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Question: Mass \({m_1}\) hits and sticks with \({m_2}\) while sliding horizontally with velocity \(v\) along t...

Mass m1{m_1} hits and sticks with m2{m_2} while sliding horizontally with velocity vv along the common line of centres of three equal masses (m1=m2=m3=m)\left( {{m_1} = {m_2} = {m_3} = m} \right). Initially masses m2{m_2} and m3{m_3} are stationary and the spring is unstretched. Minimum kinetic energy of m2{m_2} is ymv236\dfrac{{ym{v^2}}}{{36}}. Find yy.

Explanation

Solution

Hint
In this problem, there are two types of energy. First the kinetic energy of the first mass transferred to the second mass. So, the kinetic energy of the two masses gives the potential energy to the spring, so by equating the kinetic energy of the two masses with the potential energy of the spring, the solution can be determined.
The kinetic energy is given by,
KE=12mv2\Rightarrow KE = \dfrac{1}{2}m{v^2}
Where, KEKE is the kinetic energy, mm is the mass of the block and vv is the velocity of the block.
The potential energy of the spring is,
PE=12kx2\Rightarrow PE = \dfrac{1}{2}k{x^2}
Where, PEPE is the potential energy of the spring, kk is the spring constant and xx is the compression or expansion of the spring.

Complete step by step answer
Given that, The mass m1{m_1} hits the mass m2{m_2}, so that the velocity of the mass m1{m_1} is reduced by half.
Three masses are equal (m1=m2=m3=m)\left( {{m_1} = {m_2} = {m_3} = m} \right).
When the mass m1{m_1} hits the mass m2{m_2} and the mass m2{m_2} compresses the spring, then the kinetic energy of the two masses is equal to the potential energy of the spring. Then,
12(2m)(v2)2=12kx2\Rightarrow \dfrac{1}{2}\left( {2m} \right){\left( {\dfrac{v}{2}} \right)^2} = \dfrac{1}{2}k{x^2}
Both the masses are equal so 2m2m and the velocity is reduced by half when it hits the second mass so v2\dfrac{v}{2}.
From the above equation,
12(2m)v24×2k=x2\Rightarrow \dfrac{1}{2}\left( {2m} \right)\dfrac{{{v^2}}}{4} \times \dfrac{2}{k} = {x^2}
By cancelling the terms, then
mv22k=x2\Rightarrow \dfrac{{m{v^2}}}{{2k}} = {x^2}
The above equation is written as,
x=mv22k\Rightarrow x = \sqrt {\dfrac{{m{v^2}}}{{2k}}}
Since, the compression of the spring is 23\dfrac{2}{3} times of maximum of m3{m_3} from the centre of mass, then the kinetic energy of the m3{m_3} gets from the potential energy of the spring, then
Kinetic energy of m3{m_3} is 12k(23x)2 \Rightarrow \dfrac{1}{2}k{\left( {\dfrac{2}{3}x} \right)^2}
Substituting the value of xx in the above equation, then
12k(23×(mv22k))2\Rightarrow \dfrac{1}{2}k{\left( {\dfrac{2}{3} \times \left( {\sqrt {\dfrac{{m{v^2}}}{{2k}}} } \right)} \right)^2}
By squaring the terms, then
12×k×49×mv22k\Rightarrow \dfrac{1}{2} \times k \times \dfrac{4}{9} \times \dfrac{{m{v^2}}}{{2k}}
By cancelling the terms, then
mv29\Rightarrow \dfrac{{m{v^2}}}{9}
Here multiplying and dividing by 44, then
mv29×44\Rightarrow \dfrac{{m{v^2}}}{9} \times \dfrac{4}{4}
On multiplying the above equation, then
4mv236\Rightarrow \dfrac{{4m{v^2}}}{{36}}
By comparing the term given in the question ymv236\dfrac{{ym{v^2}}}{{36}}, then the value of yy is, y=4y = 4.

Note
The energy can neither be created nor destroyed. The kinetic energy of the first mass gives the kinetic energy to the second mass, and the second mass gives the potential energy to the spring. That potential energy of the spring gives again kinetic energy to the third mass. Thus, the energy gets transferred but not created.