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3 years ago

15

The motion of a particle is described by x = 10 sin (πt + π/3), where x is in meters and t is in seconds. At what time in second

s is the potential energy equal to the kinetic energy?

1 answer:

OverLord2011 [107]3 years ago

7 0

Answer:

5/12

Explanation:

Given

x = 10 sin (πt + π/3)

v = distance/time

So, v = dx/dt

Differentiating x with respect to t

v = 10π cos(πt + π/3)

Also,

½kx² = ½mv²

Substituting values for x and v in the above equation

½k(10sin (πt + π/3))² = ½m(10πcos(πt + π/3))²

Divide through by ½

k(10sin (πt + π/3))² = m(10πcos(πt + π/3))²

Open both bracket

100ksin²(πt + π/3) = 100mπ²cos²(πt + π/3)

Divide through by 100

ksin²(πt + π/3) = mπ²cos²(πt + π/3)

Divide through by kcos²(πt + π/3)

ksin²(πt + π/3) ÷ kcos²(πt + π/3) = mπ²cos²(πt + π/3) ÷ kcos²(πt + π/3)

tan²(πt + π/3) = mπ²/k

tan²(πt + π/3) = (m/k)π²

But w² = k/m and w = 2π/T

(2π/T)² = k/m

(2π)²/T² = k/m

1/T² = k/m ÷ (2π)²

1/T² = k/m*(2π)²

T² = m(2π)²/k

From the Question, T is when πt = 2π or T = 2

Substitute 2 for T in the above equation

2² = m(2π)²/k

4 = m(2π)²/k

4 = 4π²m/k

m/k = 1/π²

(m/k)π² = 1

Remember that tan²(πt + π/3) = (m/k)π²

So, tan²(πt + π/3) = 1

This gives

πt + π/3 = 45° = π/4

πt + π/3 = π/4

Divide through by π

t + ⅓ = ¼

t = ¼ - ⅓

t = -1/12 --- Negative

Using the second quadrant

πt + π/3 = 3π/4

Divide through by π

t + ⅓ = ¾

t = ¾ - ⅓

t = 5/12

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a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

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a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

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$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

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Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

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$2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}$

$13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}$

Let is clear $v_{1,f}$ in first equation:

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The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

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$v_{1,f} = 5.86-2.25\cdot (3.606)$

$v_{1,f} = -2.254\,\frac{m}{s}$

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

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