JEE Main 2015PhysicsWork Power EnergyHardMCQ

JEE Main 2015 — Work Power Energy Question with Solution

JEE Main 2015 (10 Apr Online)

Question

A block of mass m=10 kg rests on a horizontal table. The coefficient of friction between the block and the table is 0.05. When hit by a bullet of mass 50 g moving with speed v, that gets embedded in it, the block moves and comes to stop after moving a distance of 2 m on the table. If a freely falling object were to acquire speed v10 after being dropped from height H, then neglecting energy losses and taking g=10 m s-2, the value of H is close to

 

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Show full solutionCorrect option: D
Correct answer
DNone of these.

Step-by-step explanation



By the conservation of linear momentum,

m1v=m+m1v1

⇒v1=m1vm+m1 ...(1)

After collision by work energy theorem, we have

Wfriction=∆k

⇒-μm+m1gx=-12 m+m1v12

⇒μgx=12v12

⇒0.05×10×2=12m1 vm+m12

⇒5100×10×4=5010002×v210+ 5010002

⇒2=1400v2×4002012

v2=2012 2

v=2×201

For a freely falling body,

v'=2gH

v10=2gH

2×20110=2×gH

2012100=10H

H=20121000=40 m

=0.04 km

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About this question

This is a previous-year question from JEE Main 2015, covering the Work Power Energy chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.