Basically, you want to take the integral of each interval and compare them. The two intervals with the same integral represent equal displacement of the particle. And since delta(x) is always 2, all you have to do is average the initial and final velocities of each interval and multiply by two to find total displacement.

Hope it helped.

Edit to show calculations:

2 * [ (0 + 10)/2 ] = 10 for interval AB
2 * [ (7 + 3)/2 ] = 10 for interval DE

8 0

3 years ago

Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three.

SpyIntel [72]

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:

$p_{i} = p_{f} = m*v_{o} (1)$

Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:

$p_{f1} = 2*m*v_{1} (2)$

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:

$p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2} = m*v (4)$

Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:

$p_{2} = 3*m*v_{2} (5)$

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:

$p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3} = m*v (7)$

Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:

$p_{3} = 4*m*v_{3} (8)$

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.

5 0

3 years ago

Waxing or waning moon?

Dennis_Churaev [7]

Answer:

waning gibbious

Explanation:

7 0

3 years ago

Electromagnetic radiation behaves both as particles (called photons) and as waves. Wavelength (λ) and frequency (ν) are related

inessss [21]

<h2>Answer: 136.363 m</h2>