Mechanics for AP Physics 1

Reading cue. Velocity is on the vertical axis, so a sloping line is a changing speed, not a change of direction; and the fall meets the axis at 9.0 s, not at the last gridline at 10 s.

1. aDetermine Determine the acceleration of the object during the first stage of the motion shown.

   recall2 marks

   Check answer · 2 marks

   1. reads the gradient of the first segment as (8.0 − 0)/(4.0 − 0)
   2. a = 2.0 m/s², with unit

2. bCalculate Calculate the distance travelled by the object between t = 4.0 s and t = 9.0 s.

   routine3 marks

   Check answer · 3 marks

   1. rectangle from 4.0 s to 7.0 s = 8.0 × 3.0 = 24 m
   2. triangle from 7.0 s to 9.0 s = ½ × 2.0 × 8.0 = 8.0 m
   3. total distance = 32 m

3. cDetermine Determine the time at which the object has travelled exactly 20 m from its starting point.

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   Check answer · 3 marks

   1. area under the first segment = ½ × 4.0 × 8.0 = 16 m, so the object is still short of 20 m at t = 4.0 s
   2. the remaining 4.0 m is covered at the constant 8.0 m/s, taking 4.0/8.0 = 0.50 s
   3. t = 4.5 s

4. dExplain A student looks at the last stage of the graph and says the object must be moving backwards, because the line is sloping downwards. Explain why the graph does not show this, and state what the drawing would look like if the object did reverse.

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   Check answer · 3 marks

   1. the plotted line stays above the time axis for the whole 9.0 s, so the velocity is positive throughout and the direction never changes
   2. a line falling towards the axis while still above it shows the speed decreasing in an unchanged direction
   3. a reversal would be drawn as the line crossing the time axis into negative velocity

Reading cue. h is the vertical dimension line at the side of the wedge, not the distance the block slides: along the face that is h/sin θ. μ is printed twice because both surfaces share it.

1. aDetermine Determine the normal force exerted on the block by the sloping face, in terms of m, θ and physical constants.

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   Check answer · 2 marks

   1. resolves the weight mg perpendicular to the sloping face
   2. N = mg cos θ, since the block has no acceleration perpendicular to the face

2. bCalculate Calculate the acceleration of the block down the sloping face for θ = 30° and μ = 0.25.

   routine3 marks

   Check answer · 3 marks

   1. along the face, ma = mg sin θ − μmg cos θ, so a = g(sin θ − μ cos θ) with m cancelling
   2. a = 9.8(0.500 − 0.25 × 0.866) = 9.8 × 0.284
   3. a = 2.8 m/s², directed down the slope

3. cDetermine Determine the time the block takes to reach the foot of the slope when h = 1.5 m, for the same θ and μ.

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   Check answer · 4 marks

   1. the dimension line gives the vertical drop, so the distance along the face is L = h/sin θ = 1.5/0.500 = 3.0 m
   2. from rest with uniform acceleration, L = ½at², so t = √(2L/a)
   3. t = √(2 × 3.0/2.78) = √2.16
   4. t = 1.5 s

4. dJustify An identical block is released from the same height h on a steeper wedge carrying the same μ on both of its surfaces. Justify whether it stops nearer to or further from the foot of the slope than the block in the figure.

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   Check answer · 4 marks

   1. the friction force on the face is μmg cos θ and the sliding length is h/sin θ, so the energy lost on the slope is μmgh cot θ
   2. cot θ falls as θ increases, so the steeper wedge takes less energy from the block
   3. the block therefore reaches the foot with more kinetic energy, while the friction force on the floor is μmg and is unchanged by the wedge
   4. so it stops further from the foot of the slope, not nearer

Reading cue. The fall is one straight segment, so a single constant acceleration spans it; and the time axis is labelled from 1 s, so the start of the graph at t = 0 sits on an unlabelled origin.

1. aDetermine Determine the displacement of the cart between t = 5.0 s and t = 7.0 s.

   recall2 marks

   Check answer · 2 marks

   1. the segment is a rectangle of height −3.0 m/s and width 2.0 s
   2. displacement = −6.0 m, that is 6.0 m in the negative direction

2. bCalculate Calculate the average velocity of the cart over the interval t = 2.0 s to t = 5.0 s.

   routine3 marks

   Check answer · 3 marks

   1. area from 2.0 s to 3.5 s = ½(1.5)(3.0) = +2.25 m, and area from 3.5 s to 5.0 s = ½(1.5)(3.0) = −2.25 m
   2. net displacement over the interval is zero
   3. average velocity = 0/3.0 = 0 m/s, even though the cart is moving throughout and its average speed is 1.5 m/s

3. cDetermine Determine both the total distance travelled by the cart and its net displacement over the whole 7.0 s shown.

   demanding4 marks

   Check answer · 4 marks

   1. areas above the axis: +6.0 m from 0 to 2.0 s and +2.25 m from 2.0 s to 3.5 s
   2. areas below the axis: −2.25 m from 3.5 s to 5.0 s and −6.0 m from 5.0 s to 7.0 s
   3. net displacement = 6.0 + 2.25 − 2.25 − 6.0 = 0 m, so the cart ends where it started
   4. total distance = 6.0 + 2.25 + 2.25 + 6.0 = 16.5 m

4. dSketch Sketch on the same axes the velocity–time graph of a second cart that starts from the same point at t = 0, has one constant acceleration for the whole 7.0 s, and at t = 7.0 s is at the same position and has the same velocity as the cart drawn. Indicate its initial velocity and the value of its acceleration.

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   Show a hint

   For a constant acceleration the average velocity over any interval is just the mean of the velocities at its two ends. Zero net displacement then tells you at once how the initial and final velocities must be related.

   Check answer · 4 marks

   1. a single straight line of constant negative gradient across the whole 7.0 s
   2. for constant acceleration the average velocity is the mean of the initial and final values, and the net displacement must be zero as in part (c), so the initial velocity must be +3.0 m/s to match the final −3.0 m/s
   3. gradient = (−3.0 − 3.0)/7.0 = −0.86 m/s²
   4. the line crosses zero at t = 3.5 s, the same instant as the drawn cart

Reading cue. Six separate items, not an assembled apparatus: there is no protractor and no force sensor, so the stopwatch and the meter stick are the only route to anything the spring scale cannot give.

1. aDetermine Each of the known masses drawn in the stack is 0.50 kg, and the block itself weighs 4.9 N. Determine the largest normal force between the block and the level board that the equipment drawn can produce.

   recall2 marks

   Check answer · 2 marks

   1. the stack in the figure holds three masses, so the greatest load is 3 × 0.50 = 1.5 kg, a weight of 14.7 N
   2. on a level board the normal force equals the total weight: N = 4.9 + 14.7 = 19.6 N

2. bExplain Explain why the spring scale must be read while the block is already sliding, and why the largest reading, taken at the instant the block first breaks away, is not the one wanted.

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   Check answer · 3 marks

   1. the quantity sought is the coefficient of kinetic friction, which applies only while the two surfaces are sliding over one another
   2. with the block moving at constant velocity the net force is zero, so the scale reading equals the friction force exactly
   3. the peak reading at break-away measures the maximum static friction, which is larger and would give too high a value

3. cDescribe The spring scale is difficult to hold at a steady reading. Describe how the stopwatch and the meter stick drawn in the figure could be used instead, with no use of the spring scale at all, and state the equation that would give μ from those measurements.

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   Check answer · 4 marks

   1. give the block a push so that it slides freely along the level board and comes to rest, so that friction is the only horizontal force acting during the slide
   2. measure with the meter stick the distance d from the point of release to the stopping point, and time that slide with the stopwatch as t
   3. for uniform deceleration ending at rest, d = ½at², so a = 2d/t²
   4. friction alone gives a = μg, so μ = 2d/(gt²); repeat the slide several times and average

4. dJustify Justify whether the value of μ obtained would change if the block were stood on its smallest face instead of its largest, and describe the measurement from this equipment that would test your answer.

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   Check answer · 4 marks

   1. the standard model gives f = μN with no area term at all, so the coefficient should come out unchanged
   2. the weight and therefore the normal force are unchanged, so the constant-speed scale reading should also be unchanged
   3. stand the block on its smallest face with the same loading masses on top and repeat the constant-speed pull, comparing the readings
   4. agreement within the spread of the repeats supports the model, while a consistent difference would show the model failing for these two surfaces

Reading cue. The arrow drawn from the block is labelled bv, not b: it shrinks as the block slows, so the acceleration is never constant and no constant-acceleration equation applies.

1. aDetermine Determine the magnitude and the direction of the block's acceleration at t = 0.

   recall3 marks

   Check answer · 3 marks

   1. magnitude bv₀/m
   2. directed opposite to the velocity, that is to the left
   3. the surface contributes nothing, being frictionless, so this is the only horizontal force

2. bDetermine Determine the time at which the block's speed has fallen to half of v₀, in terms of m and b.

   routine3 marks

   Check answer · 3 marks

   1. v = v₀e^(−bt/m) stated or derived
   2. ½ = e^(−bt/m)
   3. t = (m/b)ln2 = 0.69 m/b

3. cDerive Derive an expression for the speed of the block as a function of the distance x it has travelled, rather than of time, and state the shape of a graph of v against x.

   demanding4 marks

   Show a hint

   Acceleration can be written as v dv/dx when speed is wanted as a function of position.

   Check answer · 4 marks

   1. write the acceleration as v dv/dx, so m v dv/dx = −bv
   2. dv/dx = −b/m, a constant
   3. v = v₀ − bx/m
   4. a straight line of negative gradient b/m, reaching v = 0 at x = mv₀/b

4. dJustify Justify the claim that the block gives up three-quarters of its initial kinetic energy while covering the first half of the distance it will ever travel, and state how long that first half takes.

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   Check answer · 4 marks

   1. the total distance is mv₀/b, so the halfway point is x = mv₀/2b
   2. from v = v₀ − bx/m the speed there is exactly v₀/2
   3. kinetic energy is then ¼ of its initial value, so ¾ has been dissipated
   4. the speed reaches v₀/2 at t = (m/b)ln2, so the first half of the journey takes a finite 0.69 m/b while the second half takes forever