Forces and momentum for AP Physics C: Mechanics

Reading cue. The labels are speeds, not velocities — direction is carried only by the arrowheads — so fix a positive direction along the track before you write down a single momentum.

1. aCalculate Calculate the total kinetic energy of the two balls at the instant drawn.

   recall2 marks

   Check answer · 2 marks

   1. ½(0.50)(4.0)² = 4.0 J and ½(1.5)(2.0)² = 3.0 J
   2. total = 7.0 J, with the directions of the arrows playing no part because kinetic energy is a scalar

2. bDetermine Determine the speed the 1.5 kg ball would need, with its direction of travel unchanged, for the total momentum of the pair to be zero.

   routine3 marks

   Check answer · 3 marks

   1. momentum of the 0.50 kg ball = 0.50 × 4.0 = 2.0 kg m/s to the right
   2. for zero total the 1.5 kg ball must carry 2.0 kg m/s to the left
   3. v = 2.0/1.5 = 1.3 m/s, slower than the 2.0 m/s marked on the figure

3. cDetermine The collision is in fact perfectly elastic rather than sticking. Determine the velocity of each ball immediately afterwards.

   demanding4 marks

   Show a hint

   Two conservation statements give two equations. The quickest route is the elastic-collision result that the balls separate as fast as they approached — read the approach speed straight off the two arrows.

   Check answer · 4 marks

   1. momentum: 0.50v₁ + 1.5v₂ = 0.50(4.0) + 1.5(−2.0) = −1.0 kg m/s
   2. kinetic energy after must equal the 7.0 J before, or equivalently the separation speed equals the 6.0 m/s approach speed
   3. solving gives v₁ = −5.0 m/s for the 0.50 kg ball, that is 5.0 m/s to the left
   4. v₂ = +1.0 m/s for the 1.5 kg ball, that is 1.0 m/s to the right

4. dExplain Explain why the two arrows drawn on the figure cannot by themselves tell you whether the collision that follows is elastic, and describe one measurement made afterwards that would settle it.

   top of the paper3 marks

   Check answer · 3 marks

   1. the figure fixes only the masses and the velocities before contact, and an elastic and an inelastic collision start from exactly the same drawn state
   2. momentum is conserved either way, so the total 1.0 kg m/s to the left is no test of elasticity
   3. measuring both final speeds and comparing the total kinetic energy after with the 7.0 J before decides it: equal means elastic, less means kinetic energy was lost

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.

   recall2 marks

   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.

   demanding4 marks

   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.

   top of the paper4 marks

   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. 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.

   routine3 marks

   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.

   demanding4 marks

   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.

   top of the paper4 marks

   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.

   top of the paper4 marks

   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