Question

A body mass $M$ at rest is struck by a moving body of mass $m.$ Show that the fraction of the initial kinetic energy of moving mass m transferred to the strucked body is $\dfrac{{4mM}}{{{{(m + M)}^2}}}\,.$

Answer 1

Hint: We are provided with a statement that involves law of conservation of momentum. We have to prove the statement using the relation based on law of conservation of momentum

Complete step by step answer
Momentum: The momentum of an object is defined as the product of mass and velocity of the object. Momentum is a vector quantity. It is represented by $'P'$ and the unit of momentum is $Kg{\text{ }}m{s^{ - 1}}$
When two objects collide with each other the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. This is known as the law of conservation of momentum.
The relation that explain the law of conservation of momentum is:
${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$
Where
${m_1}{u_1} + {m_2}{u_2}$ is the total momentum of the two objects before collision.
${m_1}{v_1} + {m_2}{v_2}$ is the total momentum of the two objects after collision.
${m_1}$ is the mass of the object 1.
${m_2}$ is the mass of the object 2.
${u_1}$ is the velocity of the object 1 before collision.
${u_2}$ is the velocity of the object 2 before collision.
${v_1}$ is the velocity of the object 1 after collision.
${v_2}$ is the velocity of the object 2 after collision.
Given,
Mass of the object 1 which is moving is $m$
Mass of the object 2 in rest is $M$
Object 1 is moving, so the velocity of the object 1 before collision be ${u_1}$
Object 2 is at rest, so the velocity of the object 1 before collision, ${u_2} = 0.$
The velocity of the object 1 after collision is ${v_1}$

When the object 2 which is at rest is struck by the moving object 1 the velocity experienced by the object 2 after collision is the sum of object 2’s velocities before and after collision.
So, the velocity of the object 2 after collision is ${v_2} = {u_1} + {v_1}$
       $ \Rightarrow {v_1} = {v_2} - {u_1}{\text{ }} \to 1$
Applying the conservation of momentum concept, we get
$ \Rightarrow {m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$
Substituting the values we get,
$ \Rightarrow m{u_1} + M(0) = m{v_1} + M{v_2}$
$ \Rightarrow m{u_1} = m{v_1} + M{v_2}$
$ \Rightarrow m{u_1} - m{v_1} = M{v_2}$
$ \Rightarrow m({u_1} - {v_1}) = M{v_2}$
From 1 substitute the value of ${v_1}$
$ \Rightarrow m({u_1} - ({v_2} - {u_1})) = M{v_2}$
$ \Rightarrow m({u_1} - {v_2} + {u_1}) = M{v_2}$
$ \Rightarrow m(2{u_1} - {v_2}) = M{v_2}$
$ \Rightarrow m2{u_1} - m{v_2} = M{v_2}$
$ \Rightarrow m2{u_1} = m{v_2} + M{v_2}$
$ \Rightarrow 2m{u_1} = (m + M){v_2}$
$ \Rightarrow {v_2} = \dfrac{{2m{u_1}}}{{(m + M)}}{\text{ }} \to {\text{2}}$

To Show that the fraction of the initial kinetic energy of moving mass m transferred to the strucked body is $\dfrac{{4mM}}{{{{(m - M)}^2}}}\,.$ we have to find the initial and final kinetic energy.
 Kinetic energy $K.E = \dfrac{1}{2}m{v^2}$
Where,
$m$ is the mass of the object.
$v$ is the velocity of the object
Kinetic energy $K.E = \dfrac{1}{2}m{v^2}$
Mass of the object 1 which is moving is $m$
The velocity of the object 1 before collision be ${u_1}$
The Initial kinetic energy of the moving object is
$ \Rightarrow K.E = \dfrac{1}{2}m{u_1}^2{\text{ }} \to {\text{3}}$
The final kinetic energy of the object at rest is
Mass of the object 2 in rest is $M$
The velocity of the object 2 after collision is
$ \Rightarrow K.E = \dfrac{1}{2}M{v_2}^2$
From 2 substitute the value of ${v_2}$
$ \Rightarrow K.E = \dfrac{1}{2}M{\left( {\dfrac{{2m{u_1}}}{{(m + M)}}} \right)^2}$
\[ \Rightarrow K.E = \dfrac{{2M{m^2}{u_1}^2}}{{{{(m + M)}^2}}}{\text{ }} \to 4\]
Equating 3 and 4 we get,
The initial kinetic energy of moving mass m transferred to the strucked body is
\[ \Rightarrow kinetic\,energy\,transferred\,to\,the\,strucked\, = \,\dfrac{1}{2}m{u_1}^2 = \dfrac{{2M{m^2}{u_1}^2}}{{{{(m + M)}^2}}}\]
\[ \Rightarrow kinetic\,energy\,transferred\,to\,the\,strucked\, = \,\dfrac{{4mM}}{{{{(m + M)}^2}}}\]
Hence proved.

Note: An isolated system is defined as the one that has the zero net external force. We can apply conservation of momentum only when the net external force is zero.