The total moment in an islated system remains constant through time. a) V₂ = 109.43 m/s and V₃ = 33.33 m/s. b) h = 56.67 m. c) d = 480 m.
What is the law of conservation of momentum?
First let us remember that momentum or lineal momentum is a motion quantity. In physics it is the fundamental quantity that characterizes the motion of any object.
The momentum is a vectorial quantity that can be calculated as the product of the object mass by its lineal velocity ⇒ mv
The total lineal moment of a system constituted by a group of objects is the sum of the vectorial moments of each of the objects.
In an isolated system -the one that does not interact with the exterior environment-, the moment remains constant through time. This is the law of conservation of momentum. The initial moment is equal to the final moment ⇒ Pinitial = Pfinal
Objects involved in crushes, explosions, collitions, and others, are considered to be isolated systems.
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So, in the exposed example we know that,
- Projectile mass ⇒2kg.
- Velocity = 50 m/s.
- Angle⇒ 40°.
- Smaller masses ⇒ 1 kg with 10m/s speed, 0.7 kg, and 0.3 kg.
a) We need to calculate the speed of the 0.7 and 0.3 kg objects.
So, 0.7 kg object moves in the original forward direction, meaning that we can use the following formula to calculate its velocity,
m₀v₀ = m₂v₂
Where
- m₀ = projectile mass = 2kg
- v₀ = horizontal component of projectile velocity
- m₂ = object mass = 0.7 kg
- v₂ = object velocity = ??
So first, we need to get the horizontal component of projectile velocity, V₀.
V₀ = V cosθ
V₀ = 50 cos40º
V₀ = 38.3 m/s
Now we can calculate the velocity of the 0.7 kg object.
m₀v₀ = m₂v₂
2 kg x 38.3 m/s = 0.7 kg x v₂
v₂ = (2 x 38.3) / 0.7
v₂ = 109.43 m/s
Now we need to calculate the velocity of the 0.3 kg object. Since this part goes exactly in the opposite direction of the 10 kg object, their addition will equal zero. So we can use the following equation to calculate the velocity of the 0.3 kg object.
m₁v₁ + m₃v₃ = 0
Where
- m₁ = 1 kg object mass
- v₁ = 1 kg object velocity = 10m/s
- m₃ = 0.3 kg object mass
- v₃ = object velocity = ??
m₁v₁ + m₃v₃ = 0
(1 kg x 10m/s) + (0.3 kg x v₃ ) = 0
(0.3 kg x v₃ ) = - (1 kg x 10m/s)
v₃ = - (1 kg x 10m/s) / 0.3
v₃ = - 33.33 m/s
The - sign is indicating the opposite direction concerning the 1 kg object.
ANSWER:
- The speed of the 0.7 kg object is 109.43 m/s
- The speed of the 0.3 kg objects is 33.33 m/s
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b) We need to get the high from the break-up point where the 0.3-kg piece go before rest
At rest,
1/2 m₃v₃² = m₃gh
Where,
- m₃ = 0.3 kg object mass
- v₃ = object velocity = 33.33m/s
- g = gravity = 9.8 m/s²
- h = high = ??
1/2 m₃v₃² = m₃gh
h = 1/2 (v₃²) /g
h = 1/2 (33.33²) / 9.8
h = 56.67 m
ANSWER: The high from the break-up point where the 0.3-kg piece go before coming to rest is 56.67 m.
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c) Finally, we need to get where does the 0.7-kg piece land
ANSWER: The distance at which the 0.7 piece lands, concerning the point from which it was fired is 480 m.
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