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Tcecarenko [31]
3 years ago
12

AP Physics I, shouldn't be too hard.

Physics
1 answer:
Nana76 [90]3 years ago
5 0

Answer:

The correct option is;

D. The kinetic energy decreases by 3·m₀·v₀²

Explanation:

The given parameters are;

The mass of object X = m₀

The initial velocity of object X = v₀

The mass of object Y = 2·m₀

The initial velocity of object Y = -2·v₀

By conservation of linear momentum, we have;

The total initial momentum = The total final momentum

Therefore, we have;

The total initial momentum = m₀·v₀ - 2·m₀·2·v₀ = The total final momentum

∴ The total final momentum = -3·m₀·v₀

The total mass of the two object after sticking together = 2·m₀ + m₀ = 3·m₀

Therefore, the velocity of the two objects after collision = (The total final momentum)/(Total mass) = -3·m₀·v₀/(3·m₀) = -v₀

The kinetic energy = 1/2 × Mass × (Velocity)²

Therefore, the kinetic energy after collision = 1/2 × (3·m₀) × v₀² = 3·m₀·v₀²/2

The kinetic energy before collision = 1/2 × m₀ × v₀² + 1/2 × (2·m₀) × (2·v₀)² = (1/2 + 4) × (m₀·v₀²)

∴ The kinetic energy before collision =  9·(m₀·v₀²)/2

The change in kinetic energy = The kinetic energy after collision - The kinetic energy before collision = 3·m₀·v₀²/2 - 9·(m₀·v₀²)/2 = -3·m₀·v₀²

Therefore, the kinetic energy decreases by 3·m₀·v₀².

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Answer:

(a) v_f=2.414\ m.s^{-1}

(b) v_s=4.828\ m.s^{-1}

Explanation:

Let:

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<u>We know Kinetic Energy is given as</u>

KE=\frac{1}{2} m.v^2 .....................................(1)

where:

m = mass

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Hence, according to the initial condition the father is having kinetic energy half the kinetic energy of the son.

KE_s=2.KE_f

\frac{1}{2} m_s.v_s^2=2\times \frac{1}{2} m_f.v_f^2

(\frac{m_f}{2})\times v_s^2=2\times m_f\times v_f^2

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According to the final condition:

\frac{1}{2} m_s.v_s^2= \frac{1}{2} m_f.(v_f+1)^2

(\frac{m_f}{2})\times v_s^2=m_f.(v_f+1)^2

v_s^2=2(v_f+1)^2

v_s=\sqrt{2}(v_f+1).....................................................(3)

(a)

From eq. 2 & 3

2v_f=\sqrt{2}(v_f+1)

\sqrt{2}\ v_f=(v_f+1)

v_f(\sqrt{2}-1)=1

v_f=\frac{1}{(\sqrt{2}-1)}

v_f=2.414\ m.s^{-1}

(b)

<em>putting the above value in eq. (2)</em>

v_s=4.828\ m.s^{-1}

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