Unit 4: worked solutions

Choose right positive and include both carts and the spring.

Why this step: The spring forces are internal and horizontal external impulse is negligible.

Write 0 = 0.40(−3.0) + 0.60v and solve v = +2.0 m/s.

Why this step: Initial momentum is zero, so the outgoing cart momenta must cancel.

Calculate K_light = ½(0.40)(3.0²) = 1.8 J and K_heavy = ½(0.60)(2.0²) = 1.2 J.

Why this step: Kinetic energies add as scalars; the lighter cart has greater speed for equal momentum magnitude.

The spring supplied 3.0 J of kinetic energy in the ideal model.

Why this step: The increase in total kinetic energy comes from spring potential energy, not an increase in total momentum.

Choose upward positive and write Δp = m(v_f − v_i) = 2.0[0 − (−3.0)] = +6.0 N·s.

Why this step: The object’s initial downward velocity must have a negative sign.

Divide by time: F_net,avg = 6.0/0.40 = +15 N.

Why this step: Net external impulse changes momentum.

Draw upward hand force and downward gravity, giving F_hand − mg = 15 N.

Why this step: The requested contact force is not the same as the net force.

F_hand = 15 + 2.0(9.8) = 34.6 N upward.

Why this step: The hand must both offset gravity and produce the upward momentum change.

Select both riders and calculate p_i = (40 + 60)(2.0) = 200 kg·m/s.

Why this step: Internal pushes cannot change this total on the frictionless surface.

Write 40(5.0) + 60v = 200, giving v = 0.

Why this step: The initially moving system need not end with opposite velocity directions.

Use v_cm = p_total/M = 200/100 = +2.0 m/s.

Why this step: Zero net external force keeps the center-of-mass velocity constant before, during, and after the push.

Include both carts and write 0.40(3.0) = (0.40 + 0.80)v_f.

Why this step: Sticking gives one final velocity, with negligible external impulse.

Solve v_f = +1.0 m/s.

Why this step: The initial momentum is +1.2 kg·m/s and the combined mass is 1.20 kg.

K_i = ½(0.40)(3.0²) = 1.8 J; K_f = ½(1.20)(1.0²) = 0.60 J.

Why this step: The stationary cart has no initial kinetic energy, but both masses contribute afterward.

The decrease is 1.2 J and remaining fraction is 1/3.

Why this step: Converted internal energy completes the energy ledger while momentum stays +1.2 kg·m/s.

In the track frame, p_A = 0.40(3.0) = +1.2 kg·m/s and p_B = 0.60(−1.0) = −0.60 kg·m/s; the total arrow is +0.60 kg·m/s.

Why this step: Use signed components from the same frame. For an arrow scale of 0.20 kg·m/s per division, the arrows have lengths six right, three left, and three right for the total.

Subtract the camera velocity from both velocities: v_A′ = +2.0 m/s and v_B′ = −2.0 m/s. The new total is 0.40(2.0) + 0.60(−2.0) = −0.40 kg·m/s.

Why this step: Momentum depends on reference frame. The whole-system change between frames is −M v_camera = −1.0 kg·m/s, which agrees with the two totals.

Do not mix a track-frame cart momentum with a camera-frame cart momentum in one system sum.

Why this step: Such a sum is not the momentum of a single system state in any stated frame. Test conservation by using one frame consistently before and after an interaction.

Choose 2,000 samples/s: a 0.020 s contact contains about 40 sampling intervals, while 50 samples/s gives only about one.

Why this step: The short pulse needs enough samples to estimate its area and changing shape. One interval can miss the peak or most of the pulse, depending on sample timing.

Measure signed velocities just outside contact and signed net force throughout the same interval. Compare m(v_f − v_i) = 0.50(−1.0 − 2.0) = −1.5 N·s with the force–time area. Check the sensor’s zero before and after the trial and account for any horizontal force other than the measured contact force.

Why this step: Matched intervals test one interaction. A force offset produces a systematic area error; contact force equals net horizontal force only when other horizontal forces are negligible or included.

The first average net force is −1.5/0.020 = −75 N. The softer bumper still gives −1.5 N·s, but its average net force is −1.5/0.040 = −37.5 N.

Why this step: Equal masses and endpoint velocities give equal signed momentum changes. Twice the contact duration halves average force magnitude at fixed impulse.

Calculate x_cm = (x₁ + 3x₂)/4 at each shared time: 4.50, 4.25, 4.00, and 3.75 m.

Why this step: The unequal masses require a mass-weighted position. Synchronized data make each pair a single system state.

Plot x_cm vertically in meters against t horizontally in seconds. The four points lie on x_cm = 4.50 − 0.25t, whose slope is −0.25 m/s before and after contact.

Why this step: A straight center-of-mass position graph supports constant center-of-mass velocity even though the individual cart velocities change during sticking.

The unweighted positions average to 3.0, 3.5, 4.0, and 3.75 m, which do not lie on one constant-slope line. Interpret the weighted result within the position uncertainties, and check track level or friction before attributing any real curvature to the interaction.

Why this step: An ordinary average wrongly treats the carts as equal masses. The weighted data support negligible net horizontal external force but do not by themselves prove that all measurement errors vanish.

Compare energy intervals, not just central values: A overlaps the initial interval from 1.16 to 1.22 J; B has no overlap and even its largest final value is below the smallest initial value.

Why this step: A kinetic-energy difference must exceed the stated measurement uncertainty to support a clear decrease.

Trial B supports an inelastic collision. Trial A is consistent with an elastic model, but these data cannot rule out a small kinetic-energy decrease or prove exact elasticity.

Why this step: Separating carts do not establish elasticity, and momentum agreement is compatible with either collision type.

For example, calibrate the video distance scale and estimate endpoint velocities from several frames immediately outside contact, repeating trials with the same initial setup.

Why this step: K contains v², so improved velocity estimates reduce a major source of energy uncertainty. Measurements far from contact can include frictional changes that are not part of the collision.

Write p_total = p₁ + p₂.

Why this step: Total momentum is a vector sum, so direction belongs in each term.

Equal and opposite momenta give p_total = 0 even with both carts moving.

Why this step: Cancellation of vectors does not make each vector zero; kinetic energies remain positive.

Write Δp = m(v_f − v_i), with right positive.

Why this step: A momentum change subtracts signed velocities, not their magnitudes.

Δp = 0.20[−4.0 − (+6.0)] = −2.0 kg·m/s.

Why this step: The ball reverses, so its velocity change is −10 m/s; the impulse is leftward.

Select both carts and write p_f − p_i = J_external.

Why this step: Internal collision impulses cancel in the combined system.

If the net horizontal external impulse is zero, p_f = p_i.

Why this step: Sticking and kinetic-energy changes do not decide whether total momentum stays constant.

Elastic means K_total,f = K_total,i.

Why this step: The collision type is defined by total kinetic energy, not by whether the objects separate.

A rebound can leave less kinetic energy, so compare measured endpoint energies.

Why this step: Only sticking identifies the perfectly inelastic subtype; bouncing does not prove elasticity.

Choose right as positive and write p = mv.

Why this step: The sign convention lets us represent leftward velocity with a negative component.

Since m is positive and v is negative, p is negative and points left.

Why this step: Multiplication by mass preserves the velocity direction; no continuing force is needed for steady motion.

Find the signed force–time area: ½ × duration × peak force.

Why this step: Area under net force versus time is impulse, and the region is triangular.

J = ½(0.20 s)(8.0 N) = +0.80 N·s.

Why this step: The area is above zero, so its sign is positive. Force height alone cannot give impulse.

Use the mass-weighted velocity: v_cm = (m₁v₁ + m₂v₂)/(m₁ + m₂).

Why this step: The center of mass gives each object weight in proportion to its mass.

v_cm = [1(4) + 3(0)]/4 = +1 m/s.

Why this step: The ordinary average of velocities would incorrectly treat unequal masses as equal.

With negligible external impulse, mv = (m + 3m)v_f.

Why this step: The stationary target contributes no initial momentum and the stuck masses share v_f.

Solve 4mv_f = mv to obtain v_f = v/4.

Why this step: The combined moving mass is four times the initially moving mass.

For the light cart, p = mv; for the heavy cart, p = (2m)(v/2).

Why this step: Compare using the same direction and reference frame.

Cancel the factor of two to get p_heavy = mv = p_light.

Why this step: Momentum depends on the product of mass and velocity, while kinetic energy has a different speed dependence.

Net force is the slope of momentum against time.

Why this step: F_net = rate of change of p.

A horizontal line has zero slope, so F_net = 0.

Why this step: The nonzero momentum value describes continuing motion, not an unbalanced force.

For the isolated constant-mass system, p_total and M remain constant.

Why this step: Zero net external force prevents a change in total momentum while internal forces redistribute it.

v_cm = p_total/M stays constant, so its graph is horizontal.

Why this step: Fragment velocities and kinetic energy can change without changing the system’s center-of-mass motion.

Momentum can remain constant while total kinetic energy decreases.

Why this step: The momentum balance and energy classification track different quantities.

Classify the collision as inelastic and include the converted internal energy.

Why this step: Energy can become deformation, sound, and thermal energy without being destroyed.

Calculate p_cannon = (6)(500) = 3000 kg·m/s.

Why this step: Momentum magnitude is mass times speed.

Calculate p_motorcycle = (225)(10) = 2250 kg·m/s.

Why this step: The cannonball's much larger speed more than offsets its smaller mass.

Each constant force gives impulse magnitude F(5 s).

Why this step: Only the directions differ; their magnitudes and durations match.

Use velocity-change magnitude = impulse magnitude/m for the same mass.

Why this step: The velocity changes point differently, but have the same magnitude.

Choose both skaters and use initial momentum (50 + 70)(8) = 960 kg·m/s.

Why this step: Their push is internal and external horizontal impulse is negligible.

Solve 50(12) + 70v = 960, so v = 360/70 = 5.143 m/s.

Why this step: Momentum constrains the second skater even though their kinetic energy may increase through muscular work.

Choose puck plus goalie and conserve horizontal momentum.

Why this step: The catch is brief with negligible external horizontal impulse; kinetic energy need not be conserved.

Find v_f = (0.160·45)/(100 + 0.160) ≈ 0.07189 m/s.

Why this step: They move together after the catch.

Inertia is resistance to acceleration and is measured by mass; it does not depend on velocity.

Why this step: Inertia measures how much force is needed for a given acceleration. In this setting that resistance is the object's mass, not its speed.

Using typical masses, a boulder is heaviest, followed by a bowling ball, a football, and a bullet.

Why this step: The question asks for typical objects rather than exact supplied masses. Use their usual mass ordering to rank inertia, with that assumption kept explicit.

At the same velocity, momentum magnitude p = mv follows the same mass ranking.

Why this step: All objects have the same velocity, so multiplying their masses by one common speed preserves the same momentum-magnitude ranking.

Stopping changes momentum, but force also depends on how long that change takes.

Why this step: Average net force is Δp/Δt.

Consider energy, contact time and deformation as well.

Why this step: Equal incoming momentum can produce very different peak forces and damage.

Write (m + 5)(10) = m(9.81) + 5(13).

Why this step: All velocities must be in the same road frame.

Solve 0.19m = 15, giving m ≈ 78.95 kg.

Why this step: The given final speed is approximate, so 80 kg is the closest option.

Treat the collision as sticking before braking and conserve momentum.

Why this step: The two initial velocities point in the same direction.

Find v_f = (23·45 + 2.3·20)/(23 + 2.3) = 42.727 m/s, so 45 − v_f = 2.273 m/s.

Why this step: Using Mg throughout is fine because the common mass unit cancels.

Use east as +x and north as +y. Momentum components are p_x = 1.5(4) = 6 and p_y = 1.5(3) = 4.5 kg·m/s.

Why this step: Momentum has the same direction as velocity. Multiply each velocity component by the drone's same 1.5 kg mass.

The magnitude is √(6²+4.5²) = 7.5 kg·m/s.

Why this step: The east and north momentum components are perpendicular. Pythagoras gives the magnitude of their resultant.

The direction is tan⁻¹(4.5/6) = 36.87° north of east.

Why this step: Both components are positive, so the direction lies north of east. Their ratio gives the first-quadrant angle from east.

Choose away from the player positive: v_i = −20 and v_f = +25 m/s.

Why this step: The reversal means adding the speed magnitudes in Δv.

Find F_avg = (0.050)(25 − (−20))/0.25 = 9 N.

Why this step: This is the average impulse-producing force in the simplified contact model.