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State the law of conservation of momentum.

Physics

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5. Figure 3 below show a portion of a ticker tape from a toy car running down a runway. The dots on the tape are made at regular time intervals of 0.02s.

5.a) Describe the motion of the toy car.
The spacing between consecutive dots on the ticker tape is increasing. Since the time interval between dots is constant (0.02 s0.02 \text{ s}), an increasing distance covered in equal time intervals indicates that the toy car is accelerating.

5.b) Calculate the initial velocity, u, of the toy car?
The initial measured segment has a length of 0.02 m0.02 \text{ m}.
The time interval for one segment is 0.02 s0.02 \text{ s}.

u=DistanceTimeu = \frac{\text{Distance}}{\text{Time}} u=0.02 m0.02 su = \frac{0.02 \text{ m}}{0.02 \text{ s}} u=1 m/su = 1 \text{ m/s}

The initial velocity of the toy car is 1 m/s\boxed{\text{1 m/s}}.

5.c) What is the final velocity, v, of the toy car over the measured portion?
The final measured segment has a length of 0.06 m0.06 \text{ m}.
The time interval for one segment is 0.02 s0.02 \text{ s}.

v=DistanceTimev = \frac{\text{Distance}}{\text{Time}} v=0.06 m0.02 sv = \frac{0.06 \text{ m}}{0.02 \text{ s}} v=3 m/sv = 3 \text{ m/s}

The final velocity of the toy car is 3 m/s\boxed{\text{3 m/s}}.

5.d) Determine the acceleration of the toy car over the measured portion.
To determine the acceleration, we use the initial velocity (uu) and final velocity (vv) calculated from the average velocities of the first and last segments, respectively. The time interval (Δt\Delta t) between the midpoints of these segments is needed.
The first segment's midpoint is at 0.01 s0.01 \text{ s}.
The last segment's midpoint is at 0.01 s+(2×0.02 s)=0.05 s0.01 \text{ s} + (2 \times 0.02 \text{ s}) = 0.05 \text{ s}.
The time duration between these two points is Δt=0.05 s−0.01 s=0.04 s\Delta t = 0.05 \text{ s} - 0.01 \text{ s} = 0.04 \text{ s}.

a=v−uΔta = \frac{v - u}{\Delta t} a=3 m/s−1 m/s0.04 sa = \frac{3 \text{ m/s} - 1 \text{ m/s}}{0.04 \text{ s}} a=2 m/s0.04 sa = \frac{2 \text{ m/s}}{0.04 \text{ s}} a=50 m/s2a = 50 \text{ m/s}^2

The acceleration of the toy car is 50 m/s2\boxed{\text{50 m/s}^2}.

6. Figure 4 above is a diagram of an uncalibrated mercury in glass thermometer. A and B show the mercury levels when the bulb is suspended inside the vapour from pure boiling water and then in pure melting ice at standard atmospheric temperature respectively.

6.a) Give the name and values of the temperatures A and B in degree Celsius.

  • A represents the *upper
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