Apply Momentum Conservation to Collisions - Grade 11 Physics

1. What percentage of the system's initial kinetic energy is transformed into other forms during the collision?
   Answer: Approximately $66.7\%$ When two objects collide and stick together (a perfectly inelastic collision), momentum is conserved but kinetic energy is not — some energy converts to heat, sound, and deformation. Your job is to find how much KE is lost as a percentage of the initial KE. Start with momentum conservation to find the final velocity. Taking rightward as positive: $p_i = (2.0)(6.0) + (1.0)(-3.0) = 12.0 - 3.0 = 9.0\ \text{kg}\cdot\text{m/s}$ $v_f = \frac{9.0}{2.0 + 1.0} = 3.0\ \text{m/s (rightward)}$ Now calculate kinetic energies: $KE_i = \frac{1}{2}(2.0)(6.0)^2 + \frac{1}{2}(1.0)(3.0)^2 = 36.0 + 4.5 = 40.5\ \text{J}$ $KE_f = \frac{1}{2}(3.0)(3.0)^2 = 13.5\ \text{J}$ $\%\ \text{lost} = \frac{40.5 - 13.5}{40.5} \times 100 = \frac{27.0}{40.5} \times 100 \approx 66.7\%$ This confirms C is correct. As for the distractors: A (33.3%) likely comes from forgetting the 1.0 kg cart's initial KE entirely and only considering the larger cart, underestimating total initial energy. B (50.0%) is a tempting "half the energy is lost" guess with no calculation behind it. D (75.0%) may result from a sign error — if you accidentally add the momenta instead of subtracting (treating both velocities as positive), you get a wrong $v_f$, inflating the energy loss. Always assign a sign convention first, then carry it through every step — sign errors are the most common trap in collision problems.
2. What is the mass of the cart?
   Answer: $1.00\ \text{kg}$ Whenever you see a collision problem where external forces are negligible, your go-to tool is conservation of momentum: the total momentum before equals the total momentum after. Define a positive direction first — here, rightward is positive. Before the collision, only the ball moves: $p_i = (0.15)(+12) = +1.80\ \text{kg}\cdot\text{m/s}$. After the collision, the ball rebounds left and the cart moves right: $p_f = (0.15)(-8.0) + m_{\text{cart}}(+3.0)$. Setting them equal: $1.80 = -1.20 + 3.0\, m_{\text{cart}}$ $3.0 = 3.0\, m_{\text{cart}}$ $m_{\text{cart}} = 1.00\ \text{kg}$ That confirms D is correct. Choice A (0.20 kg) likely comes from forgetting that the ball's velocity changes sign — using $+8.0$ instead of $-8.0$ for the rebound. That gives a momentum change of only $0.60\ \text{kg}\cdot\text{m/s}$ for the cart side, yielding a much smaller mass. Choice B (0.60 kg) results from using only the ball's final momentum (ignoring the change) or a similar partial-calculation error. Choice C (1.40 kg) comes from adding both momenta after the collision instead of setting them equal to the initial momentum — a sign or setup error. A quick strategy: always write out $p_{\text{before}} = p_{\text{after}}$ explicitly and assign signs before plugging in numbers. Rebound problems are the most common place students drop a negative sign, which cascades into a completely wrong answer.
3. Which choice best describes the velocity of the combined pucks?
   Answer: $2.33\ \text{m/s}$ at $31^\circ$ north of east When two objects collide and stick together, you're dealing with a perfectly inelastic collision. Since external horizontal forces are negligible, momentum is conserved — but because the objects travel in different directions, you must treat momentum as a vector and handle east-west and north-south components separately. Start by calculating momentum in each direction. The total east momentum is $p_x = (0.40)(5.0) = 2.0\ \text{kg}\cdot\text{m/s}$, and the total north momentum is $p_y = (0.60)(2.0) = 1.2\ \text{kg}\cdot\text{m/s}$. The combined mass is $0.40 + 0.60 = 1.0\ \text{kg}$. The final velocity components are $v_x = 2.0/1.0 = 2.0\ \text{m/s}$ east and $v_y = 1.2/1.0 = 1.2\ \text{m/s}$ north. The magnitude is $v = \sqrt{2.0^2 + 1.2^2} = \sqrt{4.0 + 1.44} = \sqrt{5.44} \approx 2.33\ \text{m/s}$. The angle north of east is $\theta = \tan^{-1}(1.2/2.0) = \tan^{-1}(0.60) \approx 31°$. This confirms answer A. Choice B likely results from averaging speeds rather than applying momentum conservation — a common shortcut error. Choice C uses the correct magnitude but flips the angle to $59°$, which would mean the puck moves mostly northward — a mix-up of opposite and adjacent sides in the arctangent. Choice D assumes equal momentum contributions leading to a tidy $45°$ angle, ignoring that the two pucks have different masses and speeds. Study tip: Always split 2D momentum problems into x and y components first, then combine at the end using the Pythagorean theorem and arctangent — never try to work with the total momentum vector directly.
4. What is the cart's velocity immediately after the collision?
   Answer: $2.75\ \text{m/s}$ to the right When a collision occurs with an external force present, you can't simply use conservation of momentum — you must account for the impulse delivered by that external force. The total change in momentum of the system equals the net impulse: from internal collision forces (which cancel) plus the external friction impulse. Start by calculating the external impulse. Friction acts leftward (negative) for $0.050\ \text{s}$: $J_{ext} = -4.0 \times 0.050 = -0.20\ \text{kg}\cdot\text{m/s}$. Now apply the impulse-momentum theorem to the entire system: $p_{final} = p_{initial} + J_{ext}$ $p_{initial} = (0.20)(10) + (0.80)(0) = 2.0\ \text{kg}\cdot\text{m/s}$ $p_{final} = 2.0 + (-0.20) = 1.80\ \text{kg}\cdot\text{m/s}$ The ball's final momentum is $(0.20)(-2.0) = -0.40\ \text{kg}\cdot\text{m/s}$, so the cart's final momentum is: $p_{cart} = 1.80 - (-0.40) = 2.20\ \text{kg}\cdot\text{m/s}$ $v_{cart} = \frac{2.20}{0.80} = 2.75\ \text{m/s to the right}$ That confirms D. Choice A (3.00 m/s) comes from ignoring the external friction entirely and using pure conservation of momentum. Choice B (2.25 m/s) likely results from applying friction's impulse in the wrong direction — adding it instead of subtracting. Choice C (1.75 m/s) may come from a sign error on the ball's rebound velocity, treating it as positive rather than negative. Your key strategy: whenever a problem mentions an external force during a collision, immediately compute its impulse $(F \cdot \Delta t)$ and adjust the system's total momentum before solving for unknowns. Don't assume momentum is conserved just because you see a collision.
5. Isolated collision: cart X's momentum drops from 12 to 4 kg m/s east. Y initially has 3 kg m/s west. Y's final momentum?
   Answer: 5 kg m/s east Take east as positive. Total initial momentum is 12 + (-3) = 9 kg m/s. After the collision X has 4 kg m/s east, so Y must carry the remaining 5 kg m/s east to keep the total 9. The tempting mistake is choosing 8 kg m/s east, but that is only the amount X lost; Y's own initial 3 kg m/s west must be included.
6. Which choice best describes the final velocity of the $1.0\ \text{kg}$ puck?
   Answer: $8.1\ \text{m/s}$ at $7.1^\circ$ north of east When a question involves a collision with no external horizontal forces, your go-to tool is conservation of momentum — applied separately in the x- and y-directions. Start by tracking momentum before and after. The total x-momentum is carried entirely by the 2.0 kg puck initially: $p_x = 2.0 \times 4.0 = 8.0\ \text{kg}\cdot\text{m/s}$ east. The total y-momentum comes from both pucks: $p_y = (2.0)(0) + (1.0)(3.0) = 3.0\ \text{kg}\cdot\text{m/s}$ north. After the collision, the 2.0 kg puck moves north at 1.0 m/s, so it carries $2.0 \times 1.0 = 2.0\ \text{kg}\cdot\text{m/s}$ north and zero east momentum. Since total momentum is conserved, the 1.0 kg puck must account for the rest: $p_{1x} = 8.0\ \text{kg}\cdot\text{m/s east} \Rightarrow v_{1x} = \frac{8.0}{1.0} = 8.0\ \text{m/s east}$ $p_{1y} = 3.0 - 2.0 = 1.0\ \text{kg}\cdot\text{m/s north} \Rightarrow v_{1y} = \frac{1.0}{1.0} = 1.0\ \text{m/s north}$ The final speed is $\sqrt{8.0^2 + 1.0^2} = \sqrt{65} \approx 8.1\ \text{m/s}$, and the angle north of east is $\theta = \tan^{-1}\!\left(\frac{1.0}{8.0}\right) \approx 7.1^\circ$. That confirms C. Choice A likely comes from arithmetic errors that underestimate both components. Choice B results from incorrectly adding momenta rather than finding the remainder. Choice D uses the correct magnitude but places the angle south of east — a sign error from mishandling the y-direction. Your strategy: always set up a momentum table with x and y columns separately, and track which puck "uses up" which portion of the total momentum after the collision.
7. A 500 kg cannon on ice recoils at 2 m/s while firing a 5 kg shell 60° above horizontal. Shell's speed?
   Answer: 400 m/s Momentum is conserved horizontally because the ice has no horizontal friction. The cannon's horizontal momentum is 500 times 2 = 1,000 kg m/s, so the shell's horizontal momentum must also be 1,000 kg m/s. Its horizontal component is 5 times v times cos 60 = 2.5v, so 2.5v = 1,000 and v = 400 m/s. The 200 m/s trap comes from using 5v = 1,000 and ignoring the 60-degree angle.
8. A 0.2 kg ball moving right at 5 m/s hits a 0.4 kg ball at rest and rebounds at 1 m/s. Speed of the 0.4 kg ball?
   Answer: 3.0 m/s Momentum is conserved. Initial momentum is 0.2 times 5 = 1.0 kg m/s. The 0.2 kg ball rebounds left at 1 m/s, so its final momentum is -0.2 kg m/s. The 0.4 kg ball must then have 1.2 kg m/s, so its speed is 1.2 / 0.4 = 3.0 m/s. The tempting error is 2.0 m/s, which comes from treating the rebound as still moving right at 1 m/s instead of opposite direction.
9. What was the projectile's speed immediately before striking the block?
   Answer: Approximately $99\ \text{m/s}$ When a projectile embeds in a block, you're dealing with a perfectly inelastic collision followed by conservation of energy. The key insight is that you must solve this in two separate stages — momentum conservation during the collision, then energy conservation during the swing. Stage 1 — Energy conservation (post-collision swing): The combined system rises $h = 0.20\ \text{m}$, so its kinetic energy converts entirely to gravitational potential energy: $\frac{1}{2}(m_{\text{total}})v_{\text{after}}^2 = (m_{\text{total}})gh$ $v_{\text{after}} = \sqrt{2gh} = \sqrt{2(9.8)(0.20)} = \sqrt{3.92} \approx 1.98\ \text{m/s}$ Stage 2 — Momentum conservation (during collision): The projectile's momentum equals the combined system's momentum immediately after: $m_p v_p = (m_p + m_b)v_{\text{after}}$ $v_p = \frac{(0.020 + 0.980)(1.98)}{0.020} = \frac{(1.000)(1.98)}{0.020} \approx 99\ \text{m/s}$ This confirms answer C. A (≈ 9.9 m/s) results from forgetting to divide by the projectile mass — essentially treating the projectile as if it has the combined mass. B (≈ 49.5 m/s) comes from applying momentum conservation with incorrect mass ratios, off by a factor of 2. D (≈ 198 m/s) typically arises from doubling the answer, often from misapplying the energy equation and using $h = 0.40\ \text{m}$ or miscalculating the mass ratio. Your study tip: always split ballistic pendulum problems into two stages — energy first (to find post-collision speed), momentum second (to find pre-collision speed). Reversing the order is the most common mistake on these problems.
10. A 2 kg cart at 4 m/s rear-ends a 3 kg cart at 1 m/s the same way; they lock. Total KE lost?
    Answer: 5.4 J Momentum is conserved: 2*4+3*1=11 kg m/s, so the locked carts move at 11/5=2.2 m/s. Initial kinetic energy = 16+1.5=17.5 J; final = 0.5*5*$(2.2)^2$=12.1 J. Lost = 17.5-12.1=5.4 J. Don't use 0.5*(2+3)*(4-1)=7.5 J; that isn't kinetic energy and ignores the correct final speed.
11. Which statement correctly gives cart B's final velocity and the change in the system's kinetic energy?
    Answer: Cart B moves right at $3.0\ \text{m/s}$, and kinetic energy decreases by $7.5\ \text{J}$. When you see a collision problem asking about both final velocity and energy change, your two tools are conservation of momentum (always applies) and kinetic energy calculations (to check what type of collision occurred). First, establish a sign convention: right is positive. Cart A has initial momentum $p_A = (0.50)(+6.0) = +3.0\ \text{kg}\cdot\text{m/s}$ and cart B has $p_B = (0.50)(-2.0) = -1.0\ \text{kg}\cdot\text{m/s}$, giving a total system momentum of $+2.0\ \text{kg}\cdot\text{m/s}$. After the collision, cart A contributes $(0.50)(+1.0) = +0.5\ \text{kg}\cdot\text{m/s}$, so cart B must carry $2.0 - 0.5 = +1.5\ \text{kg}\cdot\text{m/s}$. Dividing by its mass: $v_B = \frac{1.5}{0.50} = +3.0\ \text{m/s}$ — rightward. That confirms B for the velocity part. Now check kinetic energy. Initial KE: $\frac{1}{2}(0.50)(6.0)^2 + \frac{1}{2}(0.50)(2.0)^2 = 9.0 + 1.0 = 10.0\ \text{J}$. Final KE: $\frac{1}{2}(0.50)(1.0)^2 + \frac{1}{2}(0.50)(3.0)^2 = 0.25 + 2.25 = 2.5\ \text{J}$. The decrease is $10.0 - 2.5 = 7.5\ \text{J}$, confirming answer B. Choice A gets the velocity right but incorrectly claims kinetic energy is conserved — that would be a perfectly elastic collision, which this clearly is not. Choice C gives cart B a leftward velocity, which violates momentum conservation (you can verify the numbers don't balance). Choice D invents a velocity that also fails momentum conservation, and kinetic energy can never increase in a normal collision without an external energy source. Your strategy: always solve for the unknown velocity using momentum conservation before touching kinetic energy — momentum pins down the answer, and energy just tells you what kind of collision occurred.
12. What is the velocity of all three carts immediately after the second collision?
    Answer: $1.25\ \text{m/s}$ to the right Whenever you see a perfectly inelastic collision (objects sticking together), your go-to tool is conservation of momentum: $p_{\text{before}} = p_{\text{after}}$, applied separately to each collision. First collision: The 1.0 kg cart moving right at 6.0 m/s hits the stationary 2.0 kg cart. $p_{\text{before}} = (1.0)(6.0) + (2.0)(0) = 6.0\ \text{kg}\cdot\text{m/s}$ $v_1 = \frac{6.0}{1.0 + 2.0} = 2.0\ \text{m/s (right)}$ Second collision: The combined 3.0 kg mass moving right at 2.0 m/s hits the 1.0 kg cart moving left at 1.0 m/s. Assigning rightward as positive, the third cart's velocity is $-1.0\ \text{m/s}$. $p_{\text{before}} = (3.0)(2.0) + (1.0)(-1.0) = 6.0 - 1.0 = 5.0\ \text{kg}\cdot\text{m/s}$ $v_{\text{final}} = \frac{5.0}{3.0 + 1.0} = \frac{5.0}{4.0} = 1.25\ \text{m/s (right)}$ That confirms answer A. Answer B (0.75 m/s) results from incorrectly adding the third cart's momentum as positive instead of negative, ignoring its leftward direction. Answer C (1.75 m/s) likely comes from skipping the first collision entirely and using 6.0 m/s as the incoming speed for the second collision. Answer D (1.25 m/s left) gets the magnitude right but flips the direction — a sign error in setting up the momentum equation. Study tip: Always define a positive direction before starting, and carefully assign signs to every velocity. Direction errors are the most common trap on multi-collision problems.
13. Which set of velocities does the moving observer measure for cart A before, cart B before, and the joined carts after the collision, respectively?
    Answer: $2.0\ \text{m/s}$ east, $4.0\ \text{m/s}$ west, and $1.0\ \text{m/s}$ west Whenever you see a question involving observers moving at different velocities, you're being tested on Galilean velocity transformation: to find what a moving observer measures, subtract the observer's velocity from each ground-frame velocity. Define east as positive, so the observer moves at $+3.0\ \text{m/s}$. First, find the post-collision velocity in the ground frame. Using conservation of momentum with equal masses $m$: $mv_A + mv_B = 2m \cdot v_f \implies \frac{5.0 + (-1.0)}{2} = +2.0\ \text{m/s (east)}$ Now apply the transformation (subtract observer's velocity from each): - Cart A: $+5.0 - 3.0 = +2.0\ \text{m/s}$ → 2.0 m/s east - Cart B: $-1.0 - 3.0 = -4.0\ \text{m/s}$ → 4.0 m/s west - Joined carts: $+2.0 - 3.0 = -1.0\ \text{m/s}$ → 1.0 m/s west This matches answer A, confirming it is correct. Answer B adds the observer's velocity instead of subtracting — a classic sign error. Answer C gets cart A and cart B right but incorrectly assigns the joined carts as moving east; the result is $-1.0\ \text{m/s}$, which is west, not east. Answer D simply reports the original ground-frame velocities, forgetting to apply any transformation at all. Study tip: Always set up a clear sign convention (east = positive) before transforming velocities. The moving-observer transformation is just one step — subtract the observer's velocity from every velocity — but a sign mistake at any point cascades into the wrong answer.
14. A 1 kg ball moving east at 3 m/s hits a 1 kg ball moving north at 4 m/s; they stick. Final speed?
    Answer: 2.5 m/s The momenta are perpendicular, so combine them with the Pythagorean theorem: sqrt($3^2$ + $4^2$) = 5 kg m/s for the total momentum. After they stick, total mass is 2 kg, so final speed is 5 / 2 = 2.5 m/s. The tempting wrong answer is 5.0 m/s, which forgets to divide by the combined 2 kg mass.
15. What is the velocity of the joined carts immediately after the collision?
    Answer: $1.0\ \text{m/s}$ to the right Whenever you see carts colliding and sticking together, that's your signal to apply conservation of momentum. Since friction is negligible, the total momentum before equals the total momentum after. The key step is setting up a sign convention — let's say rightward is positive. Before the collision: - Cart 1: $p_1 = (0.60)(+4.0) = +2.4\ \text{kg}\cdot\text{m/s}$ - Cart 2: $p_2 = (0.90)(-1.0) = -0.90\ \text{kg}\cdot\text{m/s}$ - Total: $p_{total} = +2.4 + (-0.90) = +1.5\ \text{kg}\cdot\text{m/s}$ After the collision, the combined mass is $0.60 + 0.90 = 1.5\ \text{kg}$. Solving for velocity: $v_f = \frac{p_{total}}{m_{total}} = \frac{+1.5}{1.5} = +1.0\ \text{m/s}$ The positive sign confirms the direction is to the right, making B the correct answer. Choice A ($0.60\ \text{m/s}$ to the left) results from ignoring signs and subtracting momenta incorrectly, then flipping the direction. Choice C ($2.2\ \text{m/s}$) likely comes from adding the two speeds directly without accounting for the opposite directions — a classic sign-convention error. Choice D ($1.5\ \text{m/s}$) confuses the total momentum ($1.5\ \text{kg}\cdot\text{m/s}$) with the final velocity — students sometimes forget to divide by the combined mass. Study tip: Always write out your sign convention explicitly before plugging in numbers. Momentum problems on this exam almost always hinge on correctly handling direction — that single step separates the right answer from every distractor.

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