Motion — Chapter review · Physical Science: Matter, Forces & Energy, Grade 8

Describe movement with measurements and graphs, then use Newton’s laws to explain changes.

Review the key ideas

Speed, Velocity & Acceleration

Speed = distance ÷ time; velocity adds direction; acceleration = change in velocity ÷ time.

Reading Motion Graphs

On a position–time graph, slope describes velocity; a horizontal segment means rest.

Newton's Three Laws of Motion

First law: objects resist changes in motion (inertia). Second law: F = ma. Third law: every action has an equal, opposite reaction.

Collisions, Deformation & Changes in Motion

It may have changed temporarily during contact; use a before–during–after record. The bumper undergoes elastic deformation in this simplified record because it regains its thickness.

Designing & Revising a Collision Protector

Judge repeated evidence, variability, damage measures and hard constraints, rather than one selected reading. A qualifies under the thickness rule; B does not despite its lower mean peak.

Problem 1

Does a 2 m/s average speed establish a constant 2 m/s at every moment?

Answer and explanation

No. Different speeds and stops can produce the same distance/time average. More detailed position or speed measurements are needed.

Response criteria

Use the supplied conditions and relevant evidence.
Explain the scientific mechanism or calculation.
State a meaningful limitation or testable next step.

Problem 2

Over 20 s, what are average speed and velocity?

Answer and explanation

Average speed is 40/20=2 m/s; average velocity is 20/20=1 m/s east.

Response criteria

Use the supplied conditions and relevant evidence.
Explain the scientific mechanism or calculation.
State a meaningful limitation or testable next step.

Problem 3

Right is positive. A cart changes velocity from −2 to −8 m/s in 3 s. Calculate average acceleration and explain whether it speeds up or slows down.

Answer and explanation

(−8−(−2))/3 = −2 m/s². It speeds up toward the left because speed rises from 2 to 8 m/s.

Subtract signed velocities before dividing.

Compare velocity direction with acceleration direction.

Response criteria

Calculate −2 m/s².
Explain leftward motion.
Reject negative-always-slows.

Problem 4

Visual-model chapter review: Right is positive. A cart’s velocity changes from −3 to −9 m/s in 2 s. Find average acceleration and explain its speed change.

Answer and explanation

Average acceleration = (−9−(−3))/2 = −3 m/s²; speed increases leftward from 3 to 9 m/s. Negative acceleration does not always mean slowing.

Subtract signed velocities.

Compare the magnitudes and directions instead of using a sign slogan.

Average acceleration = (−9−(−3))/2 = −3 m/s²; speed increases leftward from 3 to 9 m/s. Negative acceleration does not always mean slowing.

Response criteria

Use the named labels, directions or supplied quantities accurately.
Connect the visual evidence to the scientific mechanism or calculation.
State the relevant model conditions or limitation.

Problem 5

Why must both coordinate changes be used for an interval slope?

Answer and explanation

Slope uses differences between endpoints. Ignoring the initial position or time can give the wrong velocity for the selected interval.

Response criteria

Use the supplied conditions and relevant evidence.
Explain the scientific mechanism or calculation.
State a meaningful limitation or testable next step.

Problem 6

Does the final downward line prove decreasing speed?

Answer and explanation

No. A straight downward position line has constant negative velocity in the simplified interval. Its direction differs from its speed magnitude.

Response criteria

Use the supplied conditions and relevant evidence.
Explain the scientific mechanism or calculation.
State a meaningful limitation or testable next step.

Problem 7

Read the supplied +3 m/s then −2 m/s graph, each interval lasting 2 s. Calculate distance, displacement and average speed.

Answer and explanation

Distance 10 m; displacement +2 m; average speed 2.5 m/s.

Compute signed areas +6 and −4 m.

Add magnitudes for distance and divide by 4 s for average speed.

Response criteria

Use signed area for displacement.
Use magnitudes for distance.
State average-speed units.

Problem 8

Visual-model chapter review: A velocity–time record is +4 m/s for 3 s followed by −1 m/s for 2 s. Calculate distance, displacement and average velocity over the full five seconds.

Answer and explanation

Distance 12+2 = 14 m; displacement 12−2 = 10 m right; average velocity 10/5 = 2 m/s right.

Use signed rectangle areas for displacement.

Add path magnitudes for distance and divide displacement by total time for average velocity.

Distance 12+2 = 14 m; displacement 12−2 = 10 m right; average velocity 10/5 = 2 m/s right.

Response criteria

Use the named labels, directions or supplied quantities accurately.
Connect the visual evidence to the scientific mechanism or calculation.
State the relevant model conditions or limitation.

Problem 9

Which force belongs in F=ma for this calculation?

Answer and explanation

Use the net external force on the chosen cart, not a single applied force when other forces also act.

Response criteria

Use the supplied conditions and relevant evidence.
Explain the scientific mechanism or calculation.
State a meaningful limitation or testable next step.

Problem 10

What happens if mass and force both double?

Answer and explanation

The ratio Fnet/m, and thus acceleration, remains unchanged in this simple model.

Response criteria

Use the supplied conditions and relevant evidence.
Explain the scientific mechanism or calculation.
State a meaningful limitation or testable next step.

Problem 11

Plan two comparisons from the records: one isolating net force, the other mass. Specify motion measurements, repeats, controls and what outcome would challenge the predicted relation. Explain braking separately from velocity direction.

Hints

Use the illustrated study titled Investigate force and mass separately. Keep the observations, model and conditions separate.

Answer and explanation

Compare the 1 kg cases at 1 and 2 N to isolate net force; compare 1, 2 and 4 kg at 2 N to isolate mass. Match path, resisting conditions, timing, measuring procedure and initial state. Keep all repeated records. The supplied model gives a=Fnet/m; unexpected motion needs a boundary/force/method review rather than automatic rejection of conservation or hidden invented values. A leftward net force can slow rightward motion.

Response criteria

Design two separately controlled comparisons.
Specify recorded motion, units and repeats.
Explain a challenge result and braking mechanism.

Problem 12

The cart diagram shows 10 N right and 4 N left on a 2 kg cart. A student calculates 5 m/s² using the push alone. Correct the method and say whether the cart must currently move right.

Answer and explanation

Net force is 6 N right; acceleration is 3 m/s² right. The current velocity is not given: the cart could be moving left and slowing.

Subtract resistance from the applied push.

Divide net force by mass.

Keep acceleration distinct from current motion.

Response criteria

Calculate 6 N and 3 m/s².
Do not count the green net arrow again.
Avoid inferring current velocity.

Problem 13

Visual-model chapter review: A 2.5 kg cart has a 15 N applied push and 5 N resistance. Find acceleration, and explain why its equal force on the pusher does not cancel that acceleration.

Answer and explanation

Net force is 10 N in the push direction and acceleration 10/2.5 = 4 m/s². The cart-on-pusher partner acts on the pusher, so it is not part of the cart’s net force.

Compute net force before dividing by mass.

Name the receiver of each interaction partner.

Net force is 10 N in the push direction and acceleration 10/2.5 = 4 m/s². The cart-on-pusher partner acts on the pusher, so it is not part of the cart’s net force.

Response criteria

Use the named labels, directions or supplied quantities accurately.
Connect the visual evidence to the scientific mechanism or calculation.
State the relevant model conditions or limitation.

Problem 14

Does greater acceleration of the lighter cart imply a larger collision force on it?

Answer and explanation

No. Equal interaction forces can cause different acceleration because masses differ, and net-force calculations must account for other forces.

Response criteria

Use the supplied conditions and relevant evidence.
Explain the scientific mechanism or calculation.
State a meaningful limitation or testable next step.

Problem 15

Does recovered thickness establish that no energy transferred?

Answer and explanation

No. Energy can transfer through deformation, motion, sound and thermal pathways even if the original shape returns.

Response criteria

Use the supplied conditions and relevant evidence.
Explain the scientific mechanism or calculation.
State a meaningful limitation or testable next step.

Problem 16

Visual-model chapter review: Two carts receive equal opposite 12 N contact forces in a stated horizontal interval. Their masses are 3 and 1.5 kg. Compare accelerations and explain whether equal force means equal velocity change over the same duration.

Answer and explanation

Accelerations are 4 and 8 m/s² in opposite directions. Over the same duration the lighter cart’s velocity change has twice the magnitude, despite the equal force pair.

Use each receiver’s mass in Fnet/m under the supplied net-horizontal-force assumption.

Multiply acceleration by the common interval to compare velocity changes.

Accelerations are 4 and 8 m/s² in opposite directions. Over the same duration the lighter cart’s velocity change has twice the magnitude, despite the equal force pair.

Response criteria

Use the named labels, directions or supplied quantities accurately.
Connect the visual evidence to the scientific mechanism or calculation.
State the relevant model conditions or limitation.

Problem 17

Does a larger contact face automatically mean a smaller total force?

Answer and explanation

No. It can lower pressure for the same force by spreading contact area, but total force depends on the stopping interaction.

Response criteria

Use the supplied conditions and relevant evidence.
Explain the scientific mechanism or calculation.
State a meaningful limitation or testable next step.

Problem 18

Which is a qualifying starting design under the stated limit?

Answer and explanation

A qualifies at 4 cm. B at 7 cm does not, despite its lower recorded mean peak. A revision needs retesting.

Response criteria

Use the supplied conditions and relevant evidence.
Explain the scientific mechanism or calculation.
State a meaningful limitation or testable next step.

Problem 19

Visual-model chapter review: A design must stay below 30 N in every trial. Its supplied results are 22, 24 and 35 N. Does a mean of 27 N establish that it meets the criterion? Explain a proper next step.

Answer and explanation

No. The 35 N trial fails the every-trial limit even though the mean is 27 N. Preserve the result, investigate variation, revise a mechanism and retest under matched conditions.

Read the criterion as every trial, not the mean.

Use all records before recommending a tested revision.

No. The 35 N trial fails the every-trial limit even though the mean is 27 N. Preserve the result, investigate variation, revise a mechanism and retest under matched conditions.

Response criteria

Use the named labels, directions or supplied quantities accurately.
Connect the visual evidence to the scientific mechanism or calculation.
State the relevant model conditions or limitation.

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