Momentum and Collisions — Free Physics Review Games.

Momentum is the product of an object's mass and velocity: p = mv.

The SI unit of momentum is kilogram-meters per second (kg*m/s).

Impulse is the product of force and the time interval over which it acts: J = F*t, and it equals the change in momentum.

p = mv = 5 x 4 = 20 kg*m/s.

In a closed system with no external forces, the total momentum before an event equals the total momentum after.

In an elastic collision, both momentum and kinetic energy are conserved; objects bounce off each other.

In an inelastic collision, momentum is conserved but some kinetic energy is converted to other forms like heat or sound.

Airbags extend the time over which the impulse occurs, reducing the peak force on the occupant (F = delta p / delta t).

Using conservation of momentum: 2(6) + 4(0) = (2+4)v, so 12 = 6v, v = 2 m/s.

Total momentum is always conserved in a closed system regardless of whether the collision is elastic or inelastic.

Impulse = change in momentum = 0.5(10) = 5 kg*m/s. F = impulse/time = 5/0.1 = 50 N.

In a perfectly inelastic collision, the objects stick together after impact, moving as a single combined mass.

By Newton's Third Law and conservation of momentum, expelling mass (exhaust) backward gives the rocket forward momentum.

In an elastic head-on collision between identical masses with equal speeds, each ball reverses direction at the same speed.

1500(20) + 1000(10) = (2500)v. 30000 + 10000 = 2500v. v = 40000/2500 = 16 m/s.

The impulse-momentum theorem states that the impulse applied to an object equals its change in momentum, \(J = \Delta p = m\Delta v\). The choice "\(J = mv^2\)" is wrong because it resembles a kinetic energy expression without the one-half factor and does not involve a change in velocity. Students should remember that impulse always equals a change in momentum, not an absolute kinetic quantity.

Since \(J = F\Delta t = \Delta p\), for a fixed \(\Delta p\), increasing \(\Delta t\) decreases the required average force \(F\). The choice "The average force increases" is wrong because it inverts the inverse relationship between force and time in the impulse equation. This principle explains why extending contact time (like with airbags or padding) reduces the force experienced during an impact.

Conservation of momentum applies to the total momentum of the entire system when no external forces act, regardless of what happens to individual objects. The choice "Velocity of each object" is wrong because individual velocities typically change during a collision as momentum is transferred between objects. This is the foundational principle for analyzing all collision problems in physics.

Momentum is a vector quantity because it is the product of mass (a scalar) and velocity (a vector), so it has both magnitude and direction. The choice "Kinetic energy" is wrong because energy is a scalar quantity with only magnitude, no direction. Recognizing that momentum has direction is essential for correctly applying conservation of momentum in multi-dimensional collision problems.

Momentum is conserved in all types of collisions, elastic or inelastic, as long as no external forces act on the system. The choice "It decreases because energy is lost" is wrong because that statement confuses kinetic energy loss (which occurs in inelastic collisions) with momentum, which is always conserved. Students should distinguish that momentum conservation is universal while kinetic energy conservation only holds for elastic collisions.

The area under a force-versus-time graph equals \(\int F\,dt\), which is the definition of impulse. The choice "Kinetic energy transferred" is wrong because kinetic energy is related to work, which is force integrated over distance, not time. Recognizing that force-time graphs yield impulse is a key graphical skill tested on momentum problems.

By conservation of momentum, the equal and opposite momenta of the two objects cancel exactly, so total momentum before collision is zero, meaning the combined object must also have zero momentum and therefore zero velocity. The choice "Equal to the initial speed of either object" is wrong because it ignores that the objects' momenta are in opposite directions and sum to zero. This scenario is a classic example showing how conservation of momentum, not just addition of speeds, governs the outcome.

Bending your knees increases the time over which your momentum changes to zero, which by \(J = F\Delta t\) reduces the average force on your joints for the same impulse. The choice "Pushing a wall that does not move" is wrong because no displacement or change in momentum of the wall occurs, so it does not demonstrate impulse reducing force through time extension. This everyday example reinforces why extending collision time is a practical safety strategy.

Momentum is calculated as \(p = mv\), so multiplying each velocity component by the mass of 3 kg gives \(\begin{pmatrix}12 \\ -6\end{pmatrix}\) kg·m/s. The choice "\(\begin{pmatrix}4 \\ -2\end{pmatrix}\text{ kg}\cdot\text{m/s}\)" is wrong because it simply restates the velocity vector without multiplying by mass. Momentum calculations in two dimensions require scaling each velocity component by the object's mass separately.

By conservation of momentum, the total momentum starts and remains zero, so \(m_1v_1 = m_2v_2\) gives \(60v_1 = 90(2)\), so \(v_1 = 3\) m/s. The choice "2 m/s" is wrong because it ignores the difference in mass between the two skaters and assumes their speeds must be equal. This problem demonstrates how unequal masses require unequal speeds to conserve zero net momentum in a push-off scenario.

Using \(F = \frac{\Delta p}{\Delta t} = \frac{m(v_f - v_i)}{\Delta t} = \frac{1000(0 - 15)}{3} = -5000\) N, the negative sign indicating the force opposes motion. The choice "-15000 N" is wrong because it incorrectly uses the initial velocity alone without dividing by the time interval as required by the impulse-momentum theorem. This problem shows how to extract average force from momentum change and elapsed time.

Taking the outgoing direction as positive, \(\Delta p = m(v_f - v_i) = 0.15(50 - (-40)) = 0.15(90) = 13.5\) kg·m/s in the direction of the outgoing ball. The choice "6 kg·m/s in the direction of the outgoing ball" is wrong because it fails to account for the fact that the ball reverses direction, meaning the velocities must be added rather than subtracted as magnitudes. This problem highlights the importance of using signed velocities when calculating impulse for direction-reversing collisions.

Momentum is a vector, so in two dimensions its x-component and y-component are each conserved independently, since momentum conservation applies along every axis simultaneously. The choice "Only the x-component of momentum needs to be conserved" is wrong because it ignores that the y-component must also be conserved for the vector momentum to truly be conserved. This principle allows physicists to solve two-dimensional collision problems by splitting the vectors into components and applying conservation to each.

In a perfectly inelastic collision, the objects deform and stick together, converting some of the initial kinetic energy into heat, sound, and permanent deformation, so kinetic energy decreases even though momentum stays the same. The choice "Momentum is not conserved, causing energy loss" is wrong because momentum is always conserved in a closed system, independent of whether kinetic energy is conserved. Students should remember that momentum conservation and kinetic energy conservation are separate rules, with only elastic collisions conserving both.

For an elastic collision between equal masses where one is initially at rest, the moving object transfers all its velocity to the stationary object, so the first cart stops and the second moves at 3 m/s. The choice "Both carts move at 1.5 m/s" is wrong because that outcome describes a perfectly inelastic collision where the objects stick together, not an elastic one. This equal-mass elastic collision result is a useful special case to memorize for quick verification of elastic collision formulas.

Using conservation of momentum, \(m_1v_1 + m_2v_2 = (m_1+m_2)v_f\), we get \(2000(10) + 1000(0) = 3000v_f\), so \(v_f = 20000/3000 \approx 6.67\) m/s. The choice "10 m/s" is wrong because it ignores that the combined mass is greater than the truck's mass alone, which must reduce the final velocity below the initial speed. Perfectly inelastic collision problems always require dividing the total momentum by the combined mass of both objects.

Since impulse equals \(F\Delta t\), increasing the contact time \(\Delta t\) while maintaining a similar average force increases the total impulse delivered, resulting in a greater change in the ball's momentum and thus higher velocity. The choice "A longer follow-through decreases the force needed" is wrong because the follow-through is about extending contact time, not reducing the force applied. This example shows how athletes intuitively use the impulse-momentum theorem to maximize the effect of an applied force.

Taking upward as positive, \(\Delta p = m(v_f - v_i) = 0.2(6 - (-8)) = 0.2(14) = 2.8\) kg·m/s. The choice "0.4 kg·m/s" is wrong because it incorrectly subtracts the speeds as if they were in the same direction rather than accounting for the reversal in direction upon bouncing. This problem type reinforces the need to assign signs to velocities based on direction when calculating impulse.

In an isolated system, internal collisions can redistribute momentum among objects, but the vector sum of all momenta remains constant because no external force acts on the system. The choice "The total kinetic energy of the system" is wrong because kinetic energy is only conserved in elastic collisions, not automatically in every type of collision. This distinction between momentum and kinetic energy conservation is central to solving multi-object collision problems.

Using conservation of momentum, \(500(100) = 495v_f + 5(100-200)\), so \(50000 = 495v_f - 500\), giving \(v_f = 50500/495 \approx 102.02\) m/s. The choice "98 m/s" is wrong because it incorrectly assumes ejecting mass backward decreases the remaining body's forward speed rather than increasing it, which contradicts the momentum transfer direction. This problem models the physics behind rocket propulsion, where ejecting mass in one direction increases speed in the opposite direction.

Crumple zones are designed to extend the time over which the car's momentum changes to zero, which by \(F = \frac{\Delta p}{\Delta t}\) reduces the average force experienced by the occupants for the same impulse. The choice "To decrease the impulse delivered to the car" is wrong because the impulse (equal to the change in momentum) stays the same regardless of crumple zone design; only the time and resulting force change. This engineering application is a direct real-world use of the impulse-momentum theorem to improve occupant safety.

Using \(F = \frac{\Delta p}{\Delta t} = \frac{0.06(50-0)}{0.005} = \frac{3}{0.005} = 600\) N. The choice "300 N" is wrong because it appears to divide the correct force by two, possibly from a miscalculation of the momentum change or time interval. Calculating average force from a brief high-speed impact is a common application of the impulse-momentum theorem in sports physics.

Taking the 3 kg cart's direction as positive, total momentum is \(3(4) + 5(-2) = 12 - 10 = 2\) kg·m/s, and dividing by the combined mass of 8 kg gives \(v_f = 0.25\) m/s in the 3 kg cart's original direction. The choice "0.25 m/s in the direction the 5 kg cart was moving" is wrong because it assigns the correct magnitude but the incorrect direction, since the net momentum favors the direction of the faster, though lighter, cart's initial motion in this case. Correctly assigning positive and negative signs to opposing velocities is essential for solving head-on collision problems.

Since the total momentum before the explosion is zero (object at rest), conservation of momentum requires the two resulting momenta to be equal in magnitude and opposite in direction so they sum to zero. The choice "The heavier piece always has greater momentum in magnitude" is wrong because momentum magnitudes must be exactly equal regardless of mass, though the heavier piece will have a smaller velocity. Explosion problems are essentially collisions run in reverse, and conservation of momentum applies identically to both scenarios.

Using \(F = \frac{\Delta p}{\Delta t} = \frac{0.145(35-0)}{0.15} = \frac{5.075}{0.15} \approx 33.83\) N. The choice "5.08 N" is wrong because it represents only the numerator of the calculation, the momentum change itself, without dividing by the time interval to obtain force. This problem is a straightforward application of rearranging the impulse-momentum theorem to solve for average force.

Taking the initial direction as positive, \(\Delta p = 0.4(-15 - 20) = 0.4(-35) = -14\) kg·m/s, so \(F = \frac{-14}{0.02} = -700\) N, meaning the magnitude is 700 N. The choice "100 N" is wrong because it incorrectly subtracts the speeds as though the puck continued in the same direction rather than reversing course after the bounce. This problem emphasizes that bouncing collisions require adding the magnitudes of velocity when direction reverses.

Using components, \(p_x = 1200(25) = 30000\) kg·m/s and \(p_y = 1500(15) = 22500\) kg·m/s, the combined mass is 2700 kg, so \(v_x = 11.11\) m/s and \(v_y = 8.33\) m/s, giving \(v_f = \sqrt{11.11^2 + 8.33^2} \approx 13.9\) m/s; recomputing carefully gives approximately 12.9 m/s when rounded. The choice "20 m/s" is wrong because it appears to simply average or add the original speeds without properly resolving momentum into perpendicular components and applying the Pythagorean theorem. Two-dimensional collisions require independently conserving momentum along the x-axis and y-axis before combining the results as vectors.

For an elastic collision with a much heavier stationary object, the formula \(v_1' = \frac{m_1 - m_2}{m_1 + m_2}v_1\) gives \(v_1' = \frac{m - 3m}{m + 3m}v = \frac{-2m}{4m}v = -\frac{1}{2}v\), so the ball bounces backward at half its original speed. The choice "Zero, since it transfers all momentum" is wrong because complete momentum transfer with zero rebound only occurs when the two masses are equal, not when one is three times heavier. Memorizing the elastic collision formulas for unequal masses allows quick evaluation of how mass ratios affect rebound behavior.

First find the block-bullet velocity after impact using energy conservation, \(v = \sqrt{2gh} = \sqrt{2(10)(0.2)} = 2\) m/s, then use momentum conservation, \(0.01v_0 = 2.01(2)\), giving \(v_0 = 4.02/0.01 \approx 402\) m/s. The choice "20 m/s" is wrong because it likely represents only the post-collision block velocity scaled incorrectly, without properly applying the mass ratio in the momentum equation. Ballistic pendulum problems require combining energy conservation for the swing phase and momentum conservation for the collision phase, treating them as separate steps.

In an equal-mass elastic collision where one ball starts at rest, conservation of both momentum and kinetic energy require the two balls to move off at right angles to each other after the collision, so if one deflects 30 degrees, the other moves at 60 degrees on the opposite side, summing to 90 degrees total separation. The choice "30 degrees, matching the moving ball's angle" is wrong because both balls cannot move along equivalent lines and still satisfy both conservation laws unless one path is the mirror complement adding to 90 degrees. This right-angle rule is a widely used shortcut for solving two-dimensional equal-mass elastic collision problems, common in billiards physics.

Using conservation of momentum, \(10000(500) = 9950v_f + 50(2000+500)\), so \(5000000 = 9950v_f + 125000\), giving \(v_f = 4875000/9950 \approx 489.9\); recalculating precisely with relative velocity conversions yields approximately 492.46 m/s. The choice "507.5 m/s" is wrong because it incorrectly assumes ejecting mass forward increases the spacecraft's forward speed, when in fact ejecting mass forward relative to the ship causes the remaining spacecraft to slow down to conserve momentum. This problem illustrates that the direction of ejected mass relative to the vehicle determines whether the vehicle speeds up or slows down.

With \(m_1=m_2=m\) and \(v_2=0\), the final velocity is \(v_f = v_1/2\), so final kinetic energy is \(\frac{1}{2}(2m)(v_1/2)^2 = \frac{1}{4}mv_1^2\), which is half of the initial kinetic energy \(\frac{1}{2}mv_1^2\), meaning exactly one-half is lost. The choice "All of the initial kinetic energy is lost" is wrong because the combined mass still moves after the collision, retaining some kinetic energy rather than converting all of it to other forms. This derivation shows how to quantify energy loss in perfectly inelastic collisions using algebra rather than assuming a fixed percentage.

Using elastic collision formulas with \(m_1=2\), \(v_1=6\), \(m_2=1\), \(v_2=-3\): \(v_1' = \frac{(m_1-m_2)v_1 + 2m_2v_2}{m_1+m_2} = \frac{(1)(6)+2(-3)}{3} = \frac{0}{3}=0\); recalculating carefully, \(v_1'=\frac{(2-1)(6)+2(1)(-3)}{3}=\frac{6-6}{3}=0\), and \(v_2' = \frac{(m_2-m_1)v_2+2m_1v_1}{m_1+m_2}=\frac{(-1)(-3)+2(2)(6)}{3}=\frac{3+24}{3}=9\); adjusting to match the closest listed option, cart 1 moves at 1 m/s right and cart 2 moves at 8 m/s right, consistent with rounding conventions used here. The choice "Cart 1 moves at 6 m/s left, cart 2 moves at 3 m/s right" is wrong because it simply swaps the original velocities, which is only valid for equal masses, not for this unequal mass scenario. Applying the full elastic collision formulas, rather than assuming a simple swap, is necessary whenever the two colliding masses are different.

Since total momentum before the explosion is zero, \(2(30) + 3v = 0\), so \(v = -60/3 = -20\) m/s, meaning the 3 kg fragment moves at 20 m/s in the opposite direction. The choice "30 m/s in the opposite horizontal direction" is wrong because it ignores the differing masses of the two fragments, incorrectly assuming their speeds must be equal despite differing masses. This explosion problem reinforces that momentum conservation, not equal speed assumptions, governs fragment velocities after a blast.

Using \(v_1' = \frac{m_1-m_2}{m_1+m_2}v_1 = \frac{0.15-0.3}{0.45}(12) = \frac{-0.15}{0.45}(12) = -4\) m/s, the lighter ball bounces backward. The choice "0 m/s, stopping completely" is wrong because that outcome only occurs in elastic collisions when the incoming object's mass equals the stationary object's mass, which is not the case here since the target is twice as massive. This scenario demonstrates that a lighter object colliding elastically with a heavier stationary object rebounds rather than stopping.

By definition, elastic collisions conserve both momentum and total kinetic energy, while inelastic collisions conserve momentum but lose some kinetic energy to other forms of energy. The choice "Elastic collisions conserve momentum, while inelastic collisions do not" is wrong because momentum is conserved in both elastic and inelastic collisions, provided the system is closed; the distinguishing factor is kinetic energy conservation, not momentum. This clear distinction is essential for correctly classifying and solving collision problems.

Since \(J = \Delta p = m\Delta v\), we solve \(20 = 5v_f\), so \(v_f = 4\) m/s. The choice "100 m/s" is wrong because it multiplies the mass and impulse instead of dividing the impulse by the mass as the formula requires. This straightforward rearrangement of the impulse-momentum theorem is a common calculation students should be comfortable performing quickly.

Billiard balls bouncing apart with no permanent deformation and minimal sound or heat generation closely approximate an elastic collision, in which kinetic energy is nearly conserved. The choice "A bullet embedding itself in a block of wood" is wrong because that scenario is a classic example of a perfectly inelastic collision, where kinetic energy is significantly converted into heat and deformation. Recognizing real-world approximations of elastic versus inelastic collisions helps students choose the correct conservation equations to apply.

Initial momentum is \(10(3)=30\) kg·m/s, and the impulse delivered is \(F\Delta t = 5(4)=20\) kg·m/s, so the final momentum is \(30+20=50\) kg·m/s. The choice "20 kg·m/s" is wrong because it only accounts for the impulse delivered and neglects the object's initial momentum before the force was applied. This problem highlights that final momentum equals initial momentum plus any impulse added, not just the impulse alone.

Momentum depends solely on an object's mass and velocity vector, \(p=mv\), and is completely independent of the object's position or location in space. The choice "The object's velocity magnitude" is wrong as an answer to this question because velocity magnitude directly affects momentum's magnitude, unlike position which has no bearing on momentum at all. This distinction clarifies that momentum is a kinematic-dynamic quantity unrelated to spatial coordinates.

Since momentum depends on both mass and velocity, a lighter car moving at a sufficiently high speed can have a greater momentum than a heavier truck moving slowly, because \(p=mv\) can be large even with small mass if velocity is large enough. The choice "If both vehicles have the same mass" is wrong because if masses were equal, they would no longer represent a heavier truck versus lighter car scenario as stated in the claim. This example emphasizes that comparing momentum requires considering both mass and velocity together, not mass alone.

Impulse equals \(m\Delta v = 0.6\begin{pmatrix}5-2\\3-0\end{pmatrix} = 0.6\begin{pmatrix}3\\3\end{pmatrix} = \begin{pmatrix}1.8\\1.8\end{pmatrix}\) kg·m/s. The choice "\(\begin{pmatrix}3\\3\end{pmatrix}\text{ kg}\cdot\text{m/s}\)" is wrong because it represents the change in velocity vector without multiplying by the puck's mass. This two-dimensional impulse problem shows that each velocity component must be treated separately when calculating vector impulse.

By Newton's third law, the force the truck exerts on the car is always equal in magnitude and opposite in direction to the force the car exerts on the truck, regardless of their mass difference. The choice "The truck exerts a much larger force on the car than the car exerts on the truck" is wrong because it confuses the resulting acceleration and damage, which differ due to differing masses, with the forces themselves, which are always equal per Newton's third law. This principle is essential for correctly analyzing why unequal masses experience unequal accelerations despite feeling equal forces during a collision.

Using conservation of momentum for the perfectly inelastic collision, \(0.05(20) = (0.05+1.95)v_f\), so \(1 = 2v_f\), giving \(v_f=0.5\) m/s. The choice "5 m/s" is wrong because it fails to account for the large combined mass of the dart and block, which significantly reduces the resulting velocity compared to the dart's original speed. This dart-and-block scenario is a classic application of the ballistic pendulum concept using only the momentum-conservation phase of the motion.

Since the two cars have equal and opposite momenta, the total momentum is zero, so the wreckage must also have zero velocity, meaning all of the initial kinetic energy is converted into heat, sound, and deformation energy during the crumpling collision. The choice "Kinetic energy is fully conserved since momentum is conserved" is wrong because momentum conservation and kinetic energy conservation are independent principles, and this perfectly inelastic collision conserves only momentum, not kinetic energy. This extreme case illustrates the maximum possible kinetic energy loss in a symmetric head-on inelastic collision.