SATHEE: Work Energy And Power Question 322

Work Energy And Power Question 322

Question: A bullet of mass m moving vertically upwards instantaneously with a velocity ‘u’ hits the hanging block of mass ’m’ and gets embedded in it, as shown in the figure. The height through which the block rises after the collision, (assume sufficient space above block) is:

Options:

A) $u^{2} / 2 g$

B) $u^{2} / g$

C) $u^{2} / 8 g$

D) $u^{2} / 4 g$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] From momentum conservation $m u = 2 m v$

$\Rightarrow v = \frac{u}{2}$ from energy conservation $\frac{1}{2} \times 2 m \times {(\frac{u}{2})}^{2} = 2 m g h \Rightarrow h = \frac{u^{2}}{8 g}$

Work Energy And Power Question 322

Question: A bullet of mass m moving vertically upwards instantaneously with a velocity ‘u’ hits the hanging block of mass ’m’ and gets embedded in it, as shown in the figure. The height through which the block rises after the collision, (assume sufficient space above block) is:

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

Work Energy And Power Question 322

Question: A bullet of mass m moving vertically upwards instantaneously with a velocity ‘u’ hits the hanging block of mass ’m’ and gets embedded in it, as shown in the figure. The height through which the block rises after the collision, (assume sufficient space above block) is:

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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