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A girl of mass m1=60.0 kilograms springs from a trampoline with an initial upward velocity of vi=8.00 meters per second. At height h=2.00 meters above the trampoline, the girl grabs a box of mass m2=15.0 kilograms. (Figure 1)

For this problem, use g=9.80 meters per second per second for the magnitude of the acceleration due to gravity.

What is the speed vbefore of the girl immediately before she grabs the box?
Express your answer numerically in meters per second.

What is the speed vafter of the girl immediately after she grabs the box?
Express your answer numerically in meters per second.

What is the maximum height hmax that the girl (with box) reaches? Measure hmax with respect to the top of the trampoline.

A Girl Of Mass M1600 Kilograms Springs From A Trampoline With An Initial Upward Velocity Of Vi800 Meters Per Second At Height H200 Meters Above The Trampoline T class=

Sagot :

The conservation of momentum and energy allows to shorten the results for the movement of the girl on the trampoline holding the box are:

     a) the girl's speed is v = 4.98 m / s

     b) The speed of the girl + box system is: v_f = 0.996 m / s

     c) the maximum height is: y = 2.05 m

 

Kinematics studies the movement of bodies, looking for relationships between the position, velocity and acceleration of bodies.

The momentum is defined by the product of mass and the velocity, when a system is isolated the momentum is conserved.

The mechanical energy is the sum of the kinetic energy plus the potential energies, when there is no friction in the system the mechanical energy is conserved.

Let's solve this exercise in parts:

a) Let's use kinematics to find the speed of the girl before she grabs the box

              v² = v₀² - 2 g y₁

              v² = 8² - 2 9.8 2.00

              v = R 24.8 = 4.98 m / s

b) Let's use momentum conservation for when the speed of the girl and the box together. Let's write the moment in two moments.

Initial instant. Just before you grab the box.

              p₀ = M v + 0

Final moment. Right after taking the box

             [tex]p_f[/tex]  = (m + M) [tex]v_f[/tex]

         

In system this form by the girl and the box therefore it is an isolated system and the momentum is conserved.

           [tex]p_o = p_f[/tex]  

           mv = (m + M) [tex]v_f[/tex]  

           [tex]v_f = \frac{m}{m+M} \ v[/tex]

Let's calculate

           [tex]v_f = \frac{15}{15+ 60} \ 4.98[/tex]  

           [tex]v_f[/tex]  = 0.996 m / s

c) Now we use conservation of energy after the girl has the box.

Starting point. When the girl has the box

           Em₀ = K + U

           Em₀ = ½ (m + M) v² + (m + M) g y₁

Final point. At the highest point of the trajectory

          [tex]Em_f[/tex]  = U

         [tex]Em_f[/tex] = (m + M) g y₁

As there is no friction, the energy is conserved.

           [tex]Em_o = Em_f[/tex]  

          ½ (m + M) v² + (m + M) g y₁ = (m + M) g y

          y = [tex]\frac{v^2}{2g} + y_1[/tex]  

Let's calculate

           y = [tex]\frac{0.996^2}{2 \ 9.8} + 2.0[/tex]

           y = 2.05 m

In conclusion using the conservation of momentum and energy we can shorten the results for the movement of the girl on the trampoline holding the box are:

     a) the girl's speed is v = 4.98 m / s

     b) The speed of the girl + box system is: v_f = 0.996 m / s

     c) the maximum height is: y = 2.05 m

Learn more here: brainly.com/question/15095150

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