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amm1812
3 years ago
14

Consider a system of two particles: ball A with a mass m is moving to the right a speed 2v and ball B with a mass 3m is moving t

o the left at a speed v. In the time interval before the two balls collide, what is the magnitude and the direction of the velocity of the center of mass of this system?
Physics
1 answer:
arlik [135]3 years ago
3 0

Answer:

Explanation:

Answer:

Explanation:

Given that,

System of two particle

Ball A has mass

Ma = m

Ball A is moving to the right (positive x axis) with velocity of

Va = 2v •i

Ball B has a mass

Mb = 3m

Ball B is moving to left (negative x axis) with a velocity of

Vb = -v •i

Velocity of centre of mass Vcm?

Velocity of centre of mass can be calculated using

Vcm = 1/M ΣMi•Vi

Where M is sum of mass

M = M1 + M2 + M3 +...

Therefore,

Vcm=[1/(Ma + Mb)] × (Ma•Va +Mb•Vb

Rearranging for better understanding

Vcm = (Ma•Va + Mb•Vb) / ( Ma + Mb)

Vcm = (m•2v + 3m•-v) / (m + 3m)

Vcm = (2mv — 3mv) / 4m

Vcm = —mv / 4m

Vcm = —v / 4

Vcm = —¼V •i

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siniylev [52]

If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be V_e = \sqrt{\dfrac{GM}{r}}.

<h3>What is an escape velocity?</h3>

The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.

The formula to calculate the escape velocity of earth is given below:-

V_e=\sqrt{\dfrac{2GM}{r}}

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

V_e=\sqrt{\dfrac{2GM}{r\times 2}}

V_e = \sqrt{\dfrac{GM}{r}}.

Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be V_e = \sqrt{\dfrac{GM}{r}}.

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8 0
1 year ago
Two bowling balls each have a mass of 6.8 kg. They are located next to each other with their centers 21.8 cm apart. What gravita
12345 [234]

Answer:

6.49 x 10^-8 N

Explanation:

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2 years ago
Rosný bod závisí od ?
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Answer:

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3 years ago
An object of mass m is traveling in a circle with centripetal force Fc. If the velocity of the object is v, what is the radius o
borishaifa [10]

Hi there!

Recall the equation for centripetal force:
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We can rearrange the equation to solve for 'r'.

Multiply both sides by r:
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Divide both sides by Fc:
\boxed{ r= \frac{mv^2}{F_c}}

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Two bodies fall freely from different heights and reach the ground simultaneously. The time of descent for the first body is 1s
jok3333 [9.3K]
The initial height of the first body is given by:
h_1 =  \frac{1}{2}gt^2
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
h_1 =  \frac{1}{2}(9.81 m/s^2)(1 s)^2=4.9 m

The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
h_2 =  \frac{1}{2}(9.81 m/s^2)(2 s)^2=19.6 m

The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.
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