6 (a) State the principle of conservation of energy. The principle of conservation of energy states that energy cannot be created or destroyed, but can only be transformed from one form to another. (b) Fig. 6.1 shows a glider on a horizontal frictionless track. The mass of the glider is 0.25 kg. One end of a string is fixed to the glider and the other end to a 0.10 kg mass. The 0.10 kg mass is held stationary at a height of 0.60 m from the ground. The pulley is more than 0.60 m away from the front of the glider. When the 0.10 kg mass is released, the glider has a constant acceleration of 2.8 m/s² towards the pulley. The 0.10 kg mass instantaneously comes to rest when it hits the ground. (i) Calculate the loss in potential energy of the 0.10 kg mass as it falls through the distance of 0.60 m. (ii) The glider starts from rest. Show that the velocity of the glider after travelling a distance of 0.60 m is about 1.8 m s⁻¹. (iii) Calculate the kinetic energy of the glider at this velocity of 1.8 m s⁻¹. (iv) Explain why the answer to (b)(iii) is not the same as (b)(i).

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6 (a) State the principle of conservation of energy - OCR - A-Level Physics A - Question 6 - 2011 - Paper 1

Question 6

6 (a) State the principle of conservation of energy. The principle of conservation of energy states that energy cannot be created or destroyed, but can only be tran... show full transcript

Worked Solution & Example Answer:6 (a) State the principle of conservation of energy - OCR - A-Level Physics A - Question 6 - 2011 - Paper 1

Step 1

(b)(i) Calculate the loss in potential energy of the 0.10 kg mass as it falls through the distance of 0.60 m.

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Answer

To find the loss in potential energy (PE), we use the formula:

$PE = mgh$

where:

$m = 0.10 \, \text{kg}$ (mass of the object)
$g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity)
$h = 0.60 \, \text{m}$ (height)

Calculating: $PE = 0.10 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.60 \, \text{m} = 0.5886 \, \text{J}$

Thus, the loss in potential energy is approximately 0.59 J.

Step 2

(b)(ii) Show that the velocity of the glider after travelling a distance of 0.60 m is about 1.8 m s⁻¹.

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Answer

Using the equation of motion for uniformly accelerated motion:

$v^2 = u^2 + 2as$

where:

$v$ is the final velocity
$u = 0 \, \text{m/s}$ (initial velocity, starts from rest)
$a = 2.8 \, \text{m/s}^2$ (acceleration)
$s = 0.60 \, \text{m}$ (distance)

Substituting the values: $v^2 = 0 + 2(2.8)(0.60) = 3.36$ $v = \sqrt{3.36} \approx 1.83 \, \text{m/s}$

Thus, the velocity of the glider is approximately 1.8 m/s.

Step 3

(b)(iii) Calculate the kinetic energy of the glider at this velocity of 1.8 m s⁻¹.

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Answer

The kinetic energy (KE) is calculated using the formula:

$KE = \frac{1}{2} mv^2$

where:

$m = 0.25 \, \text{kg}$ (mass of the glider)
$v = 1.8 \, \text{m/s}$ (velocity)

Substituting the values: $KE = \frac{1}{2}(0.25)(1.8)^2 \approx \frac{1}{2}(0.25)(3.24) = 0.405 \, \text{J}$

Thus, the kinetic energy of the glider is approximately 0.41 J.

Step 4

(b)(iv) Explain why the answer to (b)(iii) is not the same as (b)(i).

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Answer

The answer to (b)(iii) represents the kinetic energy of the glider when it has fallen a distance of 0.60 m and is moving at a velocity of 1.8 m/s. In contrast, (b)(i) calculates the potential energy lost by the 0.10 kg mass as it falls through the same distance.

The discrepancy arises due to energy transformation; while the potential energy lost is converted into kinetic energy, not all potential energy converts into the glider's kinetic energy due to the presence of the pulley system and forces, like tension and possibly friction, that may be at play.

6 (a) State the principle of conservation of energy - OCR - A-Level Physics A - Question 6 - 2011 - Paper 1

Worked Solution & Example Answer:6 (a) State the principle of conservation of energy - OCR - A-Level Physics A - Question 6 - 2011 - Paper 1

(b)(i) Calculate the loss in potential energy of the 0.10 kg mass as it falls through the distance of 0.60 m.

(b)(ii) Show that the velocity of the glider after travelling a distance of 0.60 m is about 1.8 m s⁻¹.

(b)(iii) Calculate the kinetic energy of the glider at this velocity of 1.8 m s⁻¹.

(b)(iv) Explain why the answer to (b)(iii) is not the same as (b)(i).

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