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geniusboy [140]
3 years ago
7

Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially

stationary and rebounds with 70% of its initial kinetic energy. What is the magnitude of the change in momentum of object A?
Physics
1 answer:
AnnZ [28]3 years ago
6 0

Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.

Th magnitude of object A's momentum change is thus 25.71 kgm/s

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By the newtons Second Law:

F = ma

Solving for m:

m = F / a

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<h3>m = 50 kg → ANSWER</h3>
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A stunt cyclist rides on the interior of a cylinder 13 m in radius. The coefficient of static friction between the tires and the
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Answer:

v=13.1m/s

Explanation:

From the question we are told that:

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Explanation:

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A dentist’s drill starts from rest. After 3.20 s of constant angu-lar acceleration, it turns at a rate of 2.51 3 104 re v/m i n.
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Answer:

ΔTita = 4205.6 rad

Explanation:

w_{i} means initial angular velocity, which is 0 rev/min

w_{f} means final angular velocity, which is 2.513*10^{4}rev/min

t means time t= 3.20 s

one revolution is equivalent to 2πrad so the final angular velocity is:

w_{f} = (2π/60) *2.513*10^{4} rad/s

w_{f}= 2628.5 rad/s

so the angular acceleration, α will be:

α = 2628.5 rad/s / 3.20 s

a = 821.40 rad/s^{2}

so the rotational motion about a fixed axis is:

w^{2} _{f} =w^{2} _{i} + 2αΔTita    where ΔTita is the angle in radians

so now find the ΔTita the subject of the formula

ΔTita = \frac{w^{2} _{f}-w^{2} _{i}  }{2a}

ΔTita = ((2628.5)^{2} - (0 rev/min)^{2}) / 2* 821.40

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