M2 Collisions question

Question: Two particles P of mass 2m and Q of mass 3m are moving towards each other with speeds 4u and 2u respectively. The direction of motion of Q is reversed by the impact and its speed after impact is u. This particle then hits a smooth vertical wall perpendicular to its direction of motion. The coefficient of restitution between Q and the wall is 2/3. In the subsequent motion, there is a further collision between Q and P. Find the speeds of P and Q after this collision.

Working:

Firstly I examined the initial collision between P and Q in order to find the speed of P after collision. Using the principle of of conservation of linear momentum, I found a resultant speed for P = -u/2 ms-1.

Next I considered the collision between Q and the wall, using the information given to find a speed of rebound for Q = 2u/3 ms-1.

Using these two expressions I then tried to find the subsequent speeds of each particle after the third collision. At this point I got stuck as the question doesn't tell us the coefficient of restitution between P and Q, and as a result I could only find a single equation linking the missing speeds so couldn't solve simultaneously. Is it that I'm not fully grasping the question or is it that a value for the coefficient of restitution between P and Q is needed? As always I'd really appreciate any help given.

(N.B. the answers are given only in terms of u)

Reply 1

jacksdrobinson

5

Hey Ndifuna :smile:

You're quite right, a value for e between P and Q is needed, but we can deduce this from the first collision.

Speed of approach between P and Q: 4u + 2u = 6u
Speed of separation: u + 1/2 u = 3/2 u

The speed of separation is from your calculated value of the resultant speed of P, and the value you are told for the speed of Q, after the first collision - plus the fact that you have worked out that both directions of motion are reversed in that collision.

Hence

e_p,q = v_sep / v_app = (3/2 u) / (6u) = 3/12.

Now, you seem able to finish the question yourself :smile:

Hope this helps,
Jack

(edited 10 years ago)

Reply 2

ghostwalker

17

Original post by jacksdrobinson

= 3/12.

=1/4

+rep.

Reply 3

jacksdrobinson

5

Oh dear.

*hides head in hands*

I do that kind of thing in class all the time too :colondollar:

Thanks for the rep :biggrin:

Reply 4

Ndifuna Ukwazi

OP

Original post by jacksdrobinson

Hey Ndifuna :smile:

You're quite right, a value for e between P and Q is needed, but we can deduce this from the first collision.

Speed of approach between P and Q: 4u + 2u = 6u
Speed of separation: u + 1/2 u = 3/2 u

The speed of separation is from your calculated value of the resultant speed of P, and the value you are told for the speed of Q, after the first collision - plus the fact that you have worked out that both directions of motion are reversed in that collision.

Hence

e_p,q = v_sep / v_app = (3/2 u) / (6u) = 3/12.

Now, you seem able to finish the question yourself :smile:

Hope this helps,
Jack

Ah thanks I get it now!

If you have the time could you take a look at this as well:

Question: Two small spheres P and Q of equal radius have masses m and 3m respectively. Sphere P is moving with speed 12u on a smooth horizontal table when it collides directly with Q which is at rest on the table. The coefficient of restitution between P and Q is 2/3. After the collision Q hits a smooth vertical wall perpendicular to the direction of its motion. The coefficient of restitution between Q and the wall is 4/5. Q then collides with P a second time. Find the speeds of P and Q after the second collision between P and Q.

It seems like pretty standard stuff to me but I think I'm making a sign error somewhere towards the end and I can't work out why despite my efforts. Would appreciate it if someone could clear up my confusion.

(edited 10 years ago)

Reply 5

ghostwalker

17

Original post by Ndifuna Ukwazi

Would appreciate it if someone could clear up my confusion.

I got velocities after first collision as -3u, 5u.
After hitting wall is -3u, -4u.
After 2nd collision, -17u/4, -43u/12.

Reply 6

Ndifuna Ukwazi

OP

Original post by ghostwalker

I got velocities after first collision as -3u, 5u.
After hitting wall is -3u, -4u.
After 2nd collision, -17u/4, -43u/12.

Thanks ghostwalker. I get the same values for the first two collisions but a slightly different answer for the third:

(Using the coefficient or restitution):

(q - p)/u = 2/3
3(q - p) = 2u
3q - 3p = 2u
3q = 2u + 3p

(Using conservation of linear momentum):

3mu + 12mu = mp + 3mp
15u = p + 3q
3q = 15u - p

Therefore,

2u + 3p = 15u - p
4p = 13u
p = 13u/4

and

3q = 15u - 13u/4
3q = 47u/4
q = 47u/12

Looking at the line highlighted in bold, it is clear than if I had -2u on the LHS instead of positive 2u I would have got the right answer, but I've been through the question a few times now and get the same equation out every time. Any thoughts?

Reply 7

ghostwalker

17

Original post by Ndifuna Ukwazi

(Using the coefficient or restitution):

(q - p)/u = 2/3
3(q - p) = 2u
3q - 3p = 2u
3q = 2u + 3p

Firstly, were the values I got correct?

Then, I'm suspicious of the line in bold, and think it should be (p - q)/u = 2/3 as p > q.

I note that you've switched around +ve and -ve for this second collision, which is fine.

Reply 8

Ndifuna Ukwazi

OP

Original post by ghostwalker

Firstly, were the values I got correct?

Then, I'm suspicious of the line in bold, and think it should be (p - q)/u = 2/3 as p > q.

I note that you've switched around +ve and -ve for this second collision, which is fine.

They agree with the answers in the back of the book so I think it's fair to say they're right.

I've had a look at my diagram and I think I can see why it should be (p - q) rather than (q - p). If it was (q - p) then I think the spheres would coalesce, which in terms of the information given would just be a presumption. In hindsight I think I must have confused p and q for the speeds of approach rather than separation and therefore worked on the basis that q > p. Really helpful as always, thank you!