Question

solid iron ball A of radius $r$ collides head on with another stationary solid iron ball B of radius $2r$ . The ratio of their speeds just after the collision ($e = 0.5$) is
A. $3$
B. $4$
C. $2$
D. $1$

Answer 1

Here we have to apply the formula of mass directly proportional to radius cube and then apply the law of conservation of linear momentum.

Complete step by step solution:
The mass of an object separated by its volume is known as density and solids are involved in most of our interactions with density. For solids, since the molecules are bound to one another, the density of a single element or combination remains comparatively stable.

Law of conservation of linear momentum- According to the law of linear momentum, the overall momentum of the system is often maintained for an object or a system of objects if no external force acts on them. There are no external forces in an isolated system, so momentum is always conserved. Since, momentum is conserved. The elements will also be maintained in any direction. In solving collision problems, the application of the law of momentum conservation is important.

Given,
A solid iron ball A of radius $r$
collides head on with another stationary solid iron ball B of radius $2r$
.
$e = 0.5$

Let us consider that density is constant.
Then, mass $\alpha {r^3}$

So, mass of the first ball,
${m_1} = \dfrac{4} {3}\pi {r^3}\rho = m$

Mass of second ball,
${m_2} = \dfrac{4} {3}\pi 8{r^3}\rho = 8m$

Now applying the law of conservation of linear momentum we get-
$mv = m{v_1} + 8m{v_2}$

$0.5 = \dfrac{{{v_2} - {v_1}}} {v}$ ...... (i)
Also,
${v_1} + 8{v_2} = v \\\ {v_2} - {v_1} = 0.5v \\\$

From equation (i) we get-
$9{v_2} = 1.5v \\\ {v_2} = \dfrac{{1.5v}} {9} \\\$

${v_1} = \dfrac{{1.5v}} {9} - 0.5v \\\ = \dfrac{{ - 3v}} {9} \\\$

$\dfrac{{{v_1}}} {{{v_2}}} = \dfrac{{3v}} {{1.5v}} = 2$

Therefore, the ratio of their speeds just after the collision ($e = 0.5$) is $2$
.
Hence, option C is correct.

Note:
Here we have to be careful while calculating the mass of the first and the second ball. The mass of the first ball is taken to be $m$ and mass of the second ball is taken to be $8m$
.