Momentum and Collisions Practice Problems with Answers

Practice problems

Work each one before opening the solution. Every step is shown, with the arithmetic left in so you can find where yours diverged.

1. 1. Momentum of a soccer ball, and the runner who matches it

   A soccer ball of mass 0.42 kg travels at 18 m/s across level ground. (a) Find the ball's momentum. (b) A 65 kg runner moves along the same line in the same direction. How fast would the runner have to move to carry the same momentum as the ball?

   Show the worked solution

   1. Take the ball's direction of travel as positive and keep it positive for the whole problem. Both objects move that way here, so no minus signs appear, but write the convention down anyway: the habit is what saves you in problem 2.

   2. (a) Apply $p = mv$ directly: $p = (0.42\ \text{kg})(18\ \text{m/s}) = 7.56\ \text{kg} \cdot \text{m/s}$, which is $7.6\ \text{kg} \cdot \text{m/s}$ to two significant figures.

   3. (b) Same relationship, rearranged: $v = p/m$. Use the unrounded momentum, $v = (7.56\ \text{kg} \cdot \text{m/s})/(65\ \text{kg}) = 0.1163\ \text{m/s}$, so $0.12\ \text{m/s}$.

   4. Check by going forwards again: $(65\ \text{kg})(0.1163\ \text{m/s}) = 7.56\ \text{kg} \cdot \text{m/s}$, which is part (a) exactly.

   5. Sanity check on the size of the answer. The runner outweighs the ball by $65/0.42 = 155$, so at equal momentum the runner's speed must be about 155 times smaller: $18/155 = 0.116\ \text{m/s}$, and it is.

   (a) $p = 7.6\ \text{kg} \cdot \text{m/s}$ in the ball's direction of motion. (b) The runner needs only $0.12\ \text{m/s}$, far below a walking pace, because momentum is linear in mass and the runner has about 155 times more of it.

2. 2. Two balls, opposite directions, one system

   On a straight east-west track, a 1.4 kg ball rolls east at 3.0 m/s while a 0.60 kg ball rolls west at 9.0 m/s. Find (a) each ball's momentum, (b) the total momentum of the two-ball system, and (c) the velocity of the system's center of mass.

   Show the worked solution

   1. Declare the axis first: east is positive. Then $v_1 = +3.0\ \text{m/s}$ and $v_2 = -9.0\ \text{m/s}$. Every number after this depends on those two signs.

   2. (a) $p_1 = (1.4\ \text{kg})(+3.0\ \text{m/s}) = +4.2\ \text{kg} \cdot \text{m/s}$ and $p_2 = (0.60\ \text{kg})(-9.0\ \text{m/s}) = -5.4\ \text{kg} \cdot \text{m/s}$.

   3. (b) Add them with their signs: $p_{\text{total}} = (+4.2) + (-5.4) = -1.2\ \text{kg} \cdot \text{m/s}$. The minus sign is the answer's direction: the system's momentum points west.

   4. Adding magnitudes instead would give $4.2 + 5.4 = 9.6\ \text{kg} \cdot \text{m/s}$, eight times too large and pointing nowhere. Momentum is a vector, and on a straight line the signs are the whole of the vector algebra.

   5. (c) Total mass is $1.4 + 0.60 = 2.0\ \text{kg}$, so $v_{\text{cm}} = p_{\text{total}}/M = (-1.2\ \text{kg} \cdot \text{m/s})/(2.0\ \text{kg}) = -0.60\ \text{m/s}$, that is $0.60\ \text{m/s}$ west.

   6. Notice that the heavier ball is the one going east, and the center of mass still drifts west. Mass alone does not settle the direction; the product $mv$ does.

   (a) $p_1 = +4.2\ \text{kg} \cdot \text{m/s}$ and $p_2 = -5.4\ \text{kg} \cdot \text{m/s}$ with east positive. (b) $p_{\text{total}} = 1.2\ \text{kg} \cdot \text{m/s}$, directed west. (c) The center of mass moves west at $0.60\ \text{m/s}$.

3. 3. A constant force that reverses a crate

   A 4.0 kg crate slides to the right along a frictionless floor at 6.0 m/s. A constant 15 N force is then applied to the left for 2.0 s. Find (a) the impulse that force delivers, (b) the crate's velocity at the end of the 2.0 s, and (c) the instant at which the crate is momentarily at rest.

   Show the worked solution

   1. Rightward is positive, so the applied force is $F = -15\ \text{N}$ and the initial velocity is $+6.0\ \text{m/s}$.

   2. Initial momentum: $p_i = (4.0\ \text{kg})(+6.0\ \text{m/s}) = +24\ \text{kg} \cdot \text{m/s}$.

   3. (a) For a constant force, $J = F \Delta t = (-15\ \text{N})(2.0\ \text{s}) = -30\ \text{N} \cdot \text{s}$. Newton seconds and kg m/s are the same unit, so this impulse is directly comparable with the momentum above.

   4. (b) The impulse-momentum theorem says the impulse is the change in momentum, so add it to what you started with: $p_f = p_i + J = (+24) + (-30) = -6.0\ \text{kg} \cdot \text{m/s}$. Then $v_f = p_f/m = (-6.0)/(4.0) = -1.5\ \text{m/s}$, that is 1.5 m/s to the left.

   5. The crate reversed. That happens exactly when the impulse exceeds the original momentum in magnitude, which it does here: 30 against 24.

   6. (c) At the turning instant the momentum is zero, so the impulse delivered up to that point must be $-24\ \text{N} \cdot \text{s}$. Solve $(-15\ \text{N})\,t = -24\ \text{N} \cdot \text{s}$ to get $t = 1.6\ \text{s}$.

   7. Cross-check the whole problem with Newton's second law instead. $a = F/m = (-15)/(4.0) = -3.75\ \text{m/s}^2$, so at $t = 2.0\ \text{s}$ the velocity is $6.0 + (-3.75)(2.0) = -1.5\ \text{m/s}$, and it passes through zero at $t = 6.0/3.75 = 1.6\ \text{s}$. Both parts agree.

   (a) $J = -30\ \text{N} \cdot \text{s}$, that is $30\ \text{N} \cdot \text{s}$ to the left. (b) $v_f = 1.5\ \text{m/s}$ to the left. (c) The crate is momentarily at rest at $t = 1.6\ \text{s}$, while the force is still acting.

4. 4. Impulse as the area under a force versus time graph

   A 1.5 kg cart sits at rest on a low-friction track. A horizontal force along the track rises linearly from 0 to 12 N over the first 0.20 s, holds steady at 12 N for the next 0.30 s, then falls linearly back to 0 over the final 0.10 s. Find (a) the impulse delivered, (b) the cart's speed at the end of the push, and (c) the average force over the whole 0.60 s.

   Show the worked solution

   1. Take the direction of the push as positive. The force never crosses the axis, so every area below counts as positive and nothing cancels.

   2. (a) Impulse is the area under the force versus time graph, so cut the shape into three pieces. Rising triangle: $\frac{1}{2}(0.20\ \text{s})(12\ \text{N}) = 1.2\ \text{N} \cdot \text{s}$.

   3. Flat rectangle: $(12\ \text{N})(0.30\ \text{s}) = 3.6\ \text{N} \cdot \text{s}$. Falling triangle: $\frac{1}{2}(0.10\ \text{s})(12\ \text{N}) = 0.60\ \text{N} \cdot \text{s}$.

   4. Total: $J = 1.2 + 3.6 + 0.60 = 5.4\ \text{N} \cdot \text{s}$.

   5. Check the shape whole instead of in pieces. It is a trapezoid with parallel sides 0.60 s (the base) and 0.30 s (the flat top) and height 12 N, so $J = \frac{1}{2}(0.60 + 0.30)(12) = \frac{1}{2}(0.90)(12) = 5.4\ \text{N} \cdot \text{s}$. Same number by a different route.

   6. (b) The cart starts at rest, so its change in momentum is simply its final momentum: $v_f = J/m = (5.4\ \text{N} \cdot \text{s})/(1.5\ \text{kg}) = 3.6\ \text{m/s}$.

   7. (c) $F_{\text{avg}} = J/\Delta t = (5.4\ \text{N} \cdot \text{s})/(0.60\ \text{s}) = 9.0\ \text{N}$. That is the height of the rectangle with the same area as the real graph, and at three quarters of the 12 N peak it sits below the peak, as any average must.

   (a) $J = 5.4\ \text{N} \cdot \text{s}$. (b) The cart leaves the push at $3.6\ \text{m/s}$. (c) $F_{\text{avg}} = 9.0\ \text{N}$, three quarters of the 12 N peak.

5. 5. Two carts that collide and separate

   On a frictionless horizontal track, a 2.5 kg cart moving right at 3.6 m/s strikes a 1.5 kg cart at rest. After the collision the 2.5 kg cart is still moving right, now at 1.2 m/s. Find (a) the 1.5 kg cart's velocity after the collision, and (b) whether the collision was elastic.

   Show the worked solution

   1. Rightward is positive. Take both carts as the system: the track is frictionless and horizontal, so gravity and the normal force cancel and no net external force acts along the track. The system's total momentum is therefore constant.

   2. (a) Before: $p_i = (2.5\ \text{kg})(3.6\ \text{m/s}) + (1.5\ \text{kg})(0) = 9.0\ \text{kg} \cdot \text{m/s}$.

   3. After: $p_f = (2.5\ \text{kg})(1.2\ \text{m/s}) + (1.5\ \text{kg})\,v_{2f} = 3.0 + 1.5\,v_{2f}$.

   4. Set $p_f = p_i$: $3.0 + 1.5\,v_{2f} = 9.0$, so $1.5\,v_{2f} = 6.0$ and $v_{2f} = 4.0\ \text{m/s}$, to the right.

   5. (b) Now, and only now, bring in kinetic energy. Before: $K_i = \frac{1}{2}(2.5)(3.6)^2 = (1.25)(12.96) = 16.2\ \text{J}$.

   6. After: $K_f = \frac{1}{2}(2.5)(1.2)^2 + \frac{1}{2}(1.5)(4.0)^2 = 1.8 + 12.0 = 13.8\ \text{J}$.

   7. The system lost $16.2 - 13.8 = 2.4\ \text{J}$, which is 14.8 percent of what it had. An elastic collision, by the definition in essential knowledge 4.4.A.1, would have left the system total untouched, so this one is inelastic.

   (a) The 1.5 kg cart moves right at $4.0\ \text{m/s}$. (b) Inelastic. Momentum holds at $9.0\ \text{kg} \cdot \text{m/s}$ across the collision, but kinetic energy falls from 16.2 J to 13.8 J, a loss of 2.4 J. The carts separated, so this is inelastic without being perfectly inelastic.

6. 6. A perfectly inelastic landing on a sled

   A 40 kg child runs at 3.5 m/s across level ice and lands on a 10 kg sled that is at rest, staying on it. Friction between the sled and the ice is negligible. Find (a) the speed of child and sled together immediately afterwards, (b) the kinetic energy before and after, and (c) the fraction of the kinetic energy that survives.

   Show the worked solution

   1. Take the child's running direction as positive and track horizontal momentum only. The vertical part of the landing is absorbed by the ice pushing up, and it never enters the horizontal bookkeeping.

   2. The child stays on the sled, so the two move off with one shared velocity. That is the defining feature of a perfectly inelastic collision in essential knowledge 4.4.A.5, and its practical value is that two unknowns collapse into one.

   3. (a) $p_i = (40\ \text{kg})(3.5\ \text{m/s}) + (10\ \text{kg})(0) = 140\ \text{kg} \cdot \text{m/s}$. The combined mass is $M = 40 + 10 = 50\ \text{kg}$, so $v_f = (140\ \text{kg} \cdot \text{m/s})/(50\ \text{kg}) = 2.8\ \text{m/s}$.

   4. (b) $K_i = \frac{1}{2}(40)(3.5)^2 = (20)(12.25) = 245\ \text{J}$ and $K_f = \frac{1}{2}(50)(2.8)^2 = (25)(7.84) = 196\ \text{J}$.

   5. The loss is $245 - 196 = 49\ \text{J}$, exactly 20 percent. It went into the scrape of the landing, the flex of the sled, sound, and a very slight warming: nonconservative transformations of the kind essential knowledge 4.4.A.4 describes.

   6. (c) The surviving fraction is $196/245 = 0.80$. There is a shortcut worth carrying: when the target starts at rest, the fraction of kinetic energy a perfectly inelastic collision keeps is $m_1/(m_1 + m_2) = 40/50 = 0.80$, with no velocities needed at all.

   7. Confirm the final energy without using $v_f$: $K_f = p^2/(2M) = (140)^2/(2 \times 50) = 19600/100 = 196\ \text{J}$, the same number reached from the momentum alone.

   (a) $2.8\ \text{m/s}$ in the child's original direction. (b) 245 J before, 196 J after. (c) 80 percent survives; the missing 20 percent is 49 J. The surviving 196 J is not optional: the 50 kg pair still carries the full $140\ \text{kg} \cdot \text{m/s}$, and anything with that mass and that momentum has exactly that kinetic energy.

7. 7. Recoil from rest: an astronaut throws a tool bag

   An astronaut of mass 75 kg floats at rest relative to her spacecraft, holding a 3.0 kg tool bag. She throws the bag away from her at 4.0 m/s. Find (a) her recoil velocity, (b) the total kinetic energy after the throw, and (c) where that energy came from, given that the system had none before.

   Show the worked solution

   1. Take the direction the bag is thrown as positive. The system is astronaut plus bag. Nothing outside that system pushes on it, so its total momentum is constant, and it begins at zero because everything is at rest.

   2. (a) $p_i = 0$, so $p_f = 0$ as well: $(75\ \text{kg})\,v_a + (3.0\ \text{kg})(+4.0\ \text{m/s}) = 0$.

   3. Solve: $(75)\,v_a = -12\ \text{kg} \cdot \text{m/s}$, so $v_a = -0.16\ \text{m/s}$. The minus sign says she moves opposite the bag, which is the only option open to a system whose total momentum has to stay at zero.

   4. Check the ratio. The speeds are 4.0 and 0.16 m/s, a factor of 25, and the masses are 75 and 3.0 kg, also a factor of 25. In any recoil or explosion from rest the two speeds come out in inverse proportion to the masses.

   5. (b) $K_{\text{bag}} = \frac{1}{2}(3.0)(4.0)^2 = 24\ \text{J}$ and $K_{\text{astronaut}} = \frac{1}{2}(75)(0.16)^2 = (37.5)(0.0256) = 0.96\ \text{J}$, so $K_f = 24.96\ \text{J}$, which is 25 J to two significant figures.

   6. That split is worth staring at. The two momenta are equal in magnitude, $12\ \text{kg} \cdot \text{m/s}$ each, yet the bag holds 25 times the kinetic energy. The reason is $K = p^2/(2m)$: at equal momentum, kinetic energy goes as $1/m$, and the bag's mass is 25 times smaller.

   7. (c) Momentum stayed fixed while kinetic energy was created, which only sounds contradictory if you expect the two to travel together. The 25 J came out of chemical energy in the astronaut's muscles. Conservation of momentum constrains the vector sum of $m\vec{v}$ and says nothing whatever about how much kinetic energy a system holds.

   (a) She recoils at $0.16\ \text{m/s}$, opposite the bag. (b) $K_f = 25\ \text{J}$: 24 J in the bag, 0.96 J in the astronaut. (c) From her muscles. The system's momentum sat at zero throughout while its kinetic energy rose from zero, which is what an explosion is. At $0.16\ \text{m/s}$ she drifts 9.6 m in the following minute.

8. 8. An elastic collision, both final velocities

   A 1.2 kg glider moving right at 5.0 m/s collides head-on and elastically with a 0.80 kg glider at rest on a frictionless air track. Find both final velocities, then verify that the collision conserves momentum and kinetic energy exactly.

   Show the worked solution

   1. Rightward is positive throughout. Elastic means two conditions hold at once: the system's total momentum is unchanged, and its total kinetic energy is unchanged.

   2. Write both down. Momentum: $(1.2)(5.0) = 1.2\,v_{1f} + 0.80\,v_{2f}$. Kinetic energy: $\frac{1}{2}(1.2)(5.0)^2 = \frac{1}{2}(1.2)\,v_{1f}^2 + \frac{1}{2}(0.80)\,v_{2f}^2$.

   3. Combining the two and discarding the do-nothing solution $v_{1f} = v_{1i}$ leaves a linear relation worth memorising: in a head-on elastic collision the relative velocity reverses, so $v_{1i} - v_{2i} = v_{2f} - v_{1f}$. Pairing that with momentum conservation gives the two standard results, neither of which is printed on the AP Physics 1 equation sheet.

   4. $$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}\, v_{1i}, \qquad v_{2f} = \frac{2 m_1}{m_1 + m_2}\, v_{1i}$$

   5. Substitute for the incoming glider: $v_{1f} = \frac{1.2 - 0.80}{1.2 + 0.80}(5.0) = \frac{0.40}{2.00}(5.0) = (0.20)(5.0) = 1.0\ \text{m/s}$, still to the right but much slower.

   6. And for the target: $v_{2f} = \frac{2(1.2)}{2.00}(5.0) = \frac{2.4}{2.00}(5.0) = (1.2)(5.0) = 6.0\ \text{m/s}$, to the right.

   7. Momentum check. Before: $(1.2)(5.0) = 6.0\ \text{kg} \cdot \text{m/s}$. After: $(1.2)(1.0) + (0.80)(6.0) = 1.2 + 4.8 = 6.0\ \text{kg} \cdot \text{m/s}$. Identical.

   8. Kinetic energy check, which is the one that has to balance to the digit. Before: $\frac{1}{2}(1.2)(5.0)^2 = (0.60)(25) = 15.0\ \text{J}$. After: $\frac{1}{2}(1.2)(1.0)^2 + \frac{1}{2}(0.80)(6.0)^2 = 0.60 + 14.4 = 15.0\ \text{J}$. Identical, so the collision really is elastic.

   9. Third check, the fastest one to do in your head: the gliders approach at $5.0 - 0 = 5.0\ \text{m/s}$ and separate at $6.0 - 1.0 = 5.0\ \text{m/s}$. Approach speed equal to separation speed happens only in an elastic collision.

   10. Worth noticing before you move on: the incoming glider kept 0.60 J of the 15.0 J it arrived with, and the target picked up 14.4 J. The total held perfectly still while the individual shares moved almost all the way across, which is precisely what essential knowledge 4.4.A.2 allows.

   $v_{1f} = 1.0\ \text{m/s}$ and $v_{2f} = 6.0\ \text{m/s}$, both to the right. Momentum is $6.0\ \text{kg} \cdot \text{m/s}$ before and after, and kinetic energy is 15.0 J before and after, so the collision is elastic on both counts.

9. 9. A two-dimensional collision, one axis at a time

   On a frictionless horizontal surface, a 3.0 kg puck moving east at 4.0 m/s collides with a 2.0 kg puck moving north at 5.0 m/s. The two stick together. Find the speed and direction of the combined pucks immediately after the collision, and the kinetic energy lost.

   Show the worked solution

   1. Set the axes before anything else: east is $+x$, north is $+y$. Momentum is a vector, so conservation holds separately along each axis. That is what turns this into two independent one-dimensional problems rather than one harder problem.

   2. Components before the collision. Only the 3.0 kg puck has any eastward motion, so $p_x = (3.0\ \text{kg})(4.0\ \text{m/s}) = 12\ \text{kg} \cdot \text{m/s}$. Only the 2.0 kg puck has any northward motion, so $p_y = (2.0\ \text{kg})(5.0\ \text{m/s}) = 10\ \text{kg} \cdot \text{m/s}$.

   3. They stick, so afterwards one 5.0 kg object carries both components: $v_x = 12/5.0 = 2.4\ \text{m/s}$ and $v_y = 10/5.0 = 2.0\ \text{m/s}$.

   4. Speed: $v = \sqrt{(2.4)^2 + (2.0)^2} = \sqrt{5.76 + 4.00} = \sqrt{9.76} = 3.124\ \text{m/s}$, so $3.1\ \text{m/s}$ to two significant figures.

   5. Direction: $\theta = \tan^{-1}(v_y/v_x) = \tan^{-1}(2.0/2.4) = \tan^{-1}(0.8333) = 39.8^\circ$, about 40 degrees north of east. Note that it is not 45 degrees even though the two speeds are similar, because the masses are not equal.

   6. Kinetic energy before: $\frac{1}{2}(3.0)(4.0)^2 + \frac{1}{2}(2.0)(5.0)^2 = 24 + 25 = 49\ \text{J}$. After: $\frac{1}{2}(5.0)(3.124)^2 = \frac{1}{2}(5.0)(9.76) = 24.4\ \text{J}$.

   7. Loss: $49 - 24.4 = 24.6\ \text{J}$, almost exactly half. Carry $v^2 = 9.76$ rather than the rounded 3.1 m/s into that last line: squaring 3.1 gives 9.61 and a final energy of 24.0 J, which is wrong in the third figure and drags the loss to 25.0 J.

   8. One scope note, from the Topic 4.3 boundary statement in the AP Physics 1 course and exam description (printed page 84). The course treats conservation of momentum in one dimension both quantitatively and qualitatively, but treats it in two dimensions only semiquantitatively. The same statement rules out exam questions that require solving simultaneous equations, while explicitly allowing questions that test whether you can set the equations up properly and reason about how changing a given mass, speed or angle would affect the other quantities; the full two-dimensional treatment, for problems with one unknown final velocity, is assigned to AP Physics 2. This problem needs no simultaneous equations, only the one conservation statement written once per axis, so expect the exam to ask you to set a case like this up and reason about it more often than to grind the arithmetic out.

   The joined pucks move off at $3.1\ \text{m/s}$, about 40 degrees north of east ($39.8^\circ$). Kinetic energy falls from 49 J to 24.4 J, so 24.6 J, close to half, is lost to the sticking. Momentum, by contrast, is unchanged on both axes: $12\ \text{kg} \cdot \text{m/s}$ east and $10\ \text{kg} \cdot \text{m/s}$ north, before and after.

10. 10. Collision first, friction second

    A 0.25 kg lump of clay slides at 12 m/s along a frictionless stretch of bench and hits a 0.75 kg block sitting at rest exactly where a rough stretch begins. The clay sticks to the block. The coefficient of kinetic friction between block and rough bench is 0.35. How far does the combined object slide before it stops? Use $g = 9.8\ \text{m/s}^2$.

    Show the worked solution

    1. Take the clay's direction of travel as positive. This is two problems joined at a single number, and the join is the velocity immediately after the collision.

    2. Stage 1, the collision. Momentum is conserved through it: $p_i = (0.25\ \text{kg})(12\ \text{m/s}) = 3.0\ \text{kg} \cdot \text{m/s}$. The clay sticks, so the combined $M = 0.25 + 0.75 = 1.0\ \text{kg}$ moves off at $v_f = 3.0/1.0 = 3.0\ \text{m/s}$.

    3. Kinetic energy is not conserved through stage 1, and that is the join the rest of the problem hangs on. The clay arrived with $\frac{1}{2}(0.25)(12)^2 = 18\ \text{J}$. Immediately after the impact the pair holds $\frac{1}{2}(1.0)(3.0)^2 = 4.5\ \text{J}$. The other 13.5 J went into deforming and heating the clay, and into sound, before the block had travelled anywhere.

    4. Stage 2, the slide. Now energy is the right tool, because friction is the only horizontal force left. The surface is level, so the normal force is $F_N = Mg = (1.0\ \text{kg})(9.8\ \text{m/s}^2) = 9.8\ \text{N}$ and friction is $f = \mu_k F_N = (0.35)(9.8\ \text{N}) = 3.43\ \text{N}$.

    5. Friction removes all 4.5 J over a distance $d$, so $f d = K$ and $d = (4.5\ \text{J})/(3.43\ \text{N}) = 1.312\ \text{m}$, which is $1.3\ \text{m}$.

    6. Cross-check with kinematics rather than energy. The deceleration is $a = \mu_k g = (0.35)(9.8) = 3.43\ \text{m/s}^2$, and $0 = v_f^2 - 2ad$ gives $d = (3.0)^2/(2 \times 3.43) = 9.0/6.86 = 1.312\ \text{m}$. Same distance.

    7. The trap, priced out. Carrying the clay's original 18 J straight into stage 2 gives $d = 18/3.43 = 5.2\ \text{m}$, four times too far. The factor of 4 is exactly the reciprocal of the fraction of kinetic energy a perfectly inelastic collision keeps when the target starts at rest, $m_1/(m_1 + m_2) = 0.25/1.0 = 0.25$.

    The clay and block slide $1.3\ \text{m}$ before stopping. Momentum carries you through the collision and energy carries you through the slide, and the two stages must not be merged: only 4.5 J of the clay's original 18 J is still kinetic when the sliding begins.

11. 11. Nothing was destroyed: reading a collision that ends at rest

    Two carts approach each other on a level, low-friction track. Cart A has mass 1.2 kg and moves right at 5.0 m/s; cart B has mass 3.0 kg and moves left at 2.0 m/s. They collide, stick together, and the pair is motionless afterwards. A student concludes that the collision destroyed $12\ \text{kg} \cdot \text{m/s}$ of momentum and 21 J of kinetic energy, and that conservation of momentum therefore fails here. Which half of that is right, and what went wrong with the other half?

    Show the worked solution

    1. Rightward is positive. Write both momenta with their signs before doing anything else: $p_A = (1.2\ \text{kg})(+5.0\ \text{m/s}) = +6.0\ \text{kg} \cdot \text{m/s}$ and $p_B = (3.0\ \text{kg})(-2.0\ \text{m/s}) = -6.0\ \text{kg} \cdot \text{m/s}$.

    2. The system's total momentum before the collision is $(+6.0) + (-6.0) = 0$. Afterwards the 4.2 kg pair is at rest, so its momentum is 0 too. Momentum was conserved. There was never any net momentum to destroy.

    3. The student's $12\ \text{kg} \cdot \text{m/s}$ came from adding magnitudes, $6.0 + 6.0$. That is scalar arithmetic applied to two quantities that point opposite ways, and it discards the very information the minus sign was carrying.

    4. The kinetic energy half of the claim is correct, because kinetic energy is a scalar and nothing cancels: $K_A = \frac{1}{2}(1.2)(5.0)^2 = 15\ \text{J}$ and $K_B = \frac{1}{2}(3.0)(2.0)^2 = 6.0\ \text{J}$, so $K_i = 21\ \text{J}$ and $K_f = 0$. All 21 J really did leave as deformation, heating and sound.

    5. The general lesson is the interesting part. This is the only kind of situation in which a collision can remove every joule of kinetic energy. A perfectly inelastic collision leaves the joined object carrying the system's total momentum, so the only way it can end at rest is if that total was zero to start with. Equal and opposite momenta is exactly that condition, and the masses need not match: 1.2 kg at 5.0 m/s and 3.0 kg at 2.0 m/s both carry $6.0\ \text{kg} \cdot \text{m/s}$.

    The energy half is right; the momentum half is not. Total momentum was zero before and zero after, so it was conserved. The student added the magnitudes of two momenta pointing opposite ways instead of adding them as signed quantities. Kinetic energy did genuinely go from 21 J to zero, and a perfectly inelastic collision can only do that when the system's total momentum is zero.

12. 12. Why momentum survives a collision that destroys kinetic energy

    Explain why the total momentum of a colliding system is conserved even in a collision that converts most of the system's kinetic energy into other forms. A good answer says what makes momentum different, rather than restating that it is conserved.

    Show the worked solution

    1. Start with the force pair. While the two objects are in contact, object 1 pushes on object 2 and object 2 pushes back on object 1. Newton's third law makes those two forces equal in magnitude and opposite in direction at every instant of the contact.

    2. Both forces act for the same length of time, because contact begins and ends for both objects at the same moment. Impulse is force multiplied by the time it acts, so the two impulses are equal in magnitude and opposite in direction: $\vec{J}_{12} = -\vec{J}_{21}$. Essential knowledge 4.3.A.3.i states exactly this, and names Newton's third law as the reason.

    3. Now apply the impulse-momentum theorem to each object separately: $\Delta \vec{p}_1 = \vec{J}_{21}$ and $\Delta \vec{p}_2 = \vec{J}_{12}$. Add them: $\Delta \vec{p}_{\text{system}} = \vec{J}_{21} + \vec{J}_{12} = 0$. Whatever momentum one object gains, the other loses, in the same instant and along the same line.

    4. Notice what that argument never needed. Nothing about how far the objects deformed, how hot they got, how much noise they made, or whether they bounced apart or stuck together. All of those are energy statements, and none of them appears anywhere in the derivation, which is why a violent collision leaves momentum conservation untouched.

    5. Now try to run the same argument for kinetic energy and watch it fail at the first step. The energy analogue of impulse is work, force multiplied by the displacement through which it acts. The two contact forces are still equal and opposite, but the two objects move through different displacements during the contact, and the contact surfaces themselves deform, so the two amounts of work do not cancel.

    6. There is a second, deeper reason. The internal forces in a real collision are not conservative: essential knowledge 4.4.A.4 says that in an inelastic collision some of the initial kinetic energy is not restored to kinetic energy but is transformed by nonconservative forces into other forms. Total energy is still conserved. The kinetic share of it is not, because energy has somewhere else to go. Momentum does not. There is no thermal momentum and no elastic potential momentum, only $m\vec{v}$.

    7. One caveat the argument exposes: it assumed the internal pair is the only significant force. That is the modelling assumption hiding inside the word collision, which essential knowledge 4.1.A.3.i defines as an interaction where the forces between the objects involved are much larger than the net external force on them during the interaction. Add a large external force, or watch long enough that gravity or friction accumulates, and the system's momentum does change, by exactly the external impulse.

    Because the two collision forces are a Newton's third law pair acting for the same contact time, their impulses are equal and opposite, so the momentum one object gains the other loses exactly. That argument mentions energy nowhere, which is why it survives a collision violent enough to destroy kinetic energy. Kinetic energy has no matching protection: the equal and opposite forces act through different displacements, so their works do not cancel, and the internal forces are nonconservative, so kinetic energy drains into thermal energy, sound and permanent deformation. Momentum has no other form to drain into.