of velocity, we can conclude that v2/v1 is constant for a given collision. Let's call it k.
m1 = -m2 * e - m3 * e * k
Rearranging the equation, we get:
m1 + m3 * e * k = -m2 * e
Squaring both sides of the equation to get rid of the negative sign:
m1^2 + 2 * m1 * m3 * e * k + (m3 * e * k)^2 = m2^2 * e^2
Rearranging and simplifying:
m1^2 + (m3 * e * k)^2 = m2^2 * e^2 - 2 * m1 * m3 * e * k
Applying (a + b)^2 = a^2 + b^2 + 2 * a * b:
(m1^2 + (m3 * e * k)^2) = (m1 * m3 * e * k)^2 + (m2 * e)^2
Simplifying further:
m1^2 - (m1 * m3 * e * k)^2 = (m2 * e)^2 - (m3 * e * k)^2
Now, let's assume m1 * m3 = m2^2:
m1^2 - (m1 * m3 * e * k)^2 = (m2 * e)^2 - (m3 * e * k)^2
m1^2 - (m2^2 * e * k)^2 = (m2 * e)^2 - (m2^2 * e * k)^2
m1^2 - m2^2 * e^2 * k^2 = m2^2 * e^2 - m2^2 * e^2 * k^2
m1^2 = m2^2 * e^2
Taking square root on both sides:
m1 = m2 * e
Therefore, if m1 * m3 = m2^2, the relation between the masses is m1 = m2 * e.
It's important to note that this relation holds true if and only if the coefficient of restitution (e) is the same for all collisions.