Question

A child is bouncing an a trampoline. The child has a mass of 30.5 kg . She leaves the surface of the trampoline with an initial upwards velocity of 10.2 m/s . When she reaches height 1.60 m above the trampoline (still on the way up), she grabs a large (stationary) plastic block. The mass of the block is 6.00 kg . Find her final velocity.

Answer 1

Answer:

v = 60.7 m / s

Explanation:

This is an exercise that we can solve using the moment equations.

To begin we must create a system that is formed by the child and the block. For this system the forces during the crash are internal, for local moment it is conserved. Let's write the moment before and after the crash

Before

    p₀ = m v₀ + 0

After

   $p_{f}$ = (m + M) v

Where m and M are the masses of the child and the block, respectively. Notice that when they collide they are joined by which the mass is the sum of the two

    p₀ = pf

    m v₀ = (m + M) v

    v = v₀ m / (m + M)          (1)

We must find the speed of the child at 1.60 m, for this we use the law of conservation of energy, calculate the energy at two points: where it jumps from the trampoline and just before grabs the block

When he jumps off the trampoline

    Em₁ = ​​K = ½ m v₁²

Just before taking the block

    Em₂ = K + U = ½ m v₂² + mg y

    Em₁ = ​​Em₂

    ½ m v₁² = ½ m v₂² + mg y

Let's calculate the speed with which it reaches the point of the crash

    v₂² = v₁² - 2 g y

    v₂ = √ (v₁² - 2 g y)

    v₂ = √ (10.2² - 2 9.8 1.60)

    v₂ = 72.7 m / s

This is the speed with which it reaches the point where the block is, vo = v₂ = 72.7 m/s, with this value we can substitute and calculate in equation (1) of conservation of the moment

    v = vo  m / (m + M)

    v = 72.7 30.5 / (30.5 + 6.00)

    v = 60.7 m / s