Answer:
2.705 m/s
Explanation:
formular for momentum(p):
P = m * v; where m and v represents mass in kg and velocity in m/s respectfully
total momemtum after an elastic collision we have;
Kindly attached to this mail.
When the collision occurs between two objects in an isolated system, then their total momentum before the collision is equal to the momentum after the collision. It is called momentum conservation principal.
The final velocity of the 22 [tex]\rm g[/tex] ball is 2.70 [tex]\rm m/s[/tex].
As given that the two balls collide, they are moving in separate direction to each other. So as per the law of momentum conservation, the total momentum loss by two balls before collision is equal to the total momentum gained by the two balls after collision.
The law of momentum conservation can be expressed by the given formula.
[tex]m_{1}v_{1}\;+\;m_{2}v_{2}\;=\;m_{3}v_{3}\;+\;m_{4}v_{4}[/tex]
Here [tex]m_{1}[/tex] and [tex]m_{2}[/tex] are mass of two objects before collision.
[tex]v_{1}[/tex] and [tex]v_{2}[/tex] are velocity of two objects before collision.
[tex]m_{3}[/tex] and [tex]m_{4}[/tex] are mass of two objects after collision.
[tex]v_{3}[/tex] and [tex]v_{4}[/tex] are velocity of two objects before collision.
Before collision, for the ball 1, Mass [tex]m_{1}\;=\;15\;\rm g[/tex] and initial velocity [tex]v_{1}\;=\;3.5\;\rm m/s[/tex]. For the ball 2, Mass [tex]m_{2}\;=\;22\;\rm g[/tex] and initial velocity [tex]v_{2}\;=\;4.0\;\rm m/s[/tex].
After collision, for the ball 1, Mass [tex]m_{3}\;=\;15\;\rm g[/tex] and final velocity [tex]v_{3}\;=\;5.4\;\rm m/s[/tex]. For the ball 2, Mass [tex]m_{4}\;=\;22\;\rm g[/tex] and final velocity [tex]v_{2}\;=\;?\;\rm m/s[/tex].
Substituting all the values in above formula,
[tex]15\;\times\;3.5\;+\;22\;\times\;4.0\;=\;15\;\times\;5.4\;+\;22\;\times\;v_{4}[/tex]
[tex]v_{4}\;=\;2.70\;\rm m/s[/tex]
Hence the final velocity for 22[tex]\rm g[/tex] ball is 2.70 [tex]\rm m/s[/tex].
For more details, follow the link given below.
https://brainly.com/question/4956182.
