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fredd [130]
3 years ago
11

A wooden rod of negligible mass and length 80.0cm is pivoted about a horizontal axis through its center. A white rat with mass 0

.450kg clings to one end of the stick, and a mouse with mass 0.220kg clings to the other end. The system is released from rest with the rod horizontal
Part A
If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position?
Take free fall acceleration to be 9.80m/s2 .
Physics
1 answer:
algol133 years ago
7 0

Answer:

The speed of the animals is 1.64m/s.

Explanation:

Let us work with variables and call the mass of the two rats m_1 and m_2,  and the length of the rod L.

Using the law of conservation of energy, which says the potential energies of the rats must equal their kinetic energies, we know that when the rod swings to the vertical position,

$m_1\frac{L}{2}g -m_2\frac{L}{2}g = \frac{1}{2}m_1v^2+\frac{1}{2}m_2v^2$

$(m_1 -m_2)g\frac{L}{2} = \frac{1}{2}(m_1+m_2)v^2$,

solving for v, we get:

$\boxed{v = \sqrt{\frac{(m_1 -m_2)gL}{(m_1+m_2)}} }$

Putting in the values for m_1, m_2, g, and L we get:

$v = \sqrt{\frac{(0.450kg -0.220kg)(9.8m/s^2)(0.8m)}{(0.450kg+0.220kg)}} $

\boxed{ v= 1.64m/s}

Therefore, as the rod swings through the vertical position , the speed of the rats is 1.64 m/s.

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Hi there!

The total momentum of the system is given by the sum of the momentum vectors of each cart. The momentum is calculated as follows:

p = m · v

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Then, the momentum of the system will be the momentum of cart A plus the momentum of cart B (let´s consider the right as the positive direction):

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Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

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mass of the bullet, m₁ = 2.47 g = 0.00247 kg

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initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

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0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

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Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

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Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

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