# Thread: Newton's Law Of Restitution

1. ## Newton's Law Of Restitution

A smooth sphere P of mass $\displaystyle 2m$ is moving in a straight line with speed $\displaystyle u$ on a smooth horizontal table. Another smooth sphere Q of mass $\displaystyle m$ is at rest on the table. The sphere P collides directly with Q. The coefficient of restitution between P and Q is $\displaystyle \frac{1}{3}$. The spheres are modelled as particles.

(a) Show that, immediately after the collision, the speeds of P and Q are $\displaystyle \frac{5}{9}u$ and $\displaystyle \frac{8}{9}u$ respectively.

After the collision, Q strikes a fixed vertical wall which is perpendicular to the direction of motion of P and Q. The coefficient of restitution between Q and the wall is e. When P and Q collide again, P is brought to rest.

(b) Find the value of e.

(c) Explain why there must be a third collision between P and Q.

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I've done part (a) by forming 2 simultaneous equations of conservation of momentum and Newton's law of restitution and solving to obtain the values required. For part (c), I know that collusion must occur as particle P is rebounded so it will hit the stationary particle Q.

The only aspect I don't understand is part (b). How would I do this? Thanks in advance.

2. Originally Posted by Air
A smooth sphere P of mass $\displaystyle 2m$ is moving in a straight line with speed $\displaystyle u$ on a smooth horizontal table. Another smooth sphere Q of mass $\displaystyle m$ is at rest on the table. The sphere P collides directly with Q. The coefficient of restitution between P and Q is $\displaystyle \frac{1}{3}$. The spheres are modelled as particles.

(a) Show that, immediately after the collision, the speeds of P and Q are $\displaystyle \frac{5}{9}u$ and $\displaystyle \frac{8}{9}u$ respectively.

After the collision, Q strikes a fixed vertical wall which is perpendicular to the direction of motion of P and Q. The coefficient of restitution between Q and the wall is e. When P and Q collide again, P is brought to rest.

(b) Find the value of e.

(c) Explain why there must be a third collision between P and Q.

____________________
____________________

I've done part (a) by forming 2 simultaneous equations of conservation of momentum and Newton's law of restitution and solving to obtain the values required. For part (c), I know that collusion must occur as particle P is rebounded so it will hit the stationary particle Q.

The only aspect I don't understand is part (b). How would I do this? Thanks in advance.
We have the second impact; Q against the wall. So calling the speed of Q before it hits the wall $\displaystyle p_Q$ we have
$\displaystyle e = \frac{v_Q^{\prime}}{v_Q}$
where $\displaystyle v_Q^{\prime}$ is the magnitude of the velocity of Q after the collision with the wall. (Obviously this velocity is now opposite to the velocity of P.)

Thus
$\displaystyle v_Q^{\prime} = ev_Q = \frac{8}{9}eu$

Now we have the second collision with P. Set this up the same way as before: with a conservation of momentum equation and the coefficient of restitution is again 1/3. Combining these two equations I get
$\displaystyle 2mv_P -mv_Q^{\prime} = \frac{1}{27}(5 + 8e)mu$
which you can solve for e.

-Dan

3. I redid this question and I am slightly confused. This picture that I am seeing is this:

So, my working is:

$\displaystyle e = \frac{w}{\frac{8u}{9}} \implies \frac{8}{9}ue = w$

$\displaystyle \text{COM:} \ \left( \frac{5}{9}u \right) \left( 2m \right) + \left( \frac{8}{9}u \right) \left( m \right) = - \left( \frac{8}{9}ue \right) \left( m \right)$

$\displaystyle \frac{10}{9} + \frac{8}{9} = -\frac{8}{9}e$

$\displaystyle e = - \frac{9}{4}$.

The answer is wrong. What have I done wrong? topsquark, I don't understand why it is negative when before it is travelling in the positive direction. The correct answer is $\displaystyle e = \frac{25}{32}$.

4. Originally Posted by Air
I redid this question and I am slightly confused. This picture that I am seeing is this:

So, my working is:

$\displaystyle e = \frac{w}{\frac{8u}{9}} \implies \frac{8}{9}ue = w$

$\displaystyle \text{COM:} \ \left( \frac{5}{9}u \right) \left( 2m \right) + \left( \frac{8}{9}u \right) \left( m \right) = - \left( \frac{8}{9}ue \right) \left( m \right)$

$\displaystyle \frac{10}{9} + \frac{8}{9} = -\frac{8}{9}e$
The positive direction in this problem has been defined to be the initial direction of P: to the right. So Q is moving to the right before it hits the wall: u is positive. When it rebounds from the wall Q is moving to the left: w is negative.

-Dan

5. Originally Posted by topsquark
The positive direction in this problem has been defined to be the initial direction of P: to the right. So Q is moving to the right before it hits the wall: u is positive. When it rebounds from the wall Q is moving to the left: w is negative.

-Dan
But when I did that with conservation of momentum, it doesn't appear to work:

$\displaystyle \text{COM:} \ \left( \frac{5}{9}u \right) \left( 2m \right) + \left( \frac{8}{9}u \right) \left( m \right) = - \left( \frac{8}{9}ue \right) \left( m \right)$

^ See, I put negative $\displaystyle \left( \frac{8}{9}ue \right)$.

6. Originally Posted by Air
After the collision, Q strikes a fixed vertical wall which is perpendicular to the direction of motion of P and Q. The coefficient of restitution between Q and the wall is e. When P and Q collide again, P is brought to rest. (b) Find the value of e.
to find e we use
e = w/(8u/9)

first we should find w
therefore

2m (5u/9)+m(-w) = 2m(0)+m(v) (1)

where w is rebound velocity of Q after it hits the wall and v is its velocity after the second collision with P

and another equation

e = 1/3 (between P and Q)

therefore
1/3 = (v-0)/ ((5u/9)+w) (2)

from eq. 1
v = (10u/9) - w (3)

substituting the value of v from eq. 3 into eq. 2 and solving simultaneously...

this gives us
w = 25u/36
and
v = 5u/12

e = (25u/36)/(8u/9)

which gives e = 25/32