1. ## Inequality

Q:
Solve $\frac{2}{5} (2-3e)u < 0$ to find a range for $e$.

My Problem:
Is it $e> \frac{2}{3}$ or $e < \frac{2}{3}$. I'm getting the answer as $e < \frac{2}{3}$ but the mark scheme says that it is $e> \frac{2}{3}$. Can someone confirm the correct answer? Thanks in advance.

2. Originally Posted by Air
Q:
Solve $\frac{2}{5} (2-3e)u < 0$ to find a range for $e$.

My Problem:
Is it $e> \frac{2}{3}$ or $e < \frac{2}{3}$. I'm getting the answer as $e < \frac{2}{3}$ but the mark scheme says that it is $e> \frac{2}{3}$. Can someone confirm the correct answer? Thanks in advance.
Hey Air,

What is u? If the answer says so, then mostly u < 0

3. Originally Posted by Isomorphism
Hey Air,

What is u? If the answer says so, then mostly u < 0
$u$ doesn't have a value so it can be divided by both sides to get rid of it. The question is asking to find the range coefficient of restitution ( $e$).

[It's a follow-on part of my method]

4. Originally Posted by Air
$u$ doesn't have a value so it can be divided by both sides to get rid of it. The question is asking to find the range coefficient of restitution ( $e$).

[It's a follow-on part of my method]
Oh that means the velocity u was negative... If u is negative then while dividing both sides by it, the inequality toggles.Or see it this way:

Product of two numbers is positive if both are positive or both are negative. Mostly your velocity u is negative and thus 2 - 3e must also be negative.

5. Originally Posted by Isomorphism
Oh that means the velocity u was negative... If u is negative then while dividing both sides by it, the inequality toggles.Or see it this way:

Product of two numbers is positive if both are positive or both are negative. Mostly your velocity u is negative and thus 2 - 3e must also be negative.
So, which is the correct inequality: $e < \frac{2}{3}$ or $e > \frac{2}{3}$?

6. Originally Posted by Air
So, which is the correct inequality: $e < \frac{2}{3}$ or $e > \frac{2}{3}$?
Hmm again could you just post the problem entirely.... Or you could confirm the sign of u. Then I will be able to tell.

7. Originally Posted by Isomorphism
Hmm again could you just post the problem entirely.... Or you could confirm the sign of u. Then I will be able to tell.

Q:

A particle $P$ of mass $3m$ is moving with speed $2u$ in a straight line on a smooth horizontal table. The particle $P$ collides with a particle $Q$ of mass $2m$ moving with speed $u$ in the opposite direction to $P$.

a) Show that the speed of Q after the collisions is $\frac15 u (9e +4)$.

As a result of the collision, the direction of motion of $P$ is reversed.

(b) Find the range of possible values of $e$.

[It's part b that we are discussing, which a follow on question]

8. Originally Posted by Air

Q:

A particle $P$ of mass $3m$ is moving with speed $2u$ in a straight line on a smooth horizontal table. The particle $P$ collides with a particle $Q$ of mass $2m$ moving with speed $u$ in the opposite direction to $P$.

a) Show that the speed of Q after the collisions is $\frac15 u (9e +4)$.

As a result of the collision, the direction of motion of $P$ is reversed.

(b) Find the range of possible values of $e$.

[It's part b that we are discussing, which a follow on question]
Ya so I get $v_{P} = \frac{4u - 6ue}5 < 0$ with u positive. This means $4 - 6e < 0 \Rightarrow 2 < 3e \Rightarrow e > \frac23$

9. Thanks Isomorphism!

Also, there is another follow-on question, which is:

c) Given that the magnitude of the impulse of $P$ on $Q$ is $\frac{32}{5} mu$, find the value of $e$.

I haven't seen Impulse element ( $I = mv - mu$) in collision questions before. How would I tackle this?

10. Originally Posted by Air
Thanks Isomorphism!

Also, there is another follow-on question, which is:

c) Given that the magnitude of the impulse of $P$ on $Q$ is $\frac{32}{5} mu$, find the value of $e$.

I haven't seen Impulse element ( $I = mv - mu$) in collision questions before. How would I tackle this?
Look at the change in velocity(while keeping sign in mind) of Q and hence momentum.this will give you I.

Here the change in momentum is $mu + \frac15 mu (9e +4) = I = \frac{32}{5} mu$. Now solve for e.