# Newton's Law Of Restitution

• Apr 8th 2008, 03:31 AM
Simplicity
Newton's Law Of Restitution
I've done part (a) but cannot understand the method of how part (b) is done. I've written down the answer I obtained (which is correct) to part (a). Can someone help? Thanks in advance.

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Q:

A small smooth ball $\displaystyle A$ of mass $\displaystyle m$ is moving on a horizontal table with speed $\displaystyle u$ when it collides directly with another small smooth ball $\displaystyle B$ of mass $\displaystyle 3m$ which is at rest on the table. The balls have the same radius and the coefficient of restitution between the balls is $\displaystyle e$. The direction of motion of $\displaystyle A$ is reversed as a result of the collision.

(a) Find, in terms of $\displaystyle e$ and $\displaystyle u$. the speeds of $\displaystyle A$ and $\displaystyle B$ immediately after the collision.

In the subsequent motion $\displaystyle B$ strikes a vertical wall, which is perpendicular to the direction of motion of $\displaystyle B$, and rebounds. The coefficient of restitution between $\displaystyle B$ and the wall is $\displaystyle \frac{3}{4}$.

Given that there is a second collision between $\displaystyle A$ and $\displaystyle B$,

(b) find the range of values of $\displaystyle e$ for which the motion described is possible.

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A:

(a)
$\displaystyle v_1 = \frac{1}{4} u (3e - 1)$
$\displaystyle v_2 = \frac{1}{4}u(e + 1)$
• Apr 8th 2008, 04:02 AM
topsquark
Quote:

Originally Posted by Air
I've done part (a) but cannot understand the method of how part (b) is done. I've written down the answer I obtained (which is correct) to part (a). Can someone help? Thanks in advance.

____________________
____________________

Q:

A small smooth ball $\displaystyle A$ of mass $\displaystyle m$ is moving on a horizontal table with speed $\displaystyle u$ when it collides directly with another small smooth ball $\displaystyle B$ of mass $\displaystyle 3m$ which is at rest on the table. The balls have the same radius and the coefficient of restitution between the balls is $\displaystyle e$. The direction of motion of $\displaystyle A$ is reversed as a result of the collision.

(a) Find, in terms of $\displaystyle e$ and $\displaystyle u$. the speeds of $\displaystyle A$ and $\displaystyle B$ immediately after the collision.

In the subsequent motion $\displaystyle B$ strikes a vertical wall, which is perpendicular to the direction of motion of $\displaystyle B$, and rebounds. The coefficient of restitution between $\displaystyle B$ and the wall is $\displaystyle \frac{3}{4}$.

Given that there is a second collision between $\displaystyle A$ and $\displaystyle B$,

(b) find the range of values of $\displaystyle e$ for which the motion described is possible.

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A:

(a)
$\displaystyle v_1 = \frac{1}{4} u (3e - 1)$
$\displaystyle v_2 = \frac{1}{4}u(e + 1)$

So let the +x direction be the direction of the motion of ball A before the initial collision. Conservation of momentum says:
$\displaystyle mu = -mv_A + 3mv_B$
(I'm taking the first term on the right to be negative since the problem stipulates that A is rebounding.)

We also have
$\displaystyle e = \frac{v_A + v_B}{u}$

You can solve this for, say $\displaystyle v_B$ and get that
$\displaystyle v_B = eu - v_A$

Plug that into your momentum equation and you can find $\displaystyle v_A$, and then use either equation to find $\displaystyle v_B$. Given this method, I agree with your answer to part a).

For part b), object B strikes the wall and rebounds in the negative direction with a speed $\displaystyle v_B^{\prime}$. Since we need ball B to strike ball A again, $\displaystyle v_B^{\prime} > v_A$ is your criterion.

So what's $\displaystyle v_B^{\prime}$? Well the coefficient of restitution for the collision with the wall is 3/4. So
$\displaystyle \frac{3}{4} = \frac{v_B^{\prime}}{v_B}$

$\displaystyle v_B^{\prime} = \frac{3}{4} \cdot v_B = \frac{3}{4} \cdot \frac{1}{4}u(e + 1) = \frac{3}{16} \cdot u(e + 1)$

So solve
$\displaystyle \frac{3}{16} \cdot u(e + 1) > \frac{1}{4} u (3e - 1)$
for e.

-Dan