If two objects collid and one is initially at rest, is it possible for both to be at rest after the collision?
1 answer:
Answer:
<em>Is it NOT possible for both particles to be at rest after the collision</em>
Explanation:
<u>Law Of Conservation Of Linear Momentum</u>
The total linear momentum of a system of particles is conserved regardless of their internal interactions while no external force is applied to the system. If and
are the masses of two particles, and only one of them is at rest, there is a non-zero total linear momentum, i.e.
if at least one of the speeds is not zero, the total initial linear momentum is not zero.
If after the collision, both particles remain at rest, both speeds will be zero and the total linear momentum won't be conserved.
So, is it NOT possible for both particles to be at rest after the collision
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Answer:
ωf = 13 rad/s
Explanation:
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where ω₀ = 5 rad/s, t = 4 s, α = 2 rad/s2
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Answer:
acceleration = 1.79 m/s^2
Tension = 6817 N
Explanation:
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Her weight is greater than the normal force (N = 450 N), so the elevator is going downwards.
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Where P is the weight of Elisabeth, N is her normal force, m is her mass and a is the acceleration. Then, we have that:
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Answer:
The distance traveled is 109.58 m
Explanation:
The Speed of sound in air = 344 m/s
Let the time in which the dime dropped by the man reach an impact in the well = t₁
Let the time in which the sound travel from the well to the man = t₂
Then
1/2× 10 × t₁² = 344 × t₂ which gives;
5 × t₁² = 344 × t₂.........................(1)
Also the total time before the man heard the dime = t₁ + t₂ = 5
Therefore;
t₂ = 5 - t₁
Substituting the value of t₂ in equation (1), we have;
5 × t₁² = 344 × (5 - t₁)
5·t₁² + 344·t₁ - 1720 = 0
Using the quadratic formula, we have;
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t₁ = 4.68 s or -73.48 s
Therefore, with the positive value for t₁ = 4.68 s, we have
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The distance traveled = 109.58 m.