Question

A student releases a block of mass m from rest at the top of a slide of height h1. The block moves down the slide and off the end of a table of height h2 , landing on the floor a horizontal distance d from the edge of the table. Friction and air resistance are negligible. The overall height H of the setup is determined by the height of the room. Therefore, if h1 is increased, h2 must decrease by the same amount so that the sum h1 + h2 remains equal to H. The student wants to adjust h1 and h2 to make d as large as possible.
Without using equations, explain why making h1 very small would cause d to be small, even though h2 would be large.

Without using equations, explain why making h2 very small would cause d to be small, even though h1 would be large

Derive an equation for d in terms of h1, h2, m, and physical constants, as appropriate.

Write the equation or step in your derivation in part (b) (not your final answer) that supports your reasoning in part (a)i.

Briefly explain your choice.

Write the equation or step in your derivation in part (b) (not your final answer) that supports your reasoning in part (a)ii.

Briefly explain your choice.

If the experiment is repeated on the Moon without changing h1 or h2 , will the new landing distance d be greater than, less than, or the same as the landing distance when the experiment is performed on Earth?

_____Greater than _____Less than _____The same as

Answer 1

Answer:

B)   d = √  ( 4 h₂ ( H - 2h₂))

Explanation:

A) 1) If the height of the slide is very small, there is no speed to leave the table, therefore do not recreate almost any horizontal distance

2) If the height is very small downwards, it touches the earth a little and the horizon is small,

B) to find an equation for horizontal distance (d)

We must maximize the speed at the bottom of the slide let's use energy

Starting point Higher

         Em₀ = U = m g h₁

Final point. Lower (slide bottom)

           Emf = K + U = ½ m v² + m gh₂

As there is no friction the energy is conserved

            mgh₁ = ½ m v² + mgh₂

            v² = 2 g (h₁-h₂)

This is the speed with which the block leaves the table, bone is the horizontal speed (vₓ)

The distance traveled when leaving the table can be searched with kinematics, projectile launch

          x = v₀ₓ t

         y =  t - ½ g t²

The height is the height of the table (y = h₂), as it comes out horizontally the vertical speed is zero

        t = √ 2h₂ / g

We substitute in the other equation

        d = √ (2g (h₁-h₂))  √ 2h₂ / g

        d = √ (4 h₂ (h₁-h₂))

        H = h₁ + h₂

        h₁ = H -h₂

        d = √  ( 4 h₂ ( H - 2h₂))

Explanation: