contestada

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a 0.405kg ball that is traveling horizontally at 8.50m/s . Your mass is 69.0kg
A) If you catch the ball, with what speed do you and the ball move afterwards?
B) If the ball hits you and bounces off your chest, so that afterwards it is moving horizontally at 7.80m/s in the opposite direction, what is your speed after the collision?

Respuesta :

Answer:

a) v = 0.0496 m / s , b)  v = 0.0957 m / s

Explanation:

a) in this case we have an inelastic collision,

To solve these problems, the most important thing is to define the system formed by all the bodies, so that the forces during the crash have been internal and the amount of movement is conserved.

Therefore the system is made up of the ball and the player

initial instant. Before catching the ball

       p₀ = m v₁

final moment. After you have grabbed the ball

      [tex]p_{f}[/tex] = (m + M) v

the moment is preserved

       po = p_{f}

       m v₁ = (m + M) v

       v = m / (m + M) v₁

let's calculate

      v = 0.405 / (0.405 + 69)   8.50

      v = 0.0496 m / s

in the same direction that the ball takes

b) In this case the ball bounces with a speed of V₂ = -7.80 m / s, the negative sign is because it is going in opposite directions

let's write the moment in two times

initial instant. Before the crash

       p₀ = m v₁

try end. Right after the crash

       p_{f} = m v₂ + M v

the moment is preserved

       p₀ = p_{f}

        m v₁ = m v₂ + M v

       v = m (v₁ -v₂) / M

       v = 0.405 /69   (8.50 - (7.80))

       v = 0.0957 m / s

in the initial direction of the ball

shubhamchouhanvrVT

a) After catching a ball  v = 0.0496 m / s ,

b) If the ball hits and travel in opposite direction  v = 0.0957 m / s

What will be the speed after the two collisions?

a) in this case we have an inelastic collision,and the momentum will be conserved.

initial instant. Before catching the ball

 [tex]P=mv_{i}[/tex]

final moment. After you have grabbed the ball

   [tex]P=(m+M)v_{f}[/tex]    

Since the moment is conserved

      [tex]P_{i} =P_{f}[/tex]

     

[tex]mv_{i} =(m+M)v_{f}[/tex]

 

  [tex]v_{f} =\dfrac{mv_{i} }{m+M}[/tex]

let's calculate

     

[tex]v_{f} =\dfrac{0.405\times8.50}{0.405+69}[/tex]

[tex]v_{f} =0.0496\dfrac{m}{s}[/tex]

in the same direction that the ball takes

b) In this case the ball bounces with a speed of   [tex]v_{f} =-7.80\dfrac{m}{s}[/tex]   the negative sign is because it is going in opposite directions

let's write the moment in two times

initial instant. Before the crash

         [tex]P_{i} =mv_{i}[/tex]

try end. Right after the crash

       [tex]P_{f} =mv_{f} +Mv[/tex]

since the moment is conserved

   

                [tex]P_{i} =P_{f}[/tex]

     

           [tex]mv_{i}=mv_{f} +Mv[/tex]

           [tex]v=\dfrac{m(v_{i} -v_{f} )}{M}[/tex]

     

           [tex]v=\dfrac{0.405\times(8.50-(-7.8)}{69}[/tex]

     

            [tex]v=0.0957 \dfrac{m}{s}[/tex]

Hence

a) After catching a ball  v = 0.0496 m / s ,

b) If the ball hits and travel in opposite direction  v = 0.0957 m / s

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