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v2f =
DEFINI
A 0.16 kg billiard ball moving to the right at 1.2 m/s has a head-on
elastic
collision with another ball of equal mass moving to the left at 0.85
m/s. The
first ball moves to the left at 0.85 m/s after the collision. Find the
velocity of
the second ball after the collision, and verify your answer by
calculating the
total kinetic energy before and after the collision.
CAN YOU WORK IT OUT PLEASE I HAVE THIS TEST NEXT PERIOD


V2f DEFINI A 016 Kg Billiard Ball Moving To The Right At 12 Ms Has A Headon Elastic Collision With Another Ball Of Equal Mass Moving To The Left At 085 Ms The F class=

Sagot :

yh that is mot the correct answer great job

lashup

The velocity of the 0.16 kg second ball after the collision with the first ball of equal mass is 1.2 m/s. The first ball's velocity before and after the collision is 1.2 and 0.85 m/s, respectively, and the second ball's velocity before the collision is 0.85 m/s.    

We can find the velocity of the second ball after the collision by conservation of linear momentum:

[tex] p_{i} = p_{f} [/tex]

[tex] m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f} [/tex]

Where:

m₁ and m₂ are the masses of the first and second ball, respectively = m = 0.16 kg

[tex]v_{1}_{i}[/tex]: is the initial velocity of the first ball = 1.2 m/s

[tex]v_{1}_{f}[/tex]: is the final velocity of the first ball = 0.85 m/s

[tex]v_{2}_{i}[/tex]: is the initial velocity of the second ball = 0.85 m/s

[tex]v_{2}_{f}[/tex]: is the final velocity of the second ball =?

Taking the direction to the right as the positive direction of motion, we have:

[tex] m(v_{1}_{i} - v_{2}_{i}) = m(-v_{1}_{f} + v_{2}_{f}) [/tex]  

[tex] v_{2}_{f} = v_{1}_{i} - v_{2}_{i} + v_{1}_{f} [/tex]

[tex] v_{2}_{f} = (1.2 - 0.85 + 0.85) m/s = 1.2 m/s [/tex]

Hence, the velocity of the second ball after the collision is 1.2 m/s to the right.

We can verify the answer by calculating the total kinetic energy before and after the collision, as follows:

[tex]K_{i} = \frac{1}{2}m(v_{1}_{i})^{2} + \frac{1}{2}m(v_{2}_{i})^{2} = \frac{1}{2}0.16 kg*[(1.2 m/s)^{2} + (0.85 m/s)^{2})] = 0.173 J[/tex]

[tex]K_{f} = \frac{1}{2}m[(v_{1}_{f})^{2} + (v_{2}_{f})^{2}] = \frac{1}{2}0.16 kg*[(0.85 m/s)^{2} + (1.2 m/s)^{2})] = 0.173 J[/tex]

Since the total kinetic energy before and after the collision is the same, the final velocity of the second ball after the collision is correct.  

Therefore, the velocity of the second ball after the collision is 1.2 m/s.

Learn more here:

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I hope it helps you!