Answer

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**Hint:-**This is a question of perfectly inelastic collision. In inelastic collision both the bodies after colliding with each other sticks together and move as one body also in this type of collision the conservation of momentum is applicable as there is no external force is applied, the initial kinetic energy is greater than the final kinetic energy for inelastic collision only.

**Formula used:**

Here we would be using the formula for conservation of momentum and Kinetic energy.

${m_1}{u_1} + {m_2}{u_2} = {m_2}v + {m_1}v$;

\[\dfrac{1}{2}{m_1}u_1^2 + \dfrac{1}{2}{m_2}u_2^2 = \dfrac{1}{2}({m_1} + {m_2}){v^2}\];

Where:

${m_1}$= Mass of object 1.

${u_1}$= Velocity if object 1.

${m_2}$= Mass of object 2.

${u_2}$= Velocity if object 2.

$v$ = Final velocity.

**Complete step-by-step solution:-**

Step 1:

Here we have to find out the weight of the car Q. Apply conservation of momentum

${m_1}{u_1} + {m_2}{u_2} = {m_2}v + {m_1}v$;

Put the given values,

${m_1} \times {u_1} + {m_2} \times 0 = ({m_2} + {m_1})v$;

Solve,

${m_1} \times {u_1} = ({m_2} + {m_1})v$;

\[\dfrac{{{m_1}}}{{({m_2} + {m_1})}} \times {u_1} = v\];

Step 2:

Since the initial kinetic energy is greater than the final kinetic energy, we can calculate the change,

\[K{E_i} - K{E_f} = \dfrac{1}{2}{m_1}u_1^2 + \dfrac{1}{2}{m_2}u_2^2 = \dfrac{1}{2}({m_1} + {m_2}){v^2}\];

Take the rightmost value and put it to LHS, Here the car Q is at rest so its velocity\[{u_2}\]will be zero,

\[K{E_i} - K{E_f} = \dfrac{1}{2}{m_1}u_1^2 + \dfrac{1}{2}{m_2} \times 0 - \dfrac{1}{2}({m_1} + {m_2}){v^2}\];

Solve,

\[K{E_i} - K{E_f} = \dfrac{1}{2}{m_1}u_1^2 - \dfrac{1}{2}({m_1} + {m_2}){v^2}\];

Take the common out,

\[K{E_i} - K{E_f} = \dfrac{1}{2}({m_1}u_1^2 - ({m_1} + {m_2}){v^2})\];

We have been given the value (Difference in K.E) in percentage, so to convert it divide by the initial kinetic energy.

\[\dfrac{{K{E_i} - K{E_f}}}{{K{E_i}}} = \dfrac{{\dfrac{1}{2}({m_1}u_1^2 - ({m_1} + {m_2}){v^2})}}{{\dfrac{1}{2}{m_1}u_1^2}}\];

Solve the division,

\[\dfrac{{K{E_i} - K{E_f}}}{{K{E_i}}} = \dfrac{{({m_1}u_1^2 - ({m_1} + {m_2}){v^2})}}{{{m_1}u_1^2}}\];

Put the value of \[{v^2}\]in the above equation,

\[\dfrac{{K{E_i} - K{E_f}}}{{K{E_i}}} = \dfrac{{({m_1}u_1^2 - ({m_1} + {m_2}){{(\dfrac{{{m_1}}}{{({m_2} + {m_1})}} \times {u_1})}^2}}}{{{m_1}u_1^2}}\];

Square the inside terms,

\[\dfrac{{K{E_i} - K{E_f}}}{{K{E_i}}} = \dfrac{{({m_1}u_1^2 - ({m_1} + {m_2})(\dfrac{{m_{_1}^2}}{{{{({m_2} + {m_1})}^2}}} \times u_{_1}^2)}}{{{m_1}u_1^2}}\];

Simplify the equation,

\[\dfrac{{K{E_i} - K{E_f}}}{{K{E_i}}} = \dfrac{{({m_1}u_1^2 - \dfrac{{m_{_1}^2}}{{({m_2} + {m_1})}} \times u_{_1}^2)}}{{{m_1}u_1^2}}\];

Take the common out,

\[\dfrac{{K{E_i} - K{E_f}}}{{K{E_i}}} = \dfrac{{{m_1}u_1^2(1 - \dfrac{{{m_1}}}{{({m_2} + {m_1})}})}}{{{m_1}u_1^2}}\];

Solve,

\[\dfrac{{K{E_i} - K{E_f}}}{{K{E_i}}} = (1 - \dfrac{{{m_1}}}{{({m_2} + {m_1})}})\];

\[\dfrac{{K{E_i} - K{E_f}}}{{K{E_i}}} = (1 - \dfrac{{{m_1}}}{{({m_2} + {m_1})}})\];

Step 3:

Here \[\dfrac{{K{E_i} - K{E_f}}}{{K{E_i}}} = 0.3\], as given in the question. Put the value in the above equation and solve.

\[0.3 = (1 - \dfrac{{{m_1}}}{{({m_2} + {m_1})}})\];

Take 1 to LHS and Solve,

\[0.3 - 1 = - \dfrac{{{m_1}}}{{({m_2} + {m_1})}}\];

Solve,

\[ - 0.7 = - \dfrac{{{m_1}}}{{({m_2} + {m_1})}}\];

\[0.7 \times ({m_2} + {m_1}) = {m_1}\];

Do the needed calculations:

\[0.7 \times {m_2} + 0.7 \times {m_1} = {m_1}\];

\[0.7 \times {m_2} = {m_1} - 0.7 \times {m_1}\];

Keep \[{m_2}\]in the LHS

\[0.7 \times {m_2} = 0.3{m_1}\];

Solve,

\[{m_2} = \dfrac{{0.3 \times {m_1}}}{{0.7}}\];

\[{m_2} = \dfrac{{0.3 \times 35}}{{0.7}}\];

The weight of the car Q is:

\[{m_2} = 15kg\];

**Final Answer:**Option “3” is correct. The weight of Q is 15kg.

**Note:-**Here we have to first apply the conservation of momentum and find the value of the final velocity v and then as given in the question the difference between the initial K.E and the final K.E is 30%, we need to subtract the initial K.E from the final K.E and then put the value of v calculated by applying conservation of momentum. After the equation is simplified, put in the values and solve.

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