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sasho [114]
4 years ago
12

Two identical soccer balls are rolled towards each other. What will be true after they collide head-on

Physics
2 answers:
tresset_1 [31]4 years ago
8 0
They will go away from each other
Furkat [3]4 years ago
6 0

Answer:

(v_{i_{1} }+v_{i_{2} }) = (v_{f_{1} }+v_{f_{2} })

Explanation:

There are some things we can say:

  • If the Collision was Elastic, then the conservation of energy applies, where the kinetic energy of the system before they collide will be the same after they collide.
  • If the collision was inelastic, then we can assert that the kinetic energy of the system before the collision and after the collision will be different, because there is some energy lost in the form of heat for example, though the energy of the entire system stays the same (if we add the losses).
  • The law of conservation of momentum states that the momentum of the system before the collision is exactly the same after the collision so if m_{1} =m_{2}:

m_{i_{1} } *v_{i_{1} }+m_{i_{2} } *v_{i_{2} } = m_{f_{1} } *v_{f_{1} }+m_{f_{2} } *v_{f_{2} }

m_{i_{1} } *(v_{i_{1} }+v_{i_{2} }) = m_{f_{1} } *(v_{f_{1} }+v_{f_{2} })

Since there is conservation of mass as well (we assume that nothing breaks apart after the collision):

(v_{i_{1} }+v_{i_{2} }) = (v_{f_{1} }+v_{f_{2} })

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Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha
krek1111 [17]

Answer:

F=18.58\times 10^{-11}\ N

\theta=30.276^{\circ}

Explanation:

Given:

mass of first particle, m_1=6.7\ kg

mass of second particle, m_2=5.1\ kg

mass of third particle, m_3=3.7\ kg

coordinate position of first particle in meters, (x_1,y_1)\equiv(4.2,0)

coordinate position of second particle in meters, (x_2,y_2)\equiv(0,2.8)

coordinate position of third particle in meters, (x_3,y_3)\equiv(0,0)

<u>Now, gravitational force on particle 3 due to particle 1:</u>

F_{31}=G\frac{m_1.m_3}{r_{31}^2}

F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}

F_{31}=9.37\times 10^{-11}\ N

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

F_{32}=G\frac{m_2.m_3}{r_{21}^2}

F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}

F_{32}=16.05\times 10^{-11}\ N

towards positive X axis.

<u>Now the net force</u>

F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}

F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}

F=18.58\times 10^{-11}\ N

<em>For angle in counterclockwise direction from the +x-axis</em>

tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}

\theta=30.276^{\circ}

4 0
4 years ago
A basketball referee tosses the ball straight up for the starting tipoff. at what velocity must a basketball player leave the gr
sladkih [1.3K]
Mgh=mv²/2

v=√2gh=√2·9.8·1.25=4.95m/s
5 0
4 years ago
A toy train engine rests motionless on a track. One student begins pushing the engine to the right with a force of 2 newtons. At
Svetradugi [14.3K]
I think it’s B

Hope it helps
8 0
3 years ago
A bowling ball has a mass of 6 kg. What happens to its momentum when its speed increases from 2m/s to 4 m/s?
Vitek1552 [10]
Here, Initial momentum = mu = 6*2 = 12 Kg m/s
Final momentum = mv = 6*4 = 24 Kg m/s

In short, Your Answer would be Option C

Hope this helps!
6 0
4 years ago
Read 2 more answers
What is the velocity of a car that travels -50 meters in 5 seconds then 10 meters in 5 seconds
diamong [38]

Answer:

2 m/s

Explanation:

The first part of the question the car is going in reverse or negative along the x axis. Then the second part the car is moving forward along the x axis. So the car would only have velocity in the current direction of movement. So our equation for velocity is as follows.

v = d/t

v = 10 m/5 s

v = 2 m/s

3 0
3 years ago
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