Motion Forces And Energy All subtopics

Fill the measuring cylinder with water and record the initial volume; gently lower the stone into the water; record the new water level reading at the bottom of the meniscus at eye level; subtract the initial volume from the final volume to obtain the volume of the stone.

Viewing from above causes a parallax error; the student reads a value that is higher than the true volume because the line of sight strikes the scale above the bottom of the meniscus; therefore the recorded volume is too large.

A light gate removes human reaction time from the measurement; therefore the time recorded is more accurate, especially for short time intervals.

Human reaction time is approximately 0.2 s; this occurs both when starting and stopping the watch, so the total uncertainty could be up to 0.4 s; this is a large fraction of the 0.8 s interval (up to 50%); therefore the percentage uncertainty is very high.

Finding the thickness of one sheet $$\text{thickness}=\frac{4.2}{50}$$ $$\text{thickness}=0.084\text{ mm}$$ Each sheet has a thickness of 0.084 mm.

Human reaction time causes an uncertainty of approximately 0.2-0.4 s in the total time; when 20 oscillations are timed, this uncertainty is shared across all 20 periods, so the error per period is approximately 0.01-0.02 s; when only 1 oscillation is timed, the full reaction-time error applies to that single period; therefore the percentage uncertainty per period is much smaller when multiples are timed.

A vector quantity has magnitude and direction.

Velocity is speed in a stated direction; although the speed remains 15 m s⁻¹, the car continuously changes direction as it moves around the circle; because the direction changes, the velocity changes; therefore the velocity is not constant.

Finding the resultant magnitude Equation used $$R=\sqrt{{v}_{1}^{2}+{v}_{2}^{2}}$$ Given $${v}_{1}=8.0{\text{ m s}}^{-1}\text{ (east)}$$ $${v}_{2}=6.0{\text{ m s}}^{-1}\text{ (north)}$$ Substitution $$R=\sqrt{{8.0}^{2}+{6.0}^{2}}$$ $$R=\sqrt{64.0+36.0}$$ $$R=\sqrt{100.0}$$ $$R=10.0{\text{ m s}}^{-1}$$ Finding the direction $$tan\theta =\frac{6.0}{8.0}$$ $$tan\theta =0.75$$ $$\theta ={tan}^{-1}(0.75)$$ $$\theta =36.9^{\circ}$$ The resultant velocity is 10.0 m s⁻¹ at 37° north of east (2 s.f.).

The two forces act at right angles, not in the same direction; vector addition requires accounting for direction, so the resultant is the hypotenuse of a right-angled triangle. Using Pythagoras’ theorem: $$R=\sqrt{{5.0}^{2}+{5.0}^{2}}$$ $$R=\sqrt{50}$$ $$R\approx 7.1\text{ N}$$ This is less than 10.0 N because the forces partly oppose each other’s directional effect; simple arithmetic addition (10.0 N) applies only when both forces act in exactly the same direction.

Speed is the distance travelled per unit time.

Converting distance to SI units: $$s=\frac{9.0\times 1000}{1}=9000\text{ m}$$ Converting time to SI units: $$t=5\times 60=300\text{ s}$$ Equation used — Speed: $$v=\frac{s}{t}$$ Substitution: $$v=\frac{9000}{300}$$ $$v=30{\text{ m s}}^{-1}$$

Speed is the distance travelled per unit time and is a scalar; velocity is speed in a stated direction and is a vector.

The velocity has changed; although the speed remains 4.0 m s⁻¹, the direction has reversed from right to left. Because velocity is speed in a stated direction, a change in direction means a change in velocity.

Average speed = total distance travelled / total time taken.

Step 1: Finding the time for the first section. Converting distance to SI units: $${s}_{1}=\frac{6.0\times 1000}{1}=6000\text{ m}$$ Rearranging for time: $$t=\frac{s}{v}$$ $${t}_{1}=\frac{6000}{20}=300\text{ s}$$ Step 2: Finding the time for the second section. $${s}_{2}=\frac{4.0\times 1000}{1}=4000\text{ m}$$ $${t}_{2}=\frac{4000}{10}=400\text{ s}$$ Step 3: Finding the average speed. $$\text{total distance}=6000+4000=10 000\text{ m}$$ $$\text{total time}=300+400=700\text{ s}$$ $$\text{average speed}=\frac{10 000}{700}$$ $$\text{average speed}=14.3{\text{ m s}}^{-1}\text{ (3 s.f.)}$$

Calculate the gradient of the straight-line section of the graph; divide the change in distance (read from the y-axis) by the change in time (read from the x-axis); the gradient equals the speed.

From 0 s to 5 s the object moves with constant speed; the speed equals 20 ÷ 5 = 4.0 m s⁻¹. From 5 s to 10 s the object is at rest; the distance does not change, so the gradient is zero and the speed is zero.

The area under a speed–time graph represents the distance travelled by the object.

The area under the graph forms a trapezium. $$\text{distance}=\frac{1}{2}\times (a+b)\times h$$ where $a$ and $b$ are the two parallel sides (the two speeds) and $h$ is the time. $$\text{distance}=\frac{1}{2}\times (30+10)\times 8.0$$ $$\text{distance}=\frac{1}{2}\times 40\times 8.0$$ $$\text{distance}=160\text{ m}$$