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Grade 12th passMechanics

A steel ball of mass 0.5 kg is dropped from a height of 4m on to a horizontal heavy steel slab.
The collision is elastic and the ball rebounceto
Its original height .
(a) calculate the impulse delivered to the ball during impact .
(B) if the ball is in contact with the slab for 0.002
,find the average reaction force on the ball during impact .

Profile image of Manish kumar
6 Years agoGrade 12th pass
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1 Answer

Profile image of Arun
6 Years ago
applying equation of motion v^2=u^2 + 2as 
v=?,u=0 
a = g= 10, s = 4 put in above  
V1=4 underoot5 
as ball rebounds to same hieght changed vilocity(V2) again will be 4 underoot 5 
Now impulse = Force ( time)= change in momentum. 
Impulse = m(V2-V1) = 2mV1 = 2 (0.5) 4 underoot5 = 4underoot5(answer) 
Now  
Force = impulse/time 
= 4underoot5 /0.002=2000underoot5