Newton's Law Of Restitution

**Simplicity** · April 8th 2008, 03:31 AM

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

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

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

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

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

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A:

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

**topsquark** · April 8th 2008, 04:02 AM

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

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

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

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

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

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____________________

A:

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

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

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

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

Plug that into your momentum equation and you can find $v_A$ , and then use either equation to find $v_B$ . 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 $v_B^{\prime}$ . Since we need ball B to strike ball A again, $v_B^{\prime} > v_A$ is your criterion.

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

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

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

-Dan

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