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State the difference between a scalar quantity and a vector quantity, giving one example of each.
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a scalar has only magnitude / a vector has both magnitude and direction; example of scalar (e.g. mass / energy / time) and example of vector (e.g. force / velocity / displacement)
Identify which of the following are vector quantities: speed, weight, temperature, momentum, energy, acceleration.
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weight; momentum; acceleration (all three required for both marks)
A force of 120 N acts at 40° above the horizontal. Calculate the horizontal and vertical components of this force.
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Finding the horizontal component $${F}_{x}=120cos40^{\circ}=91.9\text{ N}$$ Finding the vertical component $${F}_{y}=120sin40^{\circ}=77.1\text{ N}$$
A velocity vector has a horizontal component of 6.0 m s⁻¹ and a vertical component of 8.0 m s⁻¹. Determine the magnitude and direction of the velocity.
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The horizontal (6.0 m s⁻¹) and vertical (8.0 m s⁻¹) components are perpendicular, so combine them with Pythagoras for the magnitude and use the tangent ratio for the direction.
Magnitude: v = √(vₓ² + v_y²) = √(6.0² + 8.0²) = √(36 + 64) = √100 = 10 m s⁻¹
Direction (angle above the horizontal): θ = tan⁻¹($v_{\mathrm{y}}$ / vₓ) = tan⁻¹(8.0 / 6.0) = tan⁻¹(1.333) = 53° above the horizontal (53.1°; equivalently 37° from the vertical)
Result: the velocity has a magnitude of 10 m s⁻¹ directed at 53° above the horizontal.
Two perpendicular forces of 9.0 N and 12.0 N act on a point. Calculate the magnitude of the resultant force.
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Because the two forces are perpendicular, they form the two legs of a right-angled triangle, so the magnitude of the resultant is found with Pythagoras' theorem:
R = √(9.0² + 12.0²) R = √(81 + 144) R = √225 R = 15 N
The magnitude of the resultant force is 15 N. (This is a 3-4-5 right triangle scaled ×3.) If a direction is required, it acts at θ = tan⁻¹(9.0/12.0) ≈ 37° from the 12.0 N force (equivalently tan⁻¹(12.0/9.0) ≈ 53° from the 9.0 N force).
Show that two perpendicular velocities of 24 m s⁻¹ and 7.0 m s⁻¹ produce a resultant velocity of magnitude 25 m s⁻¹.
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$$R=\sqrt{{24}^{2}+{7.0}^{2}}$$ $$R=\sqrt{576+49}=\sqrt{625}=25{\text{ m s}}^{-1}$$ as required.
Explain why the horizontal velocity of a projectile remains constant when air resistance is neglected.
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there is no horizontal force acting on the projectile (only gravity acts vertically); therefore the horizontal acceleration is zero, so horizontal velocity stays constant
A stone is thrown horizontally from the top of a cliff at 15 m s⁻¹. The cliff is 45 m high. Calculate the time taken to reach the ground.
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Taking downwards as positive, for the vertical motion: $$s=ut+1/2a{t}^{2} (u=0)$$ $$45=1/2\times 9.81\times {t}^{2}$$ $${t}^{2}=\frac{90}{9.81}=9.174$$ $$t=3.03\text{ s}$$
Show that the stone in question 8 lands 45 m horizontally from the base of the cliff.
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$$R={u}_{x}\times t=15\times 3.03$$ $$R=45.4\text{ m}\approx 45\text{ m}$$ as required.
Deduce whether a ball launched at 18 m s⁻¹ at 40° above the horizontal will clear a 6.0 m high wall placed 25 m from the launch point. Take g = 9.81 m s⁻².
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Resolving the initial velocity $${u}_{x}=18cos40^{\circ}=13.79{\text{ m s}}^{-1}$$ $${u}_{y}=18sin40^{\circ}=11.57{\text{ m s}}^{-1}$$ Time to reach the wall $$t=\frac{25}{13.79}=1.813\text{ s}$$ Vertical displacement at the wall $$s={u}_{y}t-1/2g{t}^{2}$$ $$s=11.57\times 1.813-1/2\times 9.81\times {1.813}^{2}$$ $$s=20.97-16.12=4.85\text{ m}$$ Since 4.85 m < 6.0 m, the ball does not clear the wall.
A projectile is launched at 30 m s⁻¹ at an angle of 60° above the horizontal from level ground. Air resistance is negligible. Discuss how the horizontal range and maximum height would change if the projectile were instead launched at the same speed but at 30° above the horizontal. In your answer you should: • calculate the maximum height and range for both launch angles • explain why the maximum height changes • explain why the range is the same for both angles.
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At 60° $${u}_{y}=30sin60^{\circ}=25.98{\text{ m s}}^{-1}$$ $${u}_{x}=30cos60^{\circ}=15.0{\text{ m s}}^{-1}$$ $$H=\frac{{u}_{y}^{2}}{2g}=\frac{{25.98}^{2}}{19.62}=34.4\text{ m}$$ $$T=\frac{2{u}_{y}}{g}=5.30\text{ s}$$ $$R={u}_{x}\times T=15.0\times 5.30=79.5\text{ m}$$ At 30° $${u}_{y}=30sin30^{\circ}=15.0{\text{ m s}}^{-1}$$ $${u}_{x}=30cos30^{\circ}=25.98{\text{ m s}}^{-1}$$ $$H=\frac{{15.0}^{2}}{19.62}=11.5\text{ m}$$ $$R=25.98\times \frac{2\times 15.0}{9.81}=25.98\times 3.058=79.5\text{ m}$$ Maximum height is greater at 60° because the vertical component u sin θ is larger, giving more time and distance to rise before gravity decelerates it to zero; range is identical because complementary angles (30° and 60°) produce the same product $u_{\mathrm{x}}$ × T = u² sin(2θ)/g, and sin 60° = sin 120°; therefore the ranges match while the trajectories differ.