Momentum solutions

Choose positive toward the wall and convert 57 g to 0.057 kg.

Why: Momentum and energy calculations use kilograms. Choosing toward the wall as positive gives the later rebound velocity a clear sign.

The initial kinetic energy is K_i = ½(0.057)(15²) = 6.4125 J.

Why: Kinetic energy depends on speed squared, so the initial 15 m/s speed gives a positive energy regardless of the coordinate sign.

After losing 3 J, K_f = 3.4125 J, so the return speed is √(2K_f/m) = 10.94 m/s.

Why: The stated 3 J loss comes out of kinetic energy. Rearranging K = ½mv² gives the remaining speed, which is a nonnegative magnitude.

The ball rebounds, so its velocity is opposite the initial direction.

Why: Energy alone cannot choose a direction because it contains v². The word 'rebounds' tells us to attach a negative sign to the return speed.

Treat the short collision along the common line of motion and neglect external impulse along that line.

Why: Choose the two crates as the system during impact. Negligible external impulse along the collision axis is what allows their total momentum to stay constant.

Before impact p_i = (2 kg)(4 m/s) + (4 kg)(0) = 8 kg·m/s.

Why: Add each crate's signed mass-times-velocity contribution. The stationary crate carries no initial momentum even though it contributes mass.

For a completely inelastic collision, p_f = (2 + 4)v_f.

Why: Sticking forces both crates to share one final velocity. Their separate momenta therefore combine into total mass times that common velocity.

Thus v_f = 8/6 = 1.33 m/s in crate A’s initial direction.

Why: Equate the final expression to the conserved 8 kg·m/s and divide by total mass. The positive result retains A's original direction.

Inertia is resistance to acceleration and is measured by mass; it does not depend on velocity.

Why: Inertia measures how much force is needed for a given acceleration. In this setting that resistance is the object's mass, not its speed.

Using typical masses, a boulder is heaviest, followed by a bowling ball, a football, and a bullet.

Why: The question asks for typical objects rather than exact supplied masses. Use their usual mass ordering to rank inertia, with that assumption kept explicit.

At the same velocity, momentum magnitude p = mv follows the same mass ranking.

Why: All objects have the same velocity, so multiplying their masses by one common speed preserves the same momentum-magnitude ranking.

Choose upward as positive: p_i = (45)(−5.10) = −229.5 kg·m/s.

Why: Upward-positive makes the initial falling velocity negative. Multiplying it by Barby's mass gives her signed starting momentum.

The first upward net impulse is J_1 = (340)(0.45) = 153 N·s; the second is J_2 = +80 N·s.

Why: An average net force supplies impulse FavgΔt. Both given impulses point upward, and gravity is already included in the net-force data.

Apply J_net = p_f − p_i: p_f = −229.5 + 153 + 80 = +3.5 kg·m/s.

Why: Impulse adds to signed momentum. The two upward impulses first cancel the downward momentum and then leave a small upward remainder.

Divide by 45 kg: v_f = +0.0778 m/s. The small positive result means upward motion.

Why: Momentum divided by mass recovers velocity. The positive remainder means Barby has just reversed into upward motion.

Take +x along the incoming cue ball and +y above it. The cue ball has mass 1 kg, initial speed 5 m/s; the 2 kg target is initially stationary.

Why: Use the incoming motion as the x direction so all initial momentum lies on that axis. The final directions then set each component's sign.

Conserve x momentum: 5 = v_1 cos34° + 2v_2 cos42°.

Why: Negligible external impulse preserves total x momentum. Project each final speed with cosine and multiply by that ball's own mass.

Conserve y momentum: 0 = v_1 sin34° − 2v_2 sin42°, so v_2 = v_1 sin34°/(2sin42°).

Why: Initial y momentum is zero. The cue ball's upward momentum must balance the target's downward momentum, which relates their two unknown speeds.

Substitution gives v_1 = 5sin42°/sin76° = 3.45 m/s and v_2 = 5sin34°/(2sin76°) = 1.44 m/s.

Why: Substitute the y-momentum relation into the x equation to eliminate one unknown. The angle sum identity condenses the two trig terms to sin 76°.

Their directions are 34° above +x and 42° below +x. The final kinetic energy is about 8.02 J versus 12.5 J initially, so these data describe an inelastic collision.

Why: Directions are part of each final velocity, not of speed alone. Comparing initial and final kinetic energies classifies the collision as inelastic rather than assuming energy conservation.

Take east as positive. The initial velocities are +4 m/s and −8 m/s.

Why: A single axis needs one consistent sign. East-positive makes the bull's westward velocity negative while the matador's remains positive.

Total momentum is p_i = 45(4) + 125(−8) = −820 kg·m/s.

Why: Add the two signed momenta. The bull's larger westward contribution outweighs the matador's eastward one.

They move together, so v_f = −820/(45+125) = −4.82 m/s.

Why: Sticking gives one velocity for the combined mass. Conserved negative momentum divided by total mass means the pair moves west.

Read the recovered preview: speed 100 m/s, stopping time 0.3 s, dummy mass 85 kg. Choose the initial motion as positive.

Why: The stopping calculation needs the dummy's mass, its initial speed and the duration. Choosing its initial motion as positive makes a stopping force negative.

The horizontal impulse is Δp = 85(0−100) = −8500 N·s.

Why: Impulse is the change from the dummy's initial momentum to zero. Its negative sign shows the required change opposes the original motion.

Assuming the belt supplies the horizontal stopping impulse, F_avg = Δp/Δt = −8500/0.3 = −28,333 N.

Why: The belt is assumed to provide that entire horizontal impulse. Divide by the stopping duration to get its average force, with the negative sign retained.

Take upward as positive, so v_i = −20 m/s and v_f = 0.

Why: The parachutist initially moves downward, so upward-positive assigns a negative initial velocity. Rest gives zero final velocity.

The change in momentum is Δp = 60[0−(−20)] = +1200 N·s.

Why: Subtract the negative initial momentum from the final zero momentum. Stopping a downward motion requires a positive upward impulse.

The average NET force is Δp/0.6 = 2000 N upward.

Why: Divide the momentum change by the contact duration to find average net force. This is the total vertical force after all contributions combine.

During contact N_avg − mg = F_net, so N_avg = 2000 + 60(9.81) = 2588.6 N.

Why: The ground force must both offset the downward weight and provide the upward net force. Adding mg converts net force into the actual contact force.

Take right as positive. Before the first collision p = 73(−10) + 64(3) = −538 kg·m/s.

Why: Right-positive gives opposite signs to the players' initial velocities. Add their mass-weighted velocities to find the first system momentum.

After the first collision the two-player speed is v_12 = −538/137 = −3.927 m/s.

Why: The first sticking collision gives one velocity for their combined 137 kg. Divide conserved momentum by that mass.

The third player, mass 55 kg, is initially at rest. Momentum remains −538 kg·m/s through the second sticking collision.

Why: Include the stationary third player in the next system. It adds mass but no initial momentum, so the second collision still starts with −538 kg·m/s.

The final total mass is 73+64+55 = 192 kg, so v_f = −538/192 = −2.802 m/s.

Why: After the second sticking collision all three share a velocity. Divide the conserved momentum by their full 192 kg mass.

Interpret the 2.5 m/s as the speed at the ramp bottom. On the frictionless rise, ½mv_patch² = ½m(2.5²) − mg(0.03).

Why: On the frictionless rise, kinetic energy supplies the added gravitational potential energy. Use 0.03 m for the height and keep the patch-entry speed distinct from the bottom speed.

Thus v_patch² = 6.25 − 0.5886 = 5.6614 m²/s² and v_patch = 2.379 m/s.

Why: Cancel the common mass and solve for the entry speed squared. The positive result confirms the block reaches the elevated patch.

On the horizontal patch, friction does work −μ_k mgd. At rest, ½mv_patch² = μ_k mgd.

Why: On the horizontal patch, gravity and the normal force do no work. Kinetic friction removes all entry kinetic energy over the stopping distance.

The stopping distance is d = 5.6614/[2(0.5)(9.81)] = 0.5771 m, shorter than the 0.6 m patch.

Why: Solve the energy balance for d and compare it with the actual patch length. A shorter stopping distance means the block cannot exit.

It stops on the patch and never reaches its far end; an attempted exit-speed calculation would give a negative squared speed.

Why: A speed squared cannot be negative. That impossible result at the far end signals that the energy ran out earlier, rather than describing an exit velocity.

Use east as +x and north as +y. Momentum components are p_x = 1.5(4) = 6 and p_y = 1.5(3) = 4.5 kg·m/s.

Why: Momentum has the same direction as velocity. Multiply each velocity component by the drone's same 1.5 kg mass.

The magnitude is √(6²+4.5²) = 7.5 kg·m/s.

Why: The east and north momentum components are perpendicular. Pythagoras gives the magnitude of their resultant.

The direction is tan⁻¹(4.5/6) = 36.87° north of east.

Why: Both components are positive, so the direction lies north of east. Their ratio gives the first-quadrant angle from east.

For each cart v = p/m: 8 m/s for 2 kg and 2 m/s for 8 kg.

Why: Rearrange p = mv separately for each mass. At equal momentum, the lighter cart must move faster.

Use K = p²/(2m): K_small = 256/4 = 64 J; K_large = 256/16 = 16 J.

Why: Substitute v = p/m into K = ½mv² to get K = p²/(2m). Equal p then gives greater kinetic energy to the smaller mass.

Inertia depends on mass, so the 8 kg cart has more inertia even though its kinetic energy is lower.

Why: Inertia is determined by mass. Neither equal momentum nor the lighter cart's greater energy changes which cart resists acceleration more.

Impulse is the triangular area under F(t): J = ½(0.04)(60) = 1.2 N·s.

Why: The force pulse forms one triangle over the full 0.04 s interval. Its signed area is the net impulse, with units N·s.

The initial momentum is 0.20(2) = 0.40 kg·m/s; final momentum is 1.60 kg·m/s.

Why: Multiply the initial velocity by mass, then add the positive impulse. The pulse increases the puck's +x momentum.

The final velocity is 1.60/0.20 = 8 m/s. Average force is 1.2/0.04 = 30 N.

Why: Divide final momentum by mass to recover velocity. Divide impulse by the full pulse duration to get the average force rather than its 60 N peak.

Add signed impulses: J = (12)(0.5)+(−9)(1.0) = −3 N·s.

Why: Impulse keeps the force signs. The later negative impulse is larger than the first positive one, giving a negative total momentum change.

Initial momentum is 3(2)=6 kg·m/s; final momentum is 6−3=3 kg·m/s.

Why: Add that −3 N·s change to the initial +6 kg·m/s momentum. The result is still positive.

Therefore v_f = 3/3 = +1 m/s. It slowed but did not reverse.

Why: Momentum divided by mass gives a positive final velocity. During the first pulse velocity increases, then decreases to +1 m/s without crossing zero.

Let launcher velocity be V and ball velocity be u. Relative speed means u−V = 300 m/s, not u = 300.

Why: The 300 m/s is measured from the recoiling launcher. Laboratory ball velocity minus laboratory launcher velocity must equal that relative speed.

Zero initial momentum gives 4V+0.020u = 0. Substitute u = V+300.

Why: Choose launcher plus ball as the system. Zero external impulse keeps their initially zero total momentum zero, even while internal forces separate them.

Thus 4.020V = −6, so V = −1.4925 m/s and u = 298.5075 m/s.

Why: Solve the two relations together. The launcher recoils backward, which makes the ball's laboratory speed slightly less than its launcher-relative speed.

Initial momentum is 12(3)=36 kg·m/s.

Why: Before separation, the full 12 kg moves together. Its initial total momentum sets the conserved value for both modules afterward.

Conservation gives 36 = 8v+4(9), so v = 0 m/s.

Why: The 4 kg module's specified velocity accounts for the entire initial momentum. The 8 kg module must therefore carry zero remaining momentum.

Initial K = ½(12)(3²)=54 J. Final K = ½(4)(9²)+½(8)(0²)=162 J.

Why: Compute and add kinetic energy for each body at each stage. The modules' different final speeds mean their energy cannot be found from the old shared speed.

The separation supplies 162−54 = 108 J from internal stored energy.

Why: Subtract initial from final kinetic energy. The increase comes from internal stored energy released in separation; momentum conservation does not require kinetic energy conservation.

Take east as positive. Initial momentum is 60(2)+3(−5)=105 kg·m/s.

Why: Choose both skater and ball as the system. East-positive makes the westbound ball's momentum negative, so it reduces the skater's initial total.

Common velocity is v = 105/63 = 1.6667 m/s east.

Why: Holding the ball gives one shared velocity for the 63 kg pair. Divide their conserved horizontal momentum by that combined mass.

Initial K = ½(60)(2²)+½(3)(5²)=157.5 J. Final K = ½(63)(1.6667²)=87.5 J.

Why: Add the separate initial kinetic energies, then use the shared final velocity for the combined final mass. Squaring removes directional signs in both energies.

Kinetic-energy decrease is 70 J, converted into internal energy during catching.

Why: Subtract final kinetic energy from initial. The lost mechanical energy becomes internal energy during the inelastic catch.

Initial momentum is 2(5)+3(−1)=7 kg·m/s.

Why: Add both signed initial momenta. With negligible external impulse, their total is the value the final momenta must reproduce.

For an elastic head-on collision, relative velocity reverses: v_2f−v_1f = 5−(−1)=6 m/s.

Why: Elastic one-dimensional collisions reverse the relative approach velocity. This adds the energy constraint to momentum conservation without choosing final velocities in advance.

Solve 2v_1f+3v_2f=7 with v_2f=v_1f+6: v_1f=−2.2 m/s and v_2f=+3.8 m/s.

Why: Use the relative-velocity equation to remove one unknown from the momentum equation. The resulting signs show the lighter cart rebounds and the heavier one moves forward.

Check energy: initial K = 26.5 J; final K = ½(2)(2.2²)+½(3)(3.8²)=26.5 J.

Why: An elastic result must preserve kinetic energy as well as momentum. Recomputing both final energies verifies that second requirement.

Momentum conservation gives 4 = −1+3v_2f; thus v_2f=5/3 = 1.6667 m/s.

Why: The rebounding first cart has negative final momentum. Conservation requires the second cart to carry the remaining positive momentum.

Initial kinetic energy is 8 J. Final kinetic energy is ½(1)(1²)+½(3)(5/3)²=4.6667 J.

Why: Compare total kinetic energies using each cart's own mass and speed. The target's new motion does not by itself prove elasticity.

The decrease is 3.3333 J, so the collision is inelastic. The carts have different final velocities and did not stick, so it is not completely inelastic.

Why: A kinetic-energy decrease makes the collision inelastic. Different final velocities show that this is not the sticking, completely inelastic case.

For equal masses with a stationary target, momentum and kinetic energy imply perpendicular outgoing velocities when both are nonzero.

Why: For equal masses, momentum says the incoming velocity is the sum of the outgoing vectors. Squaring that sum and comparing with conserved energy forces their dot product to zero.

The target therefore emerges at −60° to +x.

Why: A direction perpendicular to +30° could be +120° or −60°. The target must supply a downward component to cancel the other ball's upward momentum, selecting −60°.

The incoming momentum triangle gives v_1f = 6cos30° = 5.196 m/s and v_2f = 6sin30° = 3 m/s.

Why: The perpendicular outgoing vectors form a right triangle whose hypotenuse is the initial 6 m/s velocity. Cosine and sine give the two legs.

Check x: 5.196cos30°+3cos60°=6; check y: 5.196sin30°−3sin60°=0.

Why: Resolve the proposed velocities back onto x and y. Matching both initial components checks that the magnitudes and target direction conserve vector momentum.

Total momentum is 2(6)+4(−3)=0.

Why: The gliders' opposite mass-weighted velocities exactly cancel. Their individual motion does not require their total momentum to be nonzero.

Center-of-mass velocity is p_total/M = 0/6 = 0 m/s.

Why: Center-of-mass velocity equals total momentum divided by total mass. A zero total means the center of mass stays at rest.

Internal collision forces cannot change total momentum. If the gliders stick, their common velocity equals the unchanged center-of-mass velocity.

Why: Mutual collision forces are internal to the two-glider system and cancel in its momentum sum. Sticking therefore gives the same zero center-of-mass velocity.

Momentum conservation gives v_1f+2v_2f=6.

Why: The frictionless track supplies no horizontal impulse, so signed total momentum is unchanged. Mass weighting gives the factor 2 on the second cart.

Restitution gives v_2f−v_1f = 0.50(6)=3.

Why: The stated restitution coefficient compares relative separation speed with relative approach speed. It supplies the second relation for the unknown final velocities.

Substitute v_1f=v_2f−3: 3v_2f=9, so v_2f=3 m/s and v_1f=0.

Why: Substitute the restitution relation into momentum conservation to solve one velocity, then use it to find the other.

Final kinetic energy is 9 J versus 18 J initially; this e value describes an inelastic collision.

Why: Recalculate kinetic energy after the collision. The decrease agrees with e below one and confirms the collision is inelastic.

The short collision precedes braking, so p conservation gives v = 1(4)/(1+3)=1 m/s.

Why: Separate the impact from the later braking stage. During the short sticking collision, negligible external impulse preserves momentum and fixes the common speed.

The combined post-impact kinetic energy is ½(4)(1²)=2 J.

Why: Use the post-impact combined mass and common speed to find the energy the brake still has to remove.

The brake removes that energy: 2 N·d = 2 J, giving d = 1 m.

Why: A constant force opposite displacement does negative work of magnitude Fd. Set that removed energy equal to the post-impact kinetic energy to find the stopping distance.

Apply momentum only to the short collision; the external brake later changes momentum.

Why: The brake is external to the carts, so it changes their momentum after impact. Momentum conservation cannot be extended across the braking interval.

Thrust magnitude is the exhaust momentum rate: T = ṁu = (0.080)(2500)=200 N.

Why: With pressure thrust neglected, the rocket's thrust comes from ejecting backward momentum per unit time. Mass flow rate times relative exhaust speed gives that force.

The rocket weight is mg = 25(9.81)=245.25 N downward.

Why: Near Earth, the rocket still has a downward weight. Use its instantaneous mass for the gravity force at the requested moment.

Net vertical force is 200−245.25 = −45.25 N, so a = −45.25/25 = −1.81 m/s².

Why: Upward-positive makes thrust positive and weight negative. Their difference divided by the current rocket mass gives the instantaneous acceleration.

Upward thrust alone does not guarantee upward acceleration when it is smaller than the weight.

Why: Acceleration follows the net force rather than thrust alone. The 200 N thrust cannot outweigh 245.25 N of gravity at this instant.

The water velocity change is −8−12 = −20 m/s.

Why: Keep the jet velocities signed: turning from +12 m/s to −8 m/s changes velocity by −20 m/s, not merely by the 4 m/s speed difference.

Steady momentum rate gives F_vane→water = ṁΔv = 0.40(−20)=−8 N.

Why: For steady flow, each second's mass receives the same velocity change. Multiplying mass flow rate by that change gives the vane's force on the water.

Newton’s third law gives F_water→vane = +8 N.

Why: The water and vane exert an equal and opposite interaction pair. Reverse the force on the water to get the force on the vane.

Each design changes momentum by magnitude |Δp| = (5)(4)=20 N·s.

Why: Both pads stop the same mass from the same speed, so both must remove the same momentum. The stopping duration does not change the required impulse.

Padding A average net force is 20/0.020=1000 N; padding B is 20/0.080=250 N.

Why: Average net force is impulse divided by duration. Use each pad's own stopping time to compare its average load.

Both give the same impulse. Increasing stopping time reduces average force by a factor of four.

Why: The longer time spreads the same momentum change over four times as long. Its average force is therefore one quarter as large.

These are averages; the peak force requires the full force–time curve.

Why: An average comes from the area divided by duration. Different force curves can share that average but have different peaks, so peak force needs more information.

Initial momentum is p_i=(12 i+0 j) kg·m/s; final momentum is p_f=(0 i+12 j) kg·m/s.

Why: Multiply each initial and final velocity vector by the probe's mass. Turning changes the momentum components even though its speed stays 6 m/s.

The impulse is the vector difference J=p_f−p_i=(−12 i+12 j) N·s. Its magnitude is √(12²+12²)=16.97 N·s.

Why: Impulse is final momentum minus initial momentum, component by component. The perpendicular changes then combine by Pythagoras for its magnitude.

Average net force is J/3=(−4 i+4 j) N; its magnitude is 5.657 N.

Why: Divide every impulse component by the same 3 s interval. This gives the average net force vector; its perpendicular components also determine its magnitude.

Initial and final speeds are equal, so ΔK=½(2)(6²−6²)=0 J. Equal speeds do not imply unchanged momentum.

Why: Kinetic energy depends on speed squared, so identical endpoint speeds give zero energy change. Momentum depends on direction too, explaining the nonzero impulse.

For the combined system, p_f−p_i=J_external.

Why: For the combined carts, only external impulse changes the total momentum. Choose that system before writing the impulse–momentum relation.

Therefore p_f=10−3=7 kg·m/s.

Why: Add the actuator's signed negative impulse to the initial positive total momentum. The system loses 3 kg·m/s but still moves with positive total momentum.

The mutual cart forces give equal and opposite impulses, which cancel in total momentum. The actuator’s impulse is external and does not cancel.

Why: At every instant, the mutual collision forces are equal and opposite on different carts. Their impulses cancel in the system sum; the outside actuator's impulse remains.

Initial total momentum is p=(6 i+4 j) kg·m/s.

Why: East and north are perpendicular axes. Add the separate carts' mass-times-velocity contributions into one total momentum vector.

The common velocity is p/(2+1) = (2 i+4/3 j) m/s.

Why: Latching gives both carts one final velocity. Divide each conserved momentum component by their combined 3 kg mass.

Speed is √[2²+(4/3)²] = 2.404 m/s; direction is tan⁻¹[(4/3)/2] = 33.69° north of east.

Why: Pythagoras combines the common velocity components into speed. The positive north-to-east ratio gives the first-quadrant direction.

Required total impulse is m(v_f−v_i)=0.50(−3−1)=−2 N·s.

Why: Use signed final minus initial velocity to find the required momentum change. Reversing from +1 to −3 m/s needs a negative impulse.

The first interval supplies −5(0.20)=−1 N·s. The second must supply the remaining −1 N·s.

Why: The first force interval supplies only half the required negative impulse. Subtract it from the total to find the missing second interval's contribution.

Thus F(0.40)=−1, giving F=−2.5 N.

Why: For a constant second force, impulse equals force times its duration. Divide the remaining negative impulse by 0.40 s to obtain F.