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V125BC [204]
3 years ago
14

At the instant a 2.0-kg particle has a velocity of 4.0 m/s in the positive x direction, a 3.0-kg particle has a velocity of 5.0

m/s in the positive y direction. What is the speed of the center of mass of the two-particle system
Physics
1 answer:
satela [25.4K]3 years ago
5 0

Answer:

the speed of the center of mass of the two-particle system is 3.4 m/s.

Explanation:

Given that,

Mass of particle 1, m = 2 kg

Velocity of the particle 1, v = 4 m/s in +x direction

Mass of particle 2, m' = 3 kg

Velocity of the particle 2, v' = 5 m/s in +y direction.

The x -coordinate of velocity of the centre of mass is given by :

v_x=\dfrac{mv+m'v'}{m+m'}\\\\v_x=\dfrac{2\times 4+3\times 0}{2+3}\\\\v_x=1.6\ m/s

The y -coordinate of velocity of the centre of mass is given by :

v_y=\dfrac{mv+m'v'}{m+m'}\\\\v_y=\dfrac{2\times 0+3\times 5}{2+3}\\\\v_y=3\ m/s

So, the the speed of the center of mass of the two-particle system given by :

v=\sqrt{v_x^2+v_y^2} \\\\v=\sqrt{1.6^2+3^2} \\\\v=3.4\ m/s

So, the speed of the center of mass of the two-particle system is 3.4 m/s.

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A cat leaps into the air to catch a bird with an initial speed of 2.74 m/s at an angle of 60.0° above the ground. What is the hi
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Answer: D. 0.29 m

Explanation:

We will use the following equations to describe the leap of the cat:

y=V_{o}sin\theta t-\frac{gt^{2}}{2}   (1)

V_{y}=V_{oy}-gt   (2)

Where:

y  is the height of the cat  

V_{oy}=V_{o}sin\theta is the cat's initial velocity

\theta=60\°

g=9.8m/s^{2}  is the acceleration due gravity

t is the time

V_{y} is the y-component of the velocity

Now the cat will have its maximum height y_{max} when V_{y}=0. So equation (2) is rewritten as:

0=V_{oy}-gt   (3)

Finding t:

t=\frac{V_{oy}}{g}=\frac{V_{o}sin\theta}{g}   (4)

t=\frac{2.74 m/s sin(60\°)}{9.8m/s^{2}}   (5)

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Substituting (6) in (1):

y_{max}=(2.74 m/s)sin(60\°) (0.24 s)-\frac{(9.8m/s^{2})(0.24 s)^{2}}{2}   (7)

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