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1 Calculate h, the change in its vertical height. 4.2 Analyse the equation and calculation that you used to answer Question 4.1 and explain whether you needed to know the mass of the trolley to do the calculation.

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2 parts
4.1

0.24 m\text{0.24 m}

4.3

2.23 m/s\text{2.23 m/s}

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Here are the calculations for the questions:

Question 4: A trolley of mass 1 kg1 \text{ kg} rolls down an inclined plane. The trolley on the inclined plane has an initial velocity of 0.2 m/s0.2 \text{ m/s}. When it reaches the bottom its velocity is 2.2 m/s2.2 \text{ m/s}.

Part 4.1· 3 steps

Calculate hh, the change in its vertical height.

  1. Apply the principle of conservation of mechanical energy.

    Let the initial height be hh and the final height be 0 m0 \text{ m}.

    12mvi2+mghi=12mvf2+mghf\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f

  2. Substitute the given values (m=1 kgm = 1 \text{ kg}, vi=0.2 m/sv_i = 0.2 \text{ m/s}, hi=hh_i = h, vf=2.2 m/sv_f = 2.2 \text{ m/s}, hf=0 mh_f = 0 \text{ m}, g=9.8 m/s2g = 9.8 \text{ m/s}^2).

    12(1 kg)(0.2 m/s)2+(1 kg)(9.8 m/s2)h=12(1 kg)(2.2 m/s)2+(1 kg)(9.8 m/s2)(0 m)\frac{1}{2}(1 \text{ kg})(0.2 \text{ m/s})^2 + (1 \text{ kg})(9.8 \text{ m/s}^2)h = \frac{1}{2}(1 \text{ kg})(2.2 \text{ m/s})^2 + (1 \text{ kg})(9.8 \text{ m/s}^2)(0 \text{ m})

  3. Simplify and solve for hh.

    0.5(0.04)+9.8h=0.5(4.84)+00.5(0.04) + 9.8h = 0.5(4.84) + 0 0.02+9.8h=2.420.02 + 9.8h = 2.42 9.8h=2.42−0.029.8h = 2.42 - 0.02 9.8h=2.409.8h = 2.40 h=2.409.8h = \frac{2.40}{9.8} h≈0.24 mh \approx 0.24 \text{ m}

    The change in vertical height is 0.24 m\boxed{\text{0.24 m}}.

Part 4.2

Analyse the equation and calculation that you used to answer Question 4.1 and explain whether you needed to know the mass of the trolley to do the calculation.

The equation used is the conservation of mechanical energy: 12mvi2+mghi=12mvf2+mghf\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f. If we divide every term by the mass mm, the equation becomes 12vi2+ghi=12vf2+ghf\frac{1}{2}v_i^2 + gh_i = \frac{1}{2}v_f^2 + gh_f. This shows that the mass mm cancels out. Therefore, you did not need to know the mass of the trolley to calculate the change in vertical height.

Part 4.3

2.23 m/s\text{2.23 m/s}

Given the same situation and the same initial and final velocities, would the change in vertical height be the same for a 2 kg2 \text{ kg} trolley?

Yes, the change in vertical height would be the same for a 2 kg2 \text{ kg} trolley. As explained in 4.2, the mass cancels out in the conservation of mechanical energy equation when calculating height change based on velocity changes.

Question 5: Figure 8 shows a pendulum bob of mass 100 g100 \text{ g} that is 100 mm100 \text{ mm} above its lowest point and has a speed of 2 m/s2 \text{ m/s}. Calculate its speed when it is 50 mm50 \text{ mm} above its lowest point.

  1. Convert units to SI units.

    m=100 g=0.1 kgm = 100 \text{ g} = 0.1 \text{ kg} h1=100 mm=0.1 mh_1 = 100 \text{ mm} = 0.1 \text{ m} h2=50 mm=0.05 mh_2 = 50 \text{ mm} = 0.05 \text{ m}

  2. Apply the principle of conservation of mechanical energy.

    12mv12+mgh1=12mv22+mgh2\frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2

  3. Substitute the known values (m=0.1 kgm = 0.1 \text{ kg}, v1=2 m/sv_1 = 2 \text{ m/s}, h1=0.1 mh_1 = 0.1 \text{ m}, h2=0.05 mh_2 = 0.05 \text{ m}, g=9.8 m/s2g = 9.8 \text{ m/s}^2).

    12(0.1 kg)(2 m/s)2+(0.1 kg)(9.8 m/s2)(0.1 m)=12(0.1 kg)v22+(0.1 kg)(9.8 m/s2)(0.05 m)\frac{1}{2}(0.1 \text{ kg})(2 \text{ m/s})^2 + (0.1 \text{ kg})(9.8 \text{ m/s}^2)(0.1 \text{ m}) = \frac{1}{2}(0.1 \text{ kg})v_2^2 + (0.1 \text{ kg})(9.8 \text{ m/s}^2)(0.05 \text{ m})

  4. Simplify and solve for v22v_2^2.

    0.5(0.1)(4)+0.098=0.05v22+0.0490.5(0.1)(4) + 0.098 = 0.05v_2^2 + 0.049 0.2+0.098=0.05v22+0.0490.2 + 0.098 = 0.05v_2^2 + 0.049 0.298=0.05v22+0.0490.298 = 0.05v_2^2 + 0.049 0.05v22=0.298−0.0490.05v_2^2 = 0.298 - 0.049 0.05v22=0.2490.05v_2^2 = 0.249 v22=0.2490.05v_2^2 = \frac{0.249}{0.05} v22=4.98 m2/s2v_2^2 = 4.98 \text{ m}^2/\text{s}^2

  5. Calculate v2v_2.

    v2=4.98 m2/s2v_2 = \sqrt{4.98 \text{ m}^2/\text{s}^2} v2≈2.23 m/sv_2 \approx 2.23 \text{ m/s}

    The speed of the pendulum bob when it is 50 mm50 \text{ mm} above its lowest point is 2.23 m/s\boxed{\text{2.23 m/s}}.

    Question 6: Figure 9 shows a roller coaster of mass 500 kg500 \text{ kg} rolling down a section of track. At point A its speed is 10 m/s10 \text{ m/s} and its height above the lowest point on the track is 30 m30 \text{ m}. Point B is 15 m15 \text{ m} above the lowest point. Show by calculation whether the roller coaster has sufficient mechanical energy to reach point B. (Assume that no energy is lost in overcoming friction.)

  6. Calculate the total mechanical energy (MEAME_A) of the roller coaster at point A.

    MEA=KEA+GPEAME_A = KE_A + GPE_A MEA=12mvA2+mghAME_A = \frac{1}{2}mv_A^2 + mgh_A MEA=12(500 kg)(10 m/s)2+(500 kg)(9.8 m/s2)(30 m)ME_A = \frac{1}{2}(500 \text{ kg})(10 \text{ m/s})^2 + (500 \text{ kg})(9.8 \text{ m/s}^2)(30 \text{ m}) MEA=12(500)(100)+147000ME_A = \frac{1}{2}(500)(100) + 147000 MEA=25000 J+147000 JME_A = 25000 \text{ J} + 147000 \text{ J} MEA=172000 JME_A = 172000 \text{ J}

  7. Calculate the gravitational potential energy (GPEBGPE_B) required for the roller coaster to reach point B.

    GPEB=mghBGPE_B = mgh_B GPEB=(500 kg)(9.8 m/s2)(15 m)GPE_B = (500 \text{ kg})(9.8 \text{ m/s}^2)(15 \text{ m}) GPEB=73500 JGPE_B = 73500 \text{ J}

  8. Compare the mechanical energy at point A with the potential energy required to reach point B.

    Since MEA=172000 JME_A = 172000 \text{ J} and GPEB=73500 JGPE_B = 73500 \text{ J}, and MEA>GPEBME_A > GPE_B, the roller coaster has sufficient mechanical energy to reach point B. It will have kinetic energy remaining at point B.

    The roller coaster has sufficient mechanical energy to reach point B.

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