If the mass of a moving object is doubled, which variable will also double? *
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Answer:
the initial velocity of the ball is 104.67 m/s.
Explanation:
Given;
angle of projection, θ = 60⁰
time of flight, T = 18.5 s
let the initial velocity of the ball, = u
The time of flight is given as;
Therefore, the initial velocity of the ball is 104.67 m/s.
<u>Answer:</u> The energy gained by the air molecules is 0.011 J.
<u>Explanation:</u>
Law of conservation of energy states that energy can neither be created nor be destroyed but it can only be transformed from one form to another form.
Here, the potential energy of the coffee filter is getting converted into kinetic energy of the coffee filter and some energy is lost by it which is gained by the air molecules in the form of kinetic energy.
So, calculating the potential energy of coffee filter, we use the equation:
P = mgh
where,
m = mass of coffee filter = 1.2 g = (Conversion used: 1 kg = 1000 g)
g = acceleration due to gravity =
h = height of coffee filter = 1 m
Putting values in above equation, we get:
- Calculating the kinetic energy of coffee filter, we use the equation:
where,
m = mass of coffee filter = 1.2 g =
v = speed of coffee filter = 0.8 m/s
Putting values in above equation, we get:
As, energy lost by coffee filter = energy gained by air molecules
So, energy lost by coffee filter = Potential energy - Kinetic energy
Energy lost by coffee filter =
Hence, the energy gained by the air molecules is 0.011 J.
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: