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OLEGan [10]
3 years ago
14

Understand that, for many purposes, a system can be treated as a point-like particle with its mass concentrated at the center of

mass. a complex system of objects, both point-like and extended ones, can often be treated as a point particle, located at the system's center of mass. such an approach can greatly simplify problem solving.
Physics
1 answer:
Nataliya [291]3 years ago
4 0
Treating the system as a point-like particle allows us to assign a quantity to the object and monitor this quantity throughout any changes. The complexity of the system which includes geometry, appearance, and extensions can complicate the studying of the system.
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Under what conditions can an object have forces acting on it, but its motion does not change ?
Lady_Fox [76]

Answer: An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity. Such an object will not change its state of motion.

Explanation: I hoped that helped!!

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In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,
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Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B =\ 1.0\ Kg.

Let mass of ball A and B\ is\ m  

Final velocity of ball A\ is\ v_1

Final velocity of ball B\ is\ v_2

initial velocity of ball A\ is\ u_1

Initial velocity of ball B\ is\ u_2

Momentum after collision =mv_1+mv_2

Momentum before collision = mu_1+mu_2

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, mu_1+mu_2=mv_1+mv_2

Plugging each trial in this equation we get,

First Trial

mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3

momentum before collision \neq moment after collision

Second Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1

moment before collision = moment after collision

Third Trial

mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1

momentum before collision \neq moment after collision

Fourth Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0

momentum before collision \neq moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

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