# further mechanics collision question.Watch

#1
A smooth sphere S of mass m is moving
with speed (u) on a smooth horizontal
plane. The sphere S collides with another
smooth sphere T, of equal radius to S
but of mass km, moving in the same
straight line and in the same direction
with speed (Au), O < A < 0.5 .The coefficient
of restitution between S and Tis e.
Given that S is brought to rest by the impact,

a) Show that e= (1+kA)/k(1-A)
b)Deduce that k>1 .

Please help with part b. i have gotten close to it but i m not sure many thanks. p.s. instead of A its suppose to be lamda but thats effort.
0
3 weeks ago
#2
(Original post by Mr. sypherPB)
A smooth sphere S of mass m is moving
with speed (u) on a smooth horizontal
plane. The sphere S collides with another
smooth sphere T, of equal radius to S
but of mass km, moving in the same
straight line and in the same direction
with speed (Au), O < A < 0.5 .The coefficient
of restitution between S and Tis e.
Given that S is brought to rest by the impact,

a) Show that e= (1+kA)/k(1-A)
b)Deduce that k>1 .

Please help with part b. i have gotten close to it but i m not sure many thanks. p.s. instead of A its suppose to be lamda but thats effort.
Use the condition and go from there. Rearrange the inequality for and reach the conclusion.
0
#3
(Original post by RDKGames)
Use the condition and go from there. Rearrange the inequality for and reach the conclusion.
i did this but i don't know wht the conclusion is or where i m suppose to deduce the fact that k>1. obv k is greater than 1 as mass cant be 0.
0
3 weeks ago
#4
(Original post by Mr. sypherPB)
i did this but i don't know wht the conclusion is or where i m suppose to deduce the fact that k>1. obv k is greater than 1 as mass cant be 0.
What do you get after rearranging for k?
0
#5
e < 1
1+kA < k-kA
1/(1-2A) < k

< is the less than or equal to sign
0
3 weeks ago
#6
(Original post by Mr. sypherPB)
obv k is greater than 1 as mass cant be 0.
Tbh I’m not sure where you are pulling this sentence from. It doesn’t make sense.

(Original post by Mr. sypherPB)
e < 1
1+kA < k-kA
1/(1-2A) < k
Looks good.

What is the lower bound of 1/(1-2A) ??

You are given that 0 < A < 0.5 so use it.

< is the less than or equal to sign
No, the inequality is strict.

You can’t have e=1 since this collision is not perfectly elastic.
1
#7
0 < (A)< 0.5
0<(1-2A)<1
now how do i deduce that k>1. from here i got to this bit.
btw i really appreciate your help it means a lot. Not a lot of further maths forum that i have seen in student room so really appreciate your help.
1
3 weeks ago
#8
(Original post by Mr. sypherPB)
0 < (A)< 0.5
0<(1-2A)<1
now how do i deduce that k>1. from here i got to this bit.
btw i really appreciate your help it means a lot. Not a lot of further maths forum that i have seen in student room so really appreciate your help.
That result is two steps away.

I asked for the lower bound of 1/(1-2A) so can you figure it from what you have at the moment??
0
#9
(Original post by RDKGames)That result is two steps away.

I asked for the lower bound of 1/(1-2A) so can you figure it from what you have at the moment??

I dont know what you mean by lower bound?
0
3 weeks ago
#10
(Original post by Mr. sypherPB)
I asked for the lower bound of 1/(1-2A) so can you figure it from what you have at the moment??

I dont know what you mean by lower bound?
Really...? The lower bound is a value which bounds the function from below. Can you figure out for what value the function is always greater??

If not, then perhaps I can rephrase my question to: "Find the range of when the domain is " which hopefully you know how to do from your Pure aspect of A-Level Maths.
0
#11
(Original post by RDKGames)
Really...? The lower bound is a value which bounds the function from below. Can you figure out for what value the function is always greater??

If not, then perhaps I can rephrase my question to: "Find the range of when the domain is " which hopefully you know how to do from your Pure aspect of A-Level Maths.
umm so would it not be f(a) greater or equal to 1
Last edited by Mr. sypherPB; 3 weeks ago
0
3 weeks ago
#12
(Original post by Mr. sypherPB)
umm so would it not be f(A)> 1
Exactly. So 1 is what you'd call the lower bound. (well it's not precisely that, but I'm not gonna go into it)

The point of all that is this:

and now we know that

So the conclusion immediately follows.
1
#13
(Original post by RDKGames)
Exactly. So 1 is what you'd call the lower bound. (well it's not precisely that, but I'm not gonna go into it)

The point of all that is this:

and now we know that

So the conclusion immediately follows.
oh wow! never thought to even bring in some of normal a level maths in this. You are very smart. thank you so much. the answers at the back jumped straight to the conclusion.
0
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