Question

The centres of spheres 1, 2 and 3 lie on a single straight line. Sphere 1 moving with an (initial) velocity ${{v}_{1}}$directed along this line and hits sphere 2. Sphere 2 acquiring after the collision a velocity${{v}_{2}}$, hit sphere 3. Both collisions are absolutely elastic. What must be the mass of sphere 2 (in kg) for sphere 3 to acquire maximum velocity (The masses and of spheres 1 and 3 are 9 kg and 1 kg respectively)?

Answer 1

The concept to be used to solve this problem is the law of conservation of the momentum and the elastic collision features. The feature of the elastic collision involves the conservation of momentum along with the conservation of kinetic energy.
Formula used:
${{m}_{1}}{{v}_{1}}={{m}_{2}}{{v}_{2}}$

Complete answer:
From the given information, we have the data as follows.
The mass of sphere 1, ${{m}_{1}}=9\,kg$
The mass of sphere 3, ${{m}_{3}}=1\,kg$
The velocity of sphere 1, ${{v}_{1}}$
The velocity of sphere 3, ${{v}_{3}}$
The law of conservation of momentum states that the product of mass and velocity before the collision equals the product of mass and velocity after collision. Thus, the total momentum is conserved. An elastic collision is a collision in which there will be no loss of the kinetic energy upon collision.
Direct method:
Using the law of conservation of momentum and taking into consideration the elastic collision property, the collision between the spheres 1 and 2 and the collision between the spheres 2 and 3 is given as follows.
$\dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{{{m}_{2}}}{{{m}_{3}}}$
Let the mass of sphere 2 be represented as ‘m’.
Substitute the values.

& \dfrac{9}{m}=\dfrac{m}{1} \\\ & \Rightarrow {{m}^{2}}=9 \\\ & \therefore m=3\,kg \\\ \end{aligned}$$ $$\therefore $$The mass of the sphere 2 is 3 kg. Step by step method. According to the law of conservation of momentum, the momentum before collision equals the momentum after collision, so, the collision between the spheres 1 and 2 is given as follows. $${{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}={{m}_{1}}u$$ As the collision is elastic, so, $$\dfrac{{{v}_{2}}-{{v}_{1}}}{u-0}=1$$ $${{v}_{2}}-{{v}_{1}}=u$$ $$\therefore {{v}_{2}}=\dfrac{2{{m}_{1}}u}{{{m}_{1}}+{{m}_{2}}}$$…… (1) Similarly, the collision between the spheres 2 and 3 is given as follows. $${{m}_{2}}{{v}_{2}}+{{m}_{3}}{{v}_{3}}={{m}_{2}}u$$ As the collision is elastic, so, $$\dfrac{{{v}_{3}}-{{v}_{2}}}{u-0}=1$$ $${{v}_{3}}-{{v}_{2}}=u$$ $$\therefore {{v}_{3}}=\dfrac{2{{m}_{2}}u}{{{m}_{3}}+{{m}_{2}}}$$…… (2) Comparing the equations (1) and (2), we get, $$\dfrac{2{{m}_{1}}u}{{{m}_{1}}+{{m}_{2}}}=\dfrac{2{{m}_{2}}u}{{{m}_{3}}+{{m}_{2}}}$$ Substitute the values of masses. $$\begin{aligned} & \dfrac{9}{9+{{m}_{2}}}=\dfrac{{{m}_{2}}}{1+{{m}_{2}}} \\\ & \Rightarrow 9(1+{{m}_{2}})={{m}_{2}}(9+{{m}_{2}}) \\\ & \Rightarrow 9=m_{2}^{2} \\\ & \therefore {{m}_{2}}=3\,kg \\\ \end{aligned}$$ **Note:** Initially all the spheres were at rest, so, we have equated the difference between their velocities and even it was given that both of the collisions were elastic. Being elastic collision refers to the conservation of kinetic energy along with conservation of the momentum.