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Explain Newton's Second Law, the purpose of runway inclination, and how to calculate trolley velocity before a collision.

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3 parts
iii

0.60 m/s\text{0.60 m/s}

0.375 m/s\text{0.375 m/s}

iv

0.30 kg m/s\text{0.30 kg m/s}

i

6.20 eV\text{6.20 eV}

Step-by-step solution

8 steps

Here are the solutions for questions 6(a), 6(b), and 6(c).

6(a)
Newton's Second Law of Motion, which states that the net external force acting on an object is equal to the rate of change of its momentum (F=ΔpΔtF = \frac{\Delta p}{\Delta t}), is consistent with the law of conservation of linear momentum. If the net external force is zero, then the change in momentum is zero, meaning momentum is conserved.

6(b)

i) The runway is slightly inclined to compensate for friction. This ensures that the net force acting on the trolley along the runway is zero, allowing it to move at a constant velocity without external propulsion. It can be checked to be friction compensated by placing a trolley on the runway and giving it a slight push; if it moves at a constant velocity (or the ticker tape shows equally spaced dots), then friction is compensated.

Part ii

For momentum to be conserved in this experiment, the net external force acting on the system of trolleys must be zero.

Part iii

0.60 m/s\text{0.60 m/s}

  1. Calculate the time interval for 10 consecutive ticks.

    The ticker timer operates at a frequency of 50 Hz50 \text{ Hz}, so the time for one tick is T=1f=150 Hz=0.02 sT = \frac{1}{f} = \frac{1}{50 \text{ Hz}} = 0.02 \text{ s}.
    The time for 10 ticks is Δt=10×0.02 s=0.2 s\Delta t = 10 \times 0.02 \text{ s} = 0.2 \text{ s}.

  2. Calculate the velocity of trolley A before the collision.

    The length for 10 ticks before the collision is 12.0 cm=0.12 m12.0 \text{ cm} = 0.12 \text{ m}.

    uA=distancetime=0.12 m0.2 s=0.60 m/su_A = \frac{\text{distance}}{\text{time}} = \frac{0.12 \text{ m}}{0.2 \text{ s}} = 0.60 \text{ m/s}

    The initial velocity of trolley B is uB=0 m/su_B = 0 \text{ m/s} since it is at rest.

  3. Calculate the common velocity of the trolleys after the collision.

    The length for 10 ticks after the collision is 7.5 cm=0.075 m7.5 \text{ cm} = 0.075 \text{ m}.

    v=distancetime=0.075 m0.2 s=0.375 m/sv = \frac{\text{distance}}{\text{time}} = \frac{0.075 \text{ m}}{0.2 \text{ s}} = 0.375 \text{ m/s}

    The velocity of trolley A before collision is 0.60 m/s\boxed{\text{0.60 m/s}} and the common velocity after collision is 0.375 m/s\boxed{\text{0.375 m/s}}.

Part iv

0.30 kg m/s\text{0.30 kg m/s}

  1. Calculate the total momentum of the system before the collision.

    Mass of trolley A, mA=0.50 kgm_A = 0.50 \text{ kg}.
    Mass of trolley B, mB=0.30 kgm_B = 0.30 \text{ kg}.

    Pbefore=mAuA+mBuBP_{\text{before}} = m_A u_A + m_B u_B Pbefore=(0.50 kg)(0.60 m/s)+(0.30 kg)(0 m/s)P_{\text{before}} = (0.50 \text{ kg})(0.60 \text{ m/s}) + (0.30 \text{ kg})(0 \text{ m/s}) Pbefore=0.30 kg m/sP_{\text{before}} = 0.30 \text{ kg m/s}

  2. Calculate the total momentum of the system after the collision.

    Since the trolleys stick together, their combined mass is (mA+mB)(m_A + m_B).

    Pafter=(mA+mB)vP_{\text{after}} = (m_A + m_B) v Pafter=(0.50 kg+0.30 kg)(0.375 m/s)P_{\text{after}} = (0.50 \text{ kg} + 0.30 \text{ kg})(0.375 \text{ m/s}) Pafter=(0.80 kg)(0.375 m/s)P_{\text{after}} = (0.80 \text{ kg})(0.375 \text{ m/s}) Pafter=0.30 kg m/sP_{\text{after}} = 0.30 \text{ kg m/s}

  3. Compare the momentum before and after the collision.

    Since Pbefore=0.30 kg m/sP_{\text{before}} = 0.30 \text{ kg m/s} and Pafter=0.30 kg m/sP_{\text{after}} = 0.30 \text{ kg m/s}, the total momentum of the system before and after the collision is 0.30 kg m/s\boxed{\text{0.30 kg m/s}}. Therefore, momentum is conserved.

    6(c)

Part i

6.20 eV\text{6.20 eV}

  1. Calculate the energy of a photon in Joules.

    The wavelength of ultraviolet radiation is λ=200 nm=200×10−9 m\lambda = 200 \text{ nm} = 200 \times 10^{-9} \text{ m}.
    Planck's constant h=6.626×10−34 J sh = 6.626 \times 10^{-34} \text{ J s}.
    Speed of light c=3.00×108 m/sc = 3.00 \times 10^8 \text{ m/s}.

    E=hcλE = \frac{hc}{\lambda} E=(6.626×10−34 J s)(3.00×108 m/s)200×10−9 mE = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{200 \times 10^{-9} \text{ m}} E=1.9878×10−25 J m200×10−9 mE = \frac{1.9878 \times 10^{-25} \text{ J m}}{200 \times 10^{-9} \text{ m}} E=9.939×10−19 JE = 9.939 \times 10^{-19} \text{ J}

  2. Convert the energy from Joules to electron Volts.

    1 eV=1.602×10−19 J1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}.

    EeV=9.939×10−19 J1.602×10−19 J/eVE_{\text{eV}} = \frac{9.939 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} EeV≈6.20 eVE_{\text{eV}} \approx 6.20 \text{ eV}

    The energy of a photon of ultraviolet radiation is 6.20 eV\boxed{\text{6.20 eV}}.

    ii) The work function of the zinc plate is ϕ=4.3 eV\phi = 4.3 \text{ eV}. The energy of the incident ultraviolet photon is E=6.20 eVE = 6.20 \text{ eV}. Since E>ϕE > \phi, the incident ultraviolet radiation has enough energy to eject electrons from the zinc plate (photoelectric effect). As the electroscope is negatively charged, it has an excess of electrons. When electrons are emitted, the electroscope loses its negative charge, causing the leaf to fall.

    iii) If ultraviolet radiation is replaced by visible light whose maximum photon energy is 3.2 eV3.2 \text{ eV}, no change would be observed on the leaf of the electroscope. This is because the maximum energy of the visible light photon (3.2 eV3.2 \text{ eV}) is less than the work function of the zinc plate (4.3 eV4.3 \text{ eV}). Therefore, no electrons would be emitted from the zinc plate, and the electroscope would remain negatively charged with its leaf deflected.

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