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3 years ago

Describe the steps in solving the problem below.

Physics

1 answer:

12345 [234]3 years ago

3 0

The velocity of the red cart after the collision is 2 m/s

From the law of conservation of momentum, initial momentum of system = final momentum of system.

m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄ where m₁ = mass of red cart = 4 kg, v₁ = velocity of red cart before collision = + 4 m/s, v₃ = velocity of red cart after collision, m₂ = mass of blue cart = 1 kg, v₂ = velocity of blue cart before collision = 0 m/s (since it is initially at rest) and v₄ = velocity of blue cart after collision = + 8 m/s.

Substituting the values of the variables into the equation, we have,

m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄

4 kg × 4 m/s + 1 kg × 0 m/s = 4v₃ + 1 kg × 8 m/s

16 kgm/s + 0 kgm/s = 4v₃ + 8 kgm/s

16 kgm/s = 4v₃ + 8 kgm/s

16 kgm/s - 8 kgm/s = (4 kg)v₃

(4 kg)v₃ = 8 kgm/s

Divide both sides by 4 kg, we have

v₃ = 8 kgm/s ÷ 4 kg

v₃ = 2 m/s

The velocity of the red cart after the collision is 2 m/s.

Learn more about conservation of momentum here:

brainly.com/question/7538238

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The magnitude of the tangential velocity is $v= 0.868 m/s$

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Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago