A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 3h from the…

Question

A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height $3 h$ from the ground, as shown in the figure. A spherical ball of mass $m$ is released on the slide from rest at a height $h$ from the top of the terrace. The ball leaves the slide with a velocity ${\vec{u}}_{0} = u_{0} \overset{\land}{x}$ and falls on the ground at a distance $d$ from the building making an angle $θ$ with the horizontal. It bounces off with a velocity $\vec{v}$ and reaches a maximum height $h_{1}$ . The acceleration due to gravity is $g$ and the coefficient of restitution of the ground is $\frac{1}{\sqrt{3}}$ . Which of the following statement(s) is(are) correct?

MARKS Content Team · Accepted Answer

Correct answer: (A) u→_0=√2ghx∧; (C) θ=60^o; (D) dh_1=2√3. Using energy conservation, mgh=1/2mu0^2 ⇒u_0=√2gh u_0 is the horizontal component of the velocity throughout the flight. For vertical component of the velocity(just before collision with the ground), we can use equation of motion vz^2=0^2+2-g-3h⇒v_z=√6gh Angle θ can be written as, tanθ=v_zu=√3 ⇒θ=60^o Horizontal distance covered just before the collision will be, d=u_0T=u_0√23hg=√2gh√23hg=2√3h After collision, only velocity along z-direction changes. Therefore, v_1=ev_z=√2gh and hence v→=u_0i∧+v_1k∧ =√2ghi∧+k∧=√2ghx∧+z∧ Now, maximum height achieved after collision will be, h_1=v1^22g=h. Finally, u_0=√2gh, θ=60^o, dh_1=2√3.

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