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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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A projectile is thrown with an initial speed vo = 25.0 m/s at an initial angle with the horizontal ao = 45.0. a . Find the time
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See the explanation below

Explanation:

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v_{o}_{x}=25*cos(45)=17.67[m/s]\\v_{o}_{y}=25*sin(45)=17.67[m/s]\\

The acceleration of  gravity is equal to g = 9.81[m/s^2] downward.

The maximum height is when the velocity of the projectile is zero in the component y, that is, it will not be able to go higher, by means of the following kinematic equation we can find that time, for that specific condition.

a)

v_{y}=(v_{y})_{0}+a*t\\0 = 17.67 - 9.81*t\\17.67 = 9.81*t\\t=1.8 [s]

Note: Acceleration is taken as negative as it is directed downwards.

b)

The position in the x component can be found using the following kinematic equation

x=(v_{x})_{o}*t\\x=17.67*1.8\\x=31.82[m]

The position in the y component can be found using the following kinematic equation

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c)

Since the motion on the X-axis is at constant speed, there is no acceleration, so the only acceleration that exists is due to gravity

d)

In the attached image we can see, the projectile with the vectors of acceleration and velocity.

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