A ball of mass 0.15 kg is dropped from a height. 10 m, strikes the ground and rehounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s² ) nearly:

Option 1 - <p>0 kg m/s<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>4.2 kg m/s</p>
Option 3 - <p>2.1 kg m/s</p>
Option 4 - <p>1.4 kg m/s</p>
2 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
A
5 months ago
Correct Option - 2
Detailed Solution:

Velocity just before striking the ground
v? = √2gh
v? = √ (2*10*10) = 10√2 m/s
v? = -10√2?
If it reaches the same height, speed remains same after collision only the direction changes.
v? = 10√2 m/s
v? = 10√2?
|Impulse| = m|Δv|
= m|10√2? - (-10√2? )|
= 0.15 [2 (10√2)]
= 3√2 kg m/s
= 4.2 kg m/s

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Physics Work, Energy and Power 2021

Physics Work, Energy and Power 2021

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