3rd ball says: Ouch It Hurts!

Classical Mechanics Level 4

Three balls are kept on a straight line on a smooth surface. The first ball of mass ${ m }_{ 1 }$ is projected with velocity ${ V }_{ o }$ towards the middle ball of mass ${ m }_{ 2 }$ , which then collides with the third ball of mass ${ m }_{ 3 }$ .

If we want ${ m }_{ 3 }$ to move with the maximum possible velocity, then what should be the mass of ${ m }_{ 2 }$ in kg? All collisions are elastic.

$\textit{Details}$

Use the following data if needed :

$\bullet \quad { m }_{ 1 }\quad =\quad 4\quad kg \\ \bullet \quad { m }_{ 3 }\quad =\quad 9\quad kg \\ \bullet \quad { V }_{ o }\quad =\quad 16\quad m/s$

The answer is 6.

2 solutions

Abhishek Sharma
Dec 3, 2014

Use conservation of momentum and newton's experimental law twice to obtain the velocity of ball 3,

$v=\frac { 4{ m }_{ 1 }{ m }_{ 2 } }{ ({ m }_{ 1 }+{ m }_{ 2 })({ m }_{ 2 }+{ m }_{ 3 }) }$

Now use simple calculus , Velocity of ball 3 is maximum when mass of second ball is 6kg.

One another way to avoid Calculus ! :)

${ V }_{ o }\quad =\quad \cfrac { 16m }{ { m }^{ 2 }\quad +\quad 13m\quad +\quad 36\quad } \quad \\ \\ { V }_{ o }\quad =\quad \cfrac { 16 }{ m\quad +\quad \cfrac { 36 }{ m } \quad +\quad 13 } \\ \\ AM\quad \ge \quad GM\quad \quad (\quad \because \quad m\quad >\quad 0\quad )$

Since at minimum denominator and maximum velocity AM = GM So equality of variable is attained :

$m\quad =\quad \cfrac { 36 }{ m } \quad \Rightarrow \quad m\quad =\quad 6$ .

Deepanshu Gupta - 6 years, 6 months ago

Explain to me...... wouldn't the momentum be conserved in the collisions since the surface is smooth and the collisions are elastic, and thus the total momentum after the first collision would be 64 and after the second also 64 so no matter what the mass of the second ball is the third ball would always get the same velocity ??????????? i am really confused

Vaibhav Prasad - 6 years, 3 months ago

Imagine two masses m1 and m2. There is an elastic collision which leads to the following system: $\begin{cases} { m }_{ 1 }\quad \times \quad { v }_{ 1 }+{ m }_{ 2 }\times { v }_{ 2 }={ m }_{ 1 }\times { v' }_{ 1 }+{ m }_{ 2 }\times { v' }_{ 2 } \\ { m }_{ 1 }{ v }_{ 1 }^{ 2 }+{ m }_{ 2 }{ v }_{ 2 }^{ 2 }={ m }_{ 1 }{ v' }_{ 1 }^{ 2 }+{ m }_{ 2 }{ v' }_{ 2 }^{ 2 } \end{cases}\\ \begin{cases} { v' }_{ 1 }=\frac { m1-m2 }{ m1+m2 } { v }_{ 1 }+\frac { 2m2 }{ m1+m2 } { v }_{ 2 } \\ v'_{ 2 }=\frac { 2m1 }{ m1+m2 } { v }_{ 1 }+\frac { m2-m1 }{ m1+m2 } { v }_{ 2 } \end{cases}\\$ . Both of these systems are equivalent. In this case v'2 will be the speed of m2 after the collision because v2=0 (m2 was at rest ) you see that : $v'_{ 2 }=\frac { 2m1 }{ m1+m2 } { v }_{ 1 }$ . Therefore the speed of m2 after the collision will be related to its mass. This is why we calculate m2.

Tala Al Saleh - 6 years ago

I suppose this is a very silly question, but what are AM and GM?

Tala Al Saleh - 6 years ago

Exactly ! Btw Nice solution! :)

Keshav Tiwari - 6 years, 6 months ago

Rudraksh Sisodia
Nov 25, 2015

as the collison are all elastic hence the kinetic energy should be conserved ,,, and to get maximum velocity of last ball , initial energy is equal to final kinetic energy of third ball , get required velocity form there , and use newtons law of collision and conservation of linear momentum

3rd ball says: Ouch It Hurts!

The answer is 6.

2 solutions

0 pending reports