If #e=(1+klamda)/(k(1-lambda))#, #01#?

Question

I thought about putting this in Algebra since this is what it boils down to, but since its a collision problem and #e# is the coefficient of restitution, I thought it belongs in Physics more.

The original problem was:

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 #lamdau#, #0< lamda<1/2# . The coefficient of restitution between S and T is #e#.

Given that S is brought to rest by the impact, show that #e=(1+klamda)/(k(1-lambda))#. [I did this bit fine]

Deduce that #k>1#

Thank you!

Benjamin Asiedu · Accepted Answer

See below

Regarding this collision: # e lt 1 # #implies (1 + k lambda)/(k(1 - lambda)) lt 1# #k > 0 # and #lambda < 1/2 implies 1 - lambda > 0 #, ie the denominator is always positive, so we can make the next step whilst guaranteeing the same inequality sign: #1 + k lambda lt k(1 - lambda) # #implies 1 lt k(1 - 2 lambda) qquad square # #lambda < 1/2 implies 1 - 2 lambda > 0 #, so we can divide and guarantee the same inequality sign in #square#: #implies k > 1/(1 - 2 lambda)# Because #0 < lambda < 1/2# then: #0 <2 lambda < 1# #0 > - 2 lambda > -1# #1 >1 - 2 lambda > 0# #color(blue)(0 < 1 - 2 lambda < 1)# So: #k = (1)/("a number between 0 and 1") implies k > 1#

Adam Arnold · Answer

undefined Given the expression for the coefficient of restitution $e = \frac{1 + \lambda}{k(1 - \lambda)}$ and the condition $0 < \lambda < \frac{1}{2}$, deducing that $k > 1$ involves analyzing the denominator. For $e$ to be valid, the denominator $k(1 - \lambda)$ must be nonzero, leading to the conclusion that $k > 1$.

If #e=(1+klamda)/(k(1-lambda))#, #0&lt;lamda&lt;1/2# and e is the coefficient of restitution, deduce that #k&gt;1#?

If #e=(1+klamda)/(k(1-lambda))#, #0<lamda<1/2# and e is the coefficient of restitution, deduce that #k>1#?