2.1 Systems and Center of Mass — Worked Solutions
Work through Topic 2.1
These are the full worked solutions to the lesson questions. Try the questions first, then check each step and the reason for it.
Question 1
A 1 kg mass is at x = 0 m and a 3 kg mass at x = 4 m. Where is x_CM?
- 2 m
- 3 m
- 4 m
Step 1. Use x_CM = (m₁x₁ + m₂x₂)/(m₁ + m₂).
Why this step? The two masses differ, so a simple midpoint would give the lighter object too much influence.
Step 2. Substitute x_CM = (1·0 + 3·4)/(1 + 3) = 12/4 = 3 m.
Why this step? The numerator is the sum of mass-position products; dividing by all 4 kg leaves a position in meters.
Step 3. The answer lies nearer the 3 kg mass at 4 m.
Why this step? The heavier mass has three times the position weight, so a center at 2 m would not match the mass distribution.
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Question 2
For a uniform circular ring, where can its center of mass be?
- At the empty geometric center
- Only somewhere within the ring material
- At any point on the circumference
Step 1. Pair each small ring element with the equal-mass element opposite it. Their offsets from the geometric center cancel.
Why this step? Uniformity gives equal opposing masses, and circular symmetry provides a matching partner for every direction.
Step 2. All such pairs have their center of mass at the geometric center, so the whole ring does too.
Why this step? Adding pairs with zero mass-weighted offset keeps the total offset zero; the empty center need not contain matter.
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Question 3
You change the system from cart A alone to both carts plus their connecting spring. How is the spring interaction classified?
- It becomes internal
- It must vanish
- It becomes gravitational
Step 1. For cart A alone, the spring lies outside the system and exerts an external force on A.
Why this step? The interaction crosses the original boundary, which is what makes it external rather than its force type.
Step 2. For both carts plus spring, both sides of the spring interactions lie inside the boundary, so the interactions are internal.
Why this step? Enlarging the system reclassifies where the interaction occurs; it does not remove the spring forces or turn them into gravity.
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Question 4
Masses 2 kg, 1 kg, and 1 kg are at (0, 0), (4, 0), and (0, 4) m. Find the center-of-mass coordinates and check their location against the extreme coordinates.
Step 1. Total mass is M = 2 + 1 + 1 = 4 kg.
Why this step? Every mass contributes to the normalization of each coordinate, including a mass whose coordinate happens to be zero.
Step 2. Compute x_CM = (2·0 + 1·4 + 1·0)/4 = 1 m.
Why this step? Only the particle at x = 4 m contributes a nonzero x moment; the mass at y = 4 m still counts in M.
Step 3. Compute y_CM = (2·0 + 1·0 + 1·4)/4 = 1 m.
Why this step? Apply the same weighting separately to y; averaging radial distances would lose the coordinate directions.
Step 4. The center is (1 m, 1 m), inside the 0–4 m coordinate bounds.
Why this step? A weighted mean with positive masses cannot fall beyond the smallest or largest constituent coordinate.
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Question 5
Two carts initially at rest on a horizontal track move apart after releasing a compressed, negligible-mass spring between them. External horizontal forces are negligible. Explain why each cart can move while the center of mass remains at rest. Identify an external interaction that could change that result.
Step 1. Choose both carts and their spring as one system. The spring exerts opposite internal forces on the carts.
Why this step? Including the spring keeps its interactions inside the boundary, while still allowing each cart to feel its own force.
Step 2. Those internal forces cancel in the whole-system sum, so M a_CM = F_external = 0.
Why this step? Newton’s third-law pairing removes internal forces from the center-of-mass equation, not from each cart’s equation.
Step 3. If v_CM initially equals zero, it remains zero even as the carts separate.
Why this step? Zero external force fixes the center-of-mass velocity; it does not require each constituent velocity to be zero.
Step 4. Unequal opposing track-friction forces could give a nonzero net external force and accelerate the center of mass.
Why this step? The track is outside the chosen system; unequal friction magnitudes need not cancel when the carts move in opposite directions.
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Question 6
A 2 kg uniform board has center of mass at x = 0.50 m. A 1 kg washer is attached at x = 0.80 m. Predict the new balance point. Sketch how it changes as the washer moves farther right.
Step 1. Treat the board’s 2 kg as located at its known center, x = 0.50 m, for the position calculation.
Why this step? The board’s distributed mass has the same total mass-position moment as that equivalent point; its internal details are not needed here.
Step 2. Calculate x_CM = (2·0.50 + 1·0.80)/(2 + 1) = 1.80/3 = 0.60 m.
Why this step? The new point combines the board and washer, with the board carrying twice the weight in the average.
Step 3. For a variable washer position x_w, x_CM = (1.00 + x_w)/3.
Why this step? Only the washer position changes; keeping both masses and the board center fixed makes the relationship linear.
Step 4. The graph has slope 1/3, so a 0.30 m washer shift moves the center by 0.10 m.
Why this step? The washer is one third of the total mass, so its displacement produces one third as much center-of-mass displacement.