7. The diagram shows a mercury manometer used to measure the pressure of gas in a container.
1 answer:
You might be interested in
Answer:
The speed decreases 75%.
Explanation:
- Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
- For the first collission, only mass 1 is moving before it, so we can write the following equation:
- Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:
- From (1) and (2) we get:
- v₁ = v₀/2 (3)
- Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:
- Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:
- From (4) and (5) we get:
- v₂ = v₀/3 (6)
- Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:
- Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:
- From (7) and (8) we get:
- v₃ = v₀/4
- This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.
The speed is 0.956 m / s.
<u>Explanation</u>:
The kinetic energy is equal to the product of half of an object's mass, and the square of the velocity.
K.E = 1/2 m
where K.E represents the kinetic energy,
m represents the mass,
v represents the velocity.
K.E = 1/2 m
1.10 10^42 = 1/2
3.26
10^31
= (1.10
10^42
2) / (3.26
10^31)
v = 0.956 m / s.