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mart [117]
3 years ago
15

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

tially at rest. After the collision, puck A has velocity 0.123 m/s to the left, and puck B has velocity 0.651 m/s to the right.
What was the speed vAi of puck A before the collision?

Calculate ΔK, the change in the total kinetic energy of the system that occurs during the collision.
Physics
1 answer:
irinina [24]3 years ago
3 0

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

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Explanation:

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A wooden object (conically shaped) has a diameter of 8cm and height of 14cm. It floats in oil with 6cm of its height above oil l
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Answer:

(a) The density of the object is 316/343 × the density of the oil

(b) The fraction of oil displaced after immersing the object is 0.461 of the oil volume

Explanation:

(a) The volume, V of a cone of height, h and base diameter, D = 2×r is given by the following equation;

V = \dfrac{\pi r^{2} h}{3}

The volume of the object is therefore;

\dfrac{\pi \times 4^{2} \times 14}{3} = 74\tfrac{2}{3}\pi \, cm^3

Where 6 cm is above the oil level we have;

\dfrac{\pi \times \left (6 \times \dfrac{4}{14}   \right )^{2} \times 6}{3} = 5\tfrac{43}{49}\pi \, cm^3 above the oil level

Therefore, volume of the oil displaced = 68\tfrac{116}{147}\pi cm³ = 216.11 cm³

The density of the object is thus;

\dfrac{68\tfrac{116}{147}\pi}{ 74\tfrac{2}{3}\pi} \times  Density \ of \ the \ oil = \dfrac{316}{343}  \right ) \times  Density \ of \ the \ oil

The density of the object = 316/343 × the density of the oil.

(b) The volume of the oil = 2 × Volume of the object = 2 \times 74\tfrac{2}{3}\pi \, cm^3 = 149\tfrac{1}{3}\pi \, cm^3

The fraction of the volume displaced, x, after immersing the object is given as follows;

x = \dfrac{68\tfrac{116}{147}\pi}{ 149\tfrac{1}{3}\pi} = \dfrac{158}{343} = 0.461

The fraction of oil displaced after immersing the object = 0.461 of the volume of the oil

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