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A spring of negligible mass and force constant k = 410 N/m is hung vertically, and a 0.200 kg pan is suspended from its lower end. A butcher drops a 2.5 kg steak onto the pan from a height of 0.50 m. The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM

1. What is the speed of the pan and steak immediately after the collision?

2. What is the amplitude of the subsequent motion?

3. What is the period of that motion?

Respuesta :

Rau7star

Answer:

1) 2.9m/s

2)0.235cm

3)0.509s

Explanation:

1. moments are conserved, when the steak hits the pan:

velocity 'v' of the steak:  

v = √(2gh) = √(2*9.81*0.5) = 3.13 m/s

velocity pan + steak is

v =[tex]m_1v_1/(m_1+m_2)[/tex]= 2.5*3.13/(2.7) = 2.9 m/s

2.initial kinetic energy of pan + steak = spring energy  

1/2 mv² = 1/2 kA²  

where, A = amplitude  

2.7 x 2.9² = 410 x A²

A²=22.707/410

A = 0.235 cm

3. T = 2π√(m/k) = 2π√(2.7/410) = 0.509 s

Therefore, the period of that motion is 0.509s

1) The speed of the pan and steak immediately after the collision is 2.9m/s

2) The  amplitude of the subsequent motion is 0.235cm.

3) The period of that motion is 0.509s.

Calculation of the speed, amplitude, and the period:

1.

Since

v = √(2gh) = √(2*9.81*0.5) = 3.13 m/s

Also,

v == 2.5*3.13/(2.7)

= 2.9 m/s

2.

Now

initial kinetic energy of pan + steak = spring energy  

So,

1/2 mv² = 1/2 kA²  

Here,

A = amplitude  

2.7 x 2.9² = 410 x A²

A²=22.707/410

A = 0.235 cm

3.

Now

T = 2π√(m/k)

= 2π√(2.7/410)

= 0.509 s

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