A ball of mass m.Is released from the top pf a inclined plane of inclination theta as shown.It strikes a rigid surface at a distance 3l/4 from top elastically.Impulse imparted to ball by the rigid surface is
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initial energy of ball = potential energy at height h of ball = mgl
energy of the ball when it strikes the rigid surface = potential energy at ( l - 3l/4) + kinetic energy
= mgl/4 + 1/2mv²
from conservation of energy,
initial energy = final energy
mgl = mgl/4 + 1/2mv²
3mgl/2 = mv²
v² = 1.5gl
v = √(1.5gl)
now, impulse = change in momentum = mv - (-mv) = 2mv = 2m√(1.5gl) = m√(6gl)
hence, Impulse imparted to ball by the rigid surface is m√(6gl)
energy of the ball when it strikes the rigid surface = potential energy at ( l - 3l/4) + kinetic energy
= mgl/4 + 1/2mv²
from conservation of energy,
initial energy = final energy
mgl = mgl/4 + 1/2mv²
3mgl/2 = mv²
v² = 1.5gl
v = √(1.5gl)
now, impulse = change in momentum = mv - (-mv) = 2mv = 2m√(1.5gl) = m√(6gl)
hence, Impulse imparted to ball by the rigid surface is m√(6gl)
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