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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NTA Abhyas 2020 · Physics · Center of Mass, Momentum and Collision

A ball of mass m moving with velocity v strikes the bob of a...

A ball of mass m\text{m} moving with velocity v\text{v} strikes the bob of a pendulum at rest. The mass of the bob is also m\text{m}. If the collision is perfectly inelastic, the height to which the bob will rise is given by~<!-- MathType@Translator@5@5@MathML2 (namespace attr).tdl@MathML 2.0 (namespace attr)@ --> h =v2x.g,\text{h =}\dfrac{v^{2}}{\text{x}\text{.g}} \text{,} <!-- MathType@End@5@5@ --> then <!-- MathType@Translator@5@5@MathML2 (namespace attr).tdl@MathML 2.0 (namespace attr)@ --> x=?\text{x} = ? <!-- MathType@End@5@5@ -->

Correct answer

8

Step-by-step solution

Because the collision is perfectly inelastic, hence the two blocks stick together. By conservation of linear momentum,
2mV=mv2 \text{mV} = \text{mv} or V=v2\text{V} = \dfrac{\text{v}}{2}

By conservation of energy,

2mgh=122m(V2)2\text{mgh} = \dfrac{1}{2} \text{2m}{\left(\dfrac{\text{V}}{2}\right)}^{2}oror\text{h} = {\text{v}}^{2} / 8 \text{g}<br/><br /> <!−−MathType@End@5@5@−−><!−−MathType@Translator@5@5@MathML2(namespaceattr).tdl@MathML2.0(namespaceattr)@−−><!-- MathType@End@5@5@ --><!-- MathType@Translator@5@5@MathML2 (namespace attr).tdl@MathML 2.0 (namespace attr)@ -->\text{x} = 8
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