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M2!help...............

plz guys i need ur help... i cant understand the mark schemes!

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

(a) Show that, immediately after the collision, the speeds of P and Q are 5/9u and 8/9u 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 know part a ! but part b and c.. :confused:
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A smooth sphere P of mass m is moving in a straight line with speed u on a smooth horizontal table. Another smooth sphere Q of mass 2m is at rest on the table. The sphere P collides directly with Q. After the collision the direction of motion of P is unchanged. The spheres have the same radii and the coefficient of restitution between P and Q is e. By modelling the spheres as particles,
(a) show that the speed of Q immediately after the collision is 1/3(1 + e)u,

(b) find the range of possible values of e.

Given that e = 1/4,
(c) find the loss of kinetic energy in the collision.

(d) Give one possible form of energy into which the lost kinetic energy has been transformed.

i know part a ! but part b, c and d.. dunno!
Reply 1
You will find the speed of Q after collision with the wall is : 8/9eu and therefore
To have the other collision with P, speed of Q must be bigger the speed of P:
8/9eu> 5/9u if velocity of P is reverse.
Using the conservation of momentum and Newton's law of restitution: You can find the value of e.
The answer for part c) might be , if the direction of P doesn't change, the third collision will happen obviously.
If the direction of P changes, the velocity of Q after colliding with wall will be greater than the velocity of P.

I hope you will find the correct answer.
Reply 2
thanks.. but how can find the value of e?! i didnt understand a thing from the mark scheme..
Reply 3
After collision with the wall, speed of Q is e.8u/9

So P and Q will collide, P has velocity 5u/9, and Q : -e.8u/9
After collision of P and Q again, Q will have velocity V
2mu. 5/9 – eum.8/9 = mV
2u. 5/9 – eu.8/9 = V
and

V/(5u/9 - -8e/9) = 1/3
3V = 5u/9 + 8eu/9

==> 3[2u. 5/9 – eu.8/9] = 5u/9 + 8eu/9
30 – 24e = 5 + 8e
e = 25/32


c) as Q is moving towards the wall, then it will collide, so it will reverse it motion, i.e. go back to P, as P is at rest then they will collide!
Reply 4
after wall collision
Vp = 5u/9
Vq = -8eu/9 (now travelling in the opposite direction)

after 2nd collision

Vp = 0
Vq = V'q, say

restitution balance

e' = 1/3 = (V'q - 0)/(5u/9 + 8eu/9)
V'q = (5 + 8e)u/27
============

Momentum balance
Mp.Vp + Mq.Vq = Mp.0 + Mq.V'q
2m.5u/9 - m.8eu/9 = 0 + m.(5 + 8e)u/27
10u/9 - 8eu/9 = (5 + 8e)u/27
30 - 24e = 5 + 8e
25 = 32e
e = 25/32
=======
ossoss87
thanks.. but how can find the value of e?! i didnt understand a thing from the mark scheme..

do you want me to do the second part for you?
Reply 6
yaaaaaaaaaaaaaaay! finally i got it... thanks guys! :smile: wt about the second question ?!
A smooth sphere P of mass m is moving in a straight line with speed u on a smooth horizontal table. Another smooth sphere Q of mass 2m is at rest on the table. The sphere P collides directly with Q. After the collision the direction of motion of P is unchanged. The spheres have the same radii and the coefficient of restitution between P and Q is e. By modelling the spheres as particles,
(a) show that the speed of Q immediately after the collision is 1/3(1 + e)u,

(b) find the range of possible values of e.

Given that e = 1/4,
(c) find the loss of kinetic energy in the collision.

(d) Give one possible form of energy into which the lost kinetic energy has been transformed.


b)
sphere P
m1 = m
u1 = u
v1 = ?

sphere Q
m2 = 2m
u2 = 0
v2 = 1/3(1+e)u

mu + 0 = mv1 + 2m*1/3(1+e)u
mu = mv1 + 2m/3(1+e)u
Divide through by m as m>0
u = v1 + 2/3(1+e)u
v1 = u - 2u/3(1+e)

v1 > 0 as direction of motion is unchanged. The question basically implies that the particle P carries on moving by saying this.

=> u - 2u/3(1+e) > 0
u > 2u/3 + 2eu/3

Divide by u.

1 > 2/3 + 2e/3
1/3 > 2e/3
1 > 2e
2e < 1
e < 1/2

If you want c and d done i'll be happy to do them.