The acceleration of an object is due to the net force on the object and the object's
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Answer:
The fluids speed at a) and b)
are
and
respectively
c) Th volume of water the pipe discharges is:
Explanation:
To solve a) and b) we should use flow continuity for ideal fluids:
(1)
With Q the flux of water, but Q is using this on (1) we have:
(2)
With A the cross sectional areas and v the velocities of the fluid.
a) Here, we use that point 2 has a cross-sectional area equal to , so now we can solve (2) for
:
b) Here we use point 2 as :
c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is , so we can write:
, solving for V:
Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced
(a) The final speeds of the ice sleds is approximately 0.49 m/s each
(b) The impulse on the cat is 11.0715 kg·m/s
(c) The average force on the right sled is 922.625 N
The reason for arriving at the above values is as follows:
The given parameters are;
The masses of the two ice sleds, m₁ = m₂ = 22.7 kg
The initial speed of the ice, v₁ = v₂ = 0
The mass of the cat, m₃ = 3.63 kg
The initial speed of the cat, v₃ = 0
The horizontal speed of the cat, v₃ = 3.05 m/s
(a) The required parameter:
The final speed of the two sleds
For the first jump to the right, we have;
By the law of conservation of momentum
Initial momentum = Final momentum
∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'
Where;
v₁' = The final velocity of the ice sled on the left
v₃' = The final velocity of the cat
Plugging in the values gives;
22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05
∴ 22.7 × v₁' = -3.63 × 3.05
v₁' = -3.63 × 3.05/22.7 ≈ -0.49
The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)
The final speed ≈ 0.49 m/s
For the second jump to the left, we have;
By conservation of momentum law, m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'
Where;
v₂' = The final velocity of the ice sled on the right
v₃' = The final velocity of the cat
Plugging in the values gives;
22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05
∴ 22.7 × v₂' = -3.63 × 3.05
v₂' = -3.63 × 3.05/22.7 ≈ -0.49
The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)
The final speed ≈ 0.49 m/s
(b) The required parameter;
The impulse of the force
The impulse on the cat = Mass of the cat × Change in velocity
The change in velocity, Δv = Initial velocity - Final velocity
Where;
The initial velocity = The velocity of the cat before it lands = 3.05 m/s
The final velocity = The velocity of the cat after coming to rest =
∴ Δv = 3.05 m/s - 0 = 3.05 m/s
The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s
(c) The required information
The average velocity
Impulse = × Δt
Where;
Δt = The time of collision = The time it takes the cat to finish landing = 12 ms
12 ms = 12/1000 s = 0.012 s
We get;
∴
The average force on the right sled applied by the cat while landing, = 922.625 N
Learn more about conservation of momentum here: