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Find maximum height up to which block A of mass m rises on wedge:


A) \[h\]
B) \[\dfrac{h}{2}\]
C) \[\dfrac{h}{4}\]
D) None of these

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Last updated date: 22nd Feb 2024
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IVSAT 2024
Answer
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Hint: In this question, use the concept of the conservation of energy and the conservation of linear momentum. Apply the first energy conservation on the block and the wedge B and then apply on the block and the wedge C.

Complete step by step solution:
In this situation, without consideration of friction force, we need to balance the energies of the system in the initial and final positions only, i.e. when the block is at the top and when it is at the bottom.
First case- As we know the travel of the block from top of $B$ to bottom; by conservation of energy we get,
\[{\left( {K{E_A} + \;K{E_B} + P{E_A} + P{E_B}} \right)_{initial}} = {\left( {K{E_A} + \;K{E_B} + P{E_A} + P{E_B}} \right)_{final}}\]
Here, $KE$ represents the kinetic energy, $PE$ represents the potential energy, $A$ represents the block, $B$ and $C$ represents the wedges.
\[0 + 0 + mgh + 0 = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}m{v^2} + 0 + 0\]
After simplification we get,
\[ \Rightarrow v = \sqrt {gh} ......\left( 1 \right)\]
Which is the final speed of the block at the bottom.

Second case- Now, the block travels from bottom to reach maximum point on \[C\] where its speed will become zero. Let the distance to be travelled by the block be $x$.

Hence applying the energy conservation principal and using the value of equation \[\left( 1 \right)\], we get,
\[{\left( {K{E_A} + \;K{E_C} + P{E_A} + P{E_C}} \right)_{initial}} = {\left( {K{E_A} + \;K{E_C} + P{E_A} + P{E_C}} \right)_{final}}\]
\[\dfrac{1}{2}m{\left( {\sqrt {gh} } \right)^2} + 0 + 0 + 0 = \dfrac{1}{2}m{\left( {{v_f}} \right)^2} + \dfrac{1}{2}m{\left( {{v_f}} \right)^2} + mgx + 0......\left( 2 \right)\]
Here a part of the kinetic energy of \[A\] is getting converted into potential energy of \[A\] and the Kinetic energy of \[C\].
Now, we apply the conservation of momentum equation as,
\[{\left( {m \times {v_A} + m \times {v_C}} \right)_{initial}} = {\left( {m \times {v_A} + m \times {v_C}} \right)_{Final}}\]
Now, we substitute the values in the above equation and get,
\[ \Rightarrow m \times \sqrt {gh} + 0 = m{v_f} + m{v_f}\]
After simplification we get,
\[ \Rightarrow {v_f} = \dfrac{{\sqrt {gh} }}{2}\]
Now, we Substitute the value of ${v_f}$ in equation \[\left( 2 \right)\] as
\[\dfrac{1}{2}mgh = mv_f^2 + mgx\]
\[ \Rightarrow \dfrac{1}{2}mgh = \dfrac{1}{4}mgh + mgx\]
After simplification we get,

\[\therefore x = \dfrac{h}{4}\]

Note: In the above solution we have not considered friction in any of the surfaces. If we consider the surfaces of contact to be rough, then the calculation will contain an additional friction force term and also the energy conversion as a result of it. Also, if the wedges are kept fixed then also the calculation will be different.