A particle falling vertically from a height hits a plane surface incli

1.	A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed v0 and rebounds elastically (Fig.). Find the distance along the plane where if will hit second time.(Hint: (i) After rebound, particle still has speed v0 to start. (ii) Work out angle particle speed has with horizontal after it rebounds. (iii) Rest is similar to if particle is projected up the incline.)
Answer» As θ = 0, T = \(\frac{2v_0\,cos\,\theta}{g\,cos\,\theta}\)⇒ \(\frac{2v_0}{g}\) Considering motion along x-axis, X = L, u_x= v₀ cos θ, a_x= g sin θ, t = T = \(\frac{2v_0}{g}\) From kinematics equation- = \(u_xt+\frac{1}{2}a_xt^2\) L =\(v_0\,sin\,\theta t+\frac{1}{2}g\,sin\,\theta\, t^2\) = \((v_0sin\,\theta)(T)+\frac{1}{2}g\,sin\,\theta\, T^2\) = \((v_0 sin \,\theta)(\frac{2v_0}{g})+\frac{1}{2}g\,sin\,\theta\times(\frac{2v_0}{g})\) = \(\frac{2v^2_0}{g} sin\,\theta +\frac{1}{2}g\, sin\,\theta+\frac{4v^2_0}{g^2}\) = Distance, L = \(\frac{4v^2_0}{g} sin\,\theta\).

A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed v0 and rebounds elastically (Fig.). Find the distance along the plane where if will hit second time.(Hint: (i) After rebound, particle still has speed v0 to start. (ii) Work out angle particle speed has with horizontal after it rebounds. (iii) Rest is similar to if particle is projected up the incline.)

Answer»

As θ = 0, T = \(\frac{2v_0\,cos\,\theta}{g\,cos\,\theta}\)⇒ \(\frac{2v_0}{g}\)

Considering motion along x-axis,

X = L, u_x= v₀ cos θ, a_x= g sin θ, t = T = \(\frac{2v_0}{g}\)

From kinematics equation-

= \(u_xt+\frac{1}{2}a_xt^2\)

L =\(v_0\,sin\,\theta t+\frac{1}{2}g\,sin\,\theta\, t^2\)

= \((v_0sin\,\theta)(T)+\frac{1}{2}g\,sin\,\theta\, T^2\)

= \((v_0 sin \,\theta)(\frac{2v_0}{g})+\frac{1}{2}g\,sin\,\theta\times(\frac{2v_0}{g})\)

= \(\frac{2v^2_0}{g} sin\,\theta +\frac{1}{2}g\, sin\,\theta+\frac{4v^2_0}{g^2}\)

= Distance, L = \(\frac{4v^2_0}{g} sin\,\theta\).