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Grade 11Mechanics

A ball of mass m is released from the top of an inclined plane of inclination theta .It strikes a rigid surface at a distance 3l/4 from top elastically, l is the length of hypotenuse. Impulse imparted to ball by rigid surface is..

Profile image of Pragya
8 Years agoGrade 11
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2 Answers

Profile image of Arun
8 Years ago
Dear Pragya
 
ball's initial energy = mgh 
ball's energy when it strikes rigid surface = mgh/4 + 1/2mV² 
mgh = mgh/4 + 1/2mV² 
1/2mV² = 3mgh/4 
(0.5)V² = (0.75)gh 
V² = 1.5gh 
V = √(1.5gh) 
impulse = mV+mV = 2m√(1.5gh) = m√(6gh) 

Comment: impulse = change in (vector) momentum, thus mV's are added because V leaving rigid surface is opposite V colliding with rigid surface.
 
Regards
Arun (askIITians forum expert)
Profile image of ankit singh
5 Years ago
impulse = mV+mV = 2m√(1.5gh) = m√(6gh) V = √(1.5gh) V² = 1.5gh (0.5)V² = (0.75)gh 1/2mV² = 3mgh/4 mgh = mgh/4 + 1/2mV² ball's energy when it strikes rigid surface = mgh/4 + 1/2mV² all's initial energy = mgh