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densk [106]
2 years ago
5

The graph above shows the position x as a function of time for the center of mass of a system of particles of total mass 6. 0 kg

. For a very short time interval around 2. 0 s, an external force is exerted on an object in the system. What is the resulting change in momentum of the system?.
Physics
1 answer:
kumpel [21]2 years ago
6 0

The resulting change in momentum of the system will be +18.6 Ns. The momentum is conserved.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

m is the mass =6.0 kg

t is the time interval=2 second

From Newton's second law;

\rm \triangle P =m \triangle V \\\\ \triangle P= m(\frac{\triangle x}{\triangle t} )\\\\

From the graph;

\rm \triangle t = 2sec\\\\ \triangle x = (12-8) m

The change in the momentum is;

\rm \triangle P = m\tr(\frac{v-u}{t}) \\\\ \triangle P =9.3 \times \frac{12-8}{2} \\\\ \triangle P= +18.6 \  N.s

Hence, the resulting change in momentum of the system will be +18.6 Ns.

To learn more about the law of conservation of momentum, refer;

brainly.com/question/1113396

#SPJ1

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We have all the values to apply the law of linear momentum, however, it is necessary to define the two lines in which the study will be carried out. Being an intersection the vehicle of mass m_1 approaches through the X axis, while the vehicle of mass m_2 approaches by the y axis. In the collision equation on the X axis, we despise the velocity of object 2, since it does not come in this direction.

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(950)(16)=(950+1300)V_fcos(60.89)

V_f= \frac{(950)(16)}{(950+1300)cos(60.89)}

V_f = 13.8863 \angle 60.89\°

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