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Explain the difference between elastic and inelastic collisions. State Newton's second law of motion and apply it to calculate pressure in a hydraulic system.

Physics

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iii

2.5×105 Pa2.5 \times 10^5 \, \text{Pa}

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Part ii

This question refers to a previous context (an impact) which is not provided in the image. Therefore, it cannot be answered.

Part iii

2.5×105 Pa2.5 \times 10^5 \, \text{Pa}

  • In an elastic collision, both momentum and kinetic energy are conserved. Objects bounce off each other without any loss of kinetic energy.
  • In an inelastic collision, momentum is conserved, but kinetic energy is not conserved. Some kinetic energy is converted into other forms of energy, such as heat or sound, and the objects may stick together.

(iv) Newton's 2nd law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. It is expressed as F=maF = ma, where FF is the net force, mm is the mass, and aa is the acceleration.

  1. (a) (i)

  1. Identify the given values.

    Force applied to the smaller piston, F=50 NF = 50 \, \text{N}.
    Area of the smaller piston, A=2 cm2A = 2 \, \text{cm}^2.

  2. Convert the area to square meters.

    A=2 cm2×(1 m100 cm)2A = 2 \, \text{cm}^2 \times \left(\frac{1 \, \text{m}}{100 \, \text{cm}}\right)^2 A=2 cm2×1 m210000 cm2A = 2 \, \text{cm}^2 \times \frac{1 \, \text{m}^2}{10000 \, \text{cm}^2} A=0.0002 m2A = 0.0002 \, \text{m}^2 A=2×10−4 m2A = 2 \times 10^{-4} \, \text{m}^2

  3. Calculate the pressure produced in the oil at X using the formula P=FAP = \frac{F}{A}.

    P=50 N2×10−4 m2P = \frac{50 \, \text{N}}{2 \times 10^{-4} \, \text{m}^2} P=500.0002 PaP = \frac{50}{0.0002} \, \text{Pa} P=250000 PaP = 250000 \, \text{Pa} P=2.5×105 PaP = 2.5 \times 10^5 \, \text{Pa}

    The pressure produced in the oil at X is 2.5×105 Pa\boxed{2.5 \times 10^5 \, \text{Pa}}.

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