SATHEE: Rotational Motion Question 116

Rotational Motion Question 116

Question: A block of mass $m$ starts from rest and slides down a frictionless semi-circular track from a height h as shown. When it reaches the lowest point of the track, it collides with a stationary piece of putty. also having mass m. If the block and the putty stick together and continue to slide, the maximum height that the block-putty system could reach is

Options:

A) $h$ /4

B) $h$ /2

C) $h$

D)independent of $h$

Show Answer

Answer:

Correct Answer: B

Solution:

[b]

$v_{0} = \sqrt{2 g h}$

before collision.

Apply conservation principle of momentum, $m v_{0} = (m + m) u$

$∴ u = \frac{v_{0}}{2}$

Applying conservation principle of momentum,

$\frac{1}{2} (2 m) u^{2} = 2 m g h^{'}$

$∴ h^{'} = \frac{u^{2}}{2 g} \Rightarrow h^{'} \Rightarrow= \frac{v_{0}^{2}}{4 g}$

$h^{'} = \frac{2 g h}{4 g} \Rightarrow h^{'} = \frac{h}{2}$

Rotational Motion Question 116

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Answer:

Solution:

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Rotational Motion Question 116

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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