A ball of mass 0.15 kg hits the wall with its initial speed of 12 ms-1and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is 100 N, calculate the time during of the contact of ball with the wall.

Option 1 - <p>0.018s</p>
Option 2 - <p>0.036s</p>
Option 3 - <p>0.009s</p>
Option 4 - <p>0.072s</p>
7 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
A
7 months ago
Correct Option - 2
Detailed Solution:

F=ΔρΔt=1.8 (1.8)Δt=100

Δt=3.6100=0.036sec

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Let ‘h’ be the height at which velocity becomes equal to magnitude of Acceleration

v = g = 10

v = u + at

10 = 0 + 10t

t = 1 sec

h=ut+12at2

=0×1+12×10×1×1

= 5m

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Physics NCERT Exemplar Solutions Class 12th Chapter Five 2025

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