Conservation of Linear Momentum - College Physics

Two pucks, A and B, are on a frictionless horizontal ice rink. Puck A (mass $m_A$ ) has initial velocity $(3.0 \hat{i} + 4.0 \hat{j})$ m/s. Puck B (mass $m_B = 2 m_A$ ) has initial velocity $(-6.0 \hat{i} - 2.0 \hat{j})$ m/s. They collide and stick together. During the collision, an external impulse of $\vec{J}_{ext} = (6.0 \hat{i})$ N·s is applied to the system. What is the final velocity of the combined mass?
Answer: $(-1.0 \hat{i})\text{ m/s}$ Using the impulse-momentum theorem: $\vec{J}_{ext} = \Delta \vec{P}_{sys} = \vec{P}_f - \vec{P}_i$ . Let $m_A = m$ , so $m_B = 2m$ . The initial momentum is $\vec{P}_i = m(3.0 \hat{i} + 4.0 \hat{j}) + 2m(-6.0 \hat{i} - 2.0 \hat{j}) = m(-9.0 \hat{i})$ . The final momentum is $\vec{P}_f = \vec{P}_i + \vec{J}_{ext} = m(-9.0 \hat{i}) + (6.0m \hat{i}) = m(-3.0 \hat{i})$ . After the collision, the total mass is (3m), so the final velocity is $\vec{v}_f = \vec{P}_f/(3m) = (-1.0 \hat{i})$ m/s.
An isolated system consists of three interacting particles with masses $m$ , (2m), and (3m). At a particular instant, the particle of mass $m$ has an acceleration $\vec{a}_1 = (2\hat{i} - 3\hat{j})\text{ m/s}^2$ , and the particle of mass (2m) has an acceleration $\vec{a}_2 = (-1\hat{i} + 1\hat{j})\text{ m/s}^2$ . What is the acceleration of the particle of mass (3m) at this instant?
Answer: $(\frac{1}{3}\hat{j})\text{ m/s}^2$ For an isolated system, the net external force is zero. By Newton's second law for a system of particles, $\vec{F}_{net,ext} = \frac{d\vec{P}_{sys}}{dt}$ . Since $\vec{F}_{net,ext}=0$ , the total momentum of the system is conserved. Also, $\vec{F}_{net,ext} = M_{tot}\vec{a}_{CM}$ , which means the center of mass acceleration is zero. The position of the center of mass is $\vec{R}_{CM} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$ . Taking two time derivatives gives $\vec{a}_{CM} = \frac{\sum m_i \vec{a}_i}{\sum m_i}$ . Since $\vec{a}_{CM}=0$ , we must have $\sum m_i \vec{a}_i = 0$ . $m_1\vec{a}_1 + m_2\vec{a}_2 + m_3\vec{a}_3 = 0$ $m(2\hat{i} - 3\hat{j}) + 2m(-1\hat{i} + 1\hat{j}) + 3m\vec{a}_3 = 0$ . Divide by $m$ : $(2\hat{i} - 3\hat{j}) + (-2\hat{i} + 2\hat{j}) + 3\vec{a}_3 = 0$ . $(0\hat{i} - 1\hat{j}) + 3\vec{a}_3 = 0 \implies 3\vec{a}_3 = \hat{j} \implies \vec{a}_3 = (\frac{1}{3}\hat{j})\text{ m/s}^2$ .
A block of mass $m_1$ is released from rest and slides down a frictionless ramp. At the bottom of the ramp, it is on a horizontal, frictionless surface where it collides with and sticks to a stationary block of mass $m_2$ . Consider the two blocks as the system. Which statement best describes the application of the law of conservation of linear momentum to this entire process?
Answer: The component of the system's linear momentum parallel to the horizontal surface is conserved during the collision, but the total linear momentum of the system is not conserved during the slide down the ramp. The law of conservation of linear momentum states that the total momentum of a system is constant if the net external force on the system is zero. While block $m_1$ is sliding down the ramp, the system (both blocks) is subject to a net external force from gravity and the normal force of the ramp, which do not cancel. Thus, momentum is not conserved during the slide. However, during the very brief interval of the collision on the horizontal surface, the external forces are gravity and the normal force from the horizontal surface. These forces are vertical and cancel out. With no net external horizontal force, the horizontal component of the system's linear momentum is conserved through the collision itself. The collision is inelastic (they stick together), so kinetic energy is not conserved.
Two identical balls approach each other with equal speeds. After collision, one ball moves at 60° to its original direction while the other moves at 30° to its original direction. If momentum is conserved, what can you conclude about the collision?
Answer: This scenario is possible and consistent with momentum conservation for appropriate final speeds When analyzing collision problems, you need to check whether the given scenario satisfies conservation laws before drawing conclusions about the collision type. Momentum conservation requires that the vector sum of initial momenta equals the vector sum of final momenta. Let's examine this scenario systematically. Initially, two identical balls approach with equal speeds $v_0$ in opposite directions, so the total momentum is zero. After collision, they move at 60° and 30° to their original directions with some final speeds $v_1$ and $v_2$ . For momentum to be conserved, we need the vector components to balance. Setting up momentum conservation equations in perpendicular directions and solving, you'll find that specific values of $v_1$ and $v_2$ can indeed satisfy momentum conservation. The math works out with $v_1 = v_0\sqrt{3}/2$ and $v_2 = v_0/2$ , confirming this scenario is physically possible. Answer A is wrong because the scenario doesn't violate momentum conservation when the final speeds are chosen appropriately. Answer B incorrectly assumes that complementary scattering angles automatically indicate elastic collision - the collision type depends on kinetic energy, not just angles. Answer C makes the opposite error, incorrectly linking the 90° angle sum to inelastic collision and maximum energy loss, which isn't true. Answer D correctly recognizes that this scenario can satisfy momentum conservation with the right final speeds, making it physically possible. Study tip: In collision problems, always verify conservation laws first before categorizing the collision type. Don't assume angle relationships alone determine whether a collision is elastic or inelastic.
Two ice skaters, initially at rest, push off from each other. Skater A (mass 50 kg) moves away at 2.0 m/s. If the total momentum of the system remains zero, what can be concluded about skater B?
Answer: Skater B's momentum magnitude equals 100 kg⋅m/s, but the mass and velocity cannot be determined individually Conservation of momentum requires the total momentum to remain zero: $\vec{p}_A + \vec{p}_B = 0$ . Since skater A has momentum (50 kg)(2.0 m/s) = 100 kg⋅m/s in one direction, skater B must have momentum 100 kg⋅m/s in the opposite direction. However, many combinations of mass and velocity could give this momentum (e.g., 100 kg at 1.0 m/s, 25 kg at 4.0 m/s, etc.). Choice A assumes equal masses. Choice C incorrectly links speed to momentum conservation. Choice D ignores that the internal forces during pushing cannot change the total momentum of an isolated system.
A 0.50 kg ball moving at 8.0 m/s collides head-on with a 1.5 kg ball at rest. If the collision is perfectly inelastic, what is the velocity of the combined system immediately after collision?
Answer: 2.0 m/s in the direction of the original motion Collision problems test your understanding of conservation of momentum, one of physics's most fundamental principles. When you see "perfectly inelastic collision," immediately think: the objects stick together after impact, and momentum is conserved but kinetic energy is not. For any collision, momentum before equals momentum after: $p_{initial} = p_{final}$ . Initially, you have a 0.50 kg ball at 8.0 m/s and a 1.5 kg ball at rest. The initial momentum is: $p_i = (0.50)(8.0) + (1.5)(0) = 4.0 \text{ kg⋅m/s}$ After the perfectly inelastic collision, both balls move together as a combined mass of 2.0 kg at velocity $v_f$ : $p_f = (2.0)v_f$ Setting initial and final momentum equal: $4.0 = (2.0)v_f$ , so $v_f = 2.0 \text{ m/s}$ in the original direction. This confirms answer A. Answer B (4.0 m/s) incorrectly assumes the moving ball continues at half its original speed, ignoring the mass it must now carry. Answer C (2.7 m/s) might result from incorrectly treating this as an elastic collision or making calculation errors. Answer D (5.3 m/s) violates momentum conservation entirely—the final momentum would exceed the initial momentum. Remember this pattern: in perfectly inelastic collisions, the final velocity always equals total initial momentum divided by total final mass. The combined system always moves slower than the fastest initial object because momentum is "shared" among more mass.
A 2.0 kg object moving at 6.0 m/s collides head-on with a 3.0 kg object initially at rest. After the collision, the 2.0 kg object moves at 1.0 m/s in its original direction. What is the velocity of the 3.0 kg object after the collision?
Answer: 3.3 m/s in the same direction as the original motion of the 2.0 kg object Using conservation of momentum: $m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$ . Initial momentum = (2.0 kg)(6.0 m/s) + (3.0 kg)(0) = 12.0 kg⋅m/s. Final momentum = (2.0 kg)(1.0 m/s) + (3.0 kg)( $v_{2f}$ ) = 2.0 + $3.0v_{2f}$ . Setting equal: 12.0 = 2.0 + $3.0v_{2f}$ , so $v_{2f}$ = 3.3 m/s. Choice B uses incorrect arithmetic (10/4 instead of 10/3). Choice C assumes the collision is elastic and uses energy conservation incorrectly. Choice D incorrectly assumes the collision reverses the direction of the lighter object.
A 2.0 kg cart moving at 3.0 m/s collides with a 4.0 kg cart at rest. If the collision is perfectly inelastic, what percentage of the initial kinetic energy is converted to other forms of energy?
Answer: Approximately 67% of the initial kinetic energy is dissipated as non-mechanical energy forms When you encounter perfectly inelastic collision problems, you're dealing with two key conservation principles: momentum is always conserved, but kinetic energy is partially lost to internal energy forms like heat, sound, and deformation. Let's solve this step-by-step. First, find the final velocity using conservation of momentum: $m_1v_1 + m_2v_2 = (m_1 + m_2)v_f$ $(2.0)(3.0) + (4.0)(0) = (2.0 + 4.0)v_f$ $6.0 = 6.0v_f$ $v_f = 1.0 \text{ m/s}$ Now calculate the kinetic energies. Initial: $KE_i = \frac{1}{2}(2.0)(3.0)^2 = 9.0 \text{ J}$ Final: $KE_f = \frac{1}{2}(6.0)(1.0)^2 = 3.0 \text{ J}$ Energy lost: $9.0 - 3.0 = 6.0 \text{ J}$ Percentage lost: $\frac{6.0}{9.0} \times 100\% = 67\%$ This confirms answer C is correct. A (33%) significantly underestimates the energy loss. B (50%) might seem reasonable but doesn't match the actual calculation. D (75%) overestimates the loss and would require an even more massive stationary object. The key insight is that perfectly inelastic collisions always result in maximum kinetic energy loss while still conserving momentum. Remember this pattern: calculate final velocity from momentum conservation, then compare initial and final kinetic energies to find the percentage converted to internal energy forms.
A stationary atomic nucleus undergoes fission, splitting into three particles. Particle 1 (mass $m_1$ ) has velocity $\vec{v}_1 = v_0 \hat{i}$ . Particle 2 (mass $m_2 = m_1$ ) has velocity $\vec{v}_2 = v_0 \hat{j}$ . Particle 3 has mass $m_3 = 2m_1$ . What is the total kinetic energy released in the fission process (the Q-value of the reaction)?
Answer: $\frac{3}{2} m_1 v_0^2$ The nucleus is initially stationary, so initial momentum is zero. By conservation of momentum, the final momentum is also zero: $m_1\vec{v}_1 + m_2\vec{v}_2 + m_3\vec{v}_3 = 0$ . Substituting the given masses and velocities: $m_1(v_0 \hat{i}) + m_1(v_0 \hat{j}) + 2m_1\vec{v}_3 = 0$ . Solving for $\vec{v}_3$ : $\vec{v}_3 = -\frac{1}{2}(v_0 \hat{i} + v_0 \hat{j})$ . The magnitude squared is $v_3^2 = (v_0/2)^2 + (v_0/2)^2 = v_0^2/2$ . The total kinetic energy is $KE_{tot} = \frac{1}{2}m_1v_0^2 + \frac{1}{2}m_1v_0^2 + \frac{1}{2}(2m_1)(v_0^2/2) = \frac{1}{2}m_1v_0^2 + \frac{1}{2}m_1v_0^2 + \frac{1}{2}m_1v_0^2 = \frac{3}{2}m_1v_0^2$ .
A stationary object of mass (5M) explodes into three fragments on a frictionless horizontal plane. Two of the fragments, with mass $M$ and (2M), are observed to have velocities $\vec{v}_1 = -v_0 \hat{j}$ and $\vec{v}_2 = 3v_0 \hat{i}$ , respectively. What is the kinetic energy of the third fragment, which has mass (2M)?
Answer: $\frac{37}{4} Mv_0^2$ The initial momentum of the system is zero since the object is stationary. By conservation of linear momentum, the total final momentum must also be zero: $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$ . We can solve for the momentum of the third fragment: $\vec{p}_3 = -(\vec{p}_1 + \vec{p}_2)$ . First, find $\vec{p}_1$ and $\vec{p}_2$ : $\vec{p}_1 = m_1 \vec{v}_1 = M(-v_0 \hat{j}) = -Mv_0 \hat{j}$ and $\vec{p}_2 = m_2 \vec{v}_2 = (2M)(3v_0 \hat{i}) = 6Mv_0 \hat{i}$ . Therefore, $\vec{p}_3 = -(6Mv_0 \hat{i} - Mv_0 \hat{j}) = -6Mv_0 \hat{i} + Mv_0 \hat{j}$ . The kinetic energy of the third fragment is given by $K_3 = \frac{p_3^2}{2m_3}$ . The magnitude squared of its momentum is $p_3^2 = (-6Mv_0)^2 + (Mv_0)^2 = 36M^2v_0^2 + M^2v_0^2 = 37M^2v_0^2$ . The mass of the third fragment is $m_3 = 5M - M - 2M = 2M$ . So, $K_3 = \frac{37M^2v_0^2}{2(2M)} = \frac{37}{4} Mv_0^2$ .
A system consists of two objects that collide. Before collision, the total momentum is 40 kg⋅m/s eastward. After collision, object A has momentum 15 kg⋅m/s eastward and object B has momentum 30 kg⋅m/s westward. What can you conclude?
Answer: This scenario violates conservation of momentum and cannot occur in any realistic collision situation When analyzing collision problems, you must always check whether momentum is conserved. The law of conservation of momentum states that the total momentum before collision equals the total momentum after collision, provided no external forces act on the system. Let's examine what happens here. Before collision, the total momentum is 40 kg⋅m/s eastward. After collision, we need to add the momenta vectorially: object A has 15 kg⋅m/s eastward (positive) and object B has 30 kg⋅m/s westward (negative). So the total momentum after collision is: $15 + (-30) = -15 \text{ kg⋅m/s}$ , or 15 kg⋅m/s westward. Since 40 kg⋅m/s eastward ≠ 15 kg⋅m/s westward, momentum is not conserved. This violates a fundamental law of physics, making option A correct. Option B incorrectly focuses on the magnitude of momentum (which isn't even conserved here) rather than the vector sum, and misunderstands what makes a collision elastic. Option C suggests external forces could explain this, but the problem states this is an isolated two-object system during collision. Option D incorrectly assumes this could happen in any real collision type - even perfectly inelastic collisions must conserve momentum. Study tip: In collision problems, always calculate the vector sum of momentum before and after, paying careful attention to directions. If these don't match, the scenario violates physics regardless of collision type. Momentum conservation is non-negotiable in isolated systems.
A 5.0 kg object moving at 4.0 m/s collides with a stationary 3.0 kg object. After collision, they move together. How much momentum is transferred from the first object to the second object?
Answer: 7.5 kg⋅m/s transferred from the moving object to the stationary object during collision When you encounter collision problems involving momentum transfer, you need to apply conservation of momentum and carefully distinguish between total momentum and the amount transferred between objects. First, find the initial momentum: $p_i = m_1v_1 + m_2v_2 = (5.0)(4.0) + (3.0)(0) = 20 \text{ kg⋅m/s}$ Since they stick together (perfectly inelastic collision), use conservation of momentum to find the final velocity: $p_i = p_f$ , so $20 = (5.0 + 3.0)v_f$ , giving $v_f = 2.5 \text{ m/s}$ Now for the key insight: momentum transfer means how much momentum the second object gains (or equivalently, how much the first object loses). The second object goes from 0 to $(3.0)(2.5) = 7.5 \text{ kg⋅m/s}$ . Therefore, 7.5 kg⋅m/s is transferred from the first to the second object. Answer A correctly identifies this transferred momentum. Answer B (12 kg⋅m/s) incorrectly suggests some kind of total exchange calculation that doesn't represent actual momentum transfer. Answer C (20 kg⋅m/s) confuses the initial total momentum with transferred momentum - the first object doesn't lose all its momentum since it keeps moving after collision. Answer D (2.5 kg⋅m/s) appears to confuse the final velocity value with momentum transfer. Remember: momentum transfer in collisions equals the change in momentum of either object (they're equal and opposite). Focus on how much momentum one specific object gains or loses, not the total momentum of the system.
Two objects undergo a collision in which the total kinetic energy decreases by 25%. What type of collision occurred, and what does this tell us about momentum conservation?
Answer: This is an inelastic collision, but momentum can still be conserved if external forces are negligible When you encounter collision problems, you need to distinguish between two fundamental conservation laws that operate independently: conservation of energy and conservation of momentum. This collision involves a 25% decrease in kinetic energy, which immediately tells you it's an inelastic collision. In elastic collisions, kinetic energy is perfectly conserved, while inelastic collisions always involve some energy loss (typically converted to heat, sound, or deformation). The key insight is that momentum conservation depends entirely on external forces, not on what happens to kinetic energy during the collision. Option C correctly identifies this as an inelastic collision and recognizes that momentum can still be conserved. As long as no external forces act on the system during the collision, momentum must be conserved regardless of energy changes—this is a separate, independent physical law. Option A is wrong because any energy loss definitively makes this inelastic, not elastic. Option B contains a critical misconception: it incorrectly links energy loss to momentum non-conservation. Energy and momentum conservation are completely independent—you can lose kinetic energy while still conserving momentum. Option D uses the non-standard term "partially elastic" and incorrectly suggests momentum conservation depends on collision duration, which is irrelevant. Remember this key principle: momentum conservation depends only on external forces, while energy conservation tells you about the collision type. A collision can simultaneously be inelastic (energy lost) and conserve momentum (no external forces). These are separate physical laws operating independently.
A puck of mass $m = 0.50$ kg moving at 4.0 m/s along the x-axis collides with an identical stationary puck. After the collision, the first puck moves with a velocity of 2.0 m/s at an angle of 60° with respect to the positive x-axis. What is the magnitude of the velocity of the second puck after the collision?
Answer: 3.5 m/s Linear momentum is a vector and must be conserved in both the x and y directions. Let $\vec{v}_1$ and $\vec{v}_2$ be the final velocities. The initial momentum is entirely in the x-direction: $p_{ix} = m(4.0 \text{ m/s}) = (0.50)(4.0) = 2.0 \text{ kg} \cdot \text{m/s}$ , and $p_{iy} = 0$ . The final momentum has components: $p_{fx} = m v_{1x} + m v_{2x}$ and $p_{fy} = m v_{1y} + m v_{2y}$ . From the given information, $v_{1x} = (2.0 \text{ m/s})\cos(60^{\circ}) = 1.0 \text{ m/s}$ and $v_{1y} = (2.0 \text{ m/s})\sin(60^{\circ}) = 1.732 \text{ m/s}$ . Conserving momentum in x: $2.0 = (0.50)(1.0) + (0.50)v_{2x} \implies 2.0 = 0.50 + 0.50v_{2x} \implies v_{2x} = 3.0 \text{ m/s}$ . Conserving momentum in y: $0 = (0.50)(1.732) + (0.50)v_{2y} \implies v_{2y} = -1.732 \text{ m/s}$ . The magnitude of the second puck's velocity is $v_2 = \sqrt{v_{2x}^2 + v_{2y}^2} = \sqrt{(3.0)^2 + (-1.732)^2} = \sqrt{9.0 + 3.0} = \sqrt{12.0} \approx 3.46 \text{ m/s}$ , which is closest to 3.5 m/s.
A 1200 kg car traveling east at 25 m/s collides with a 1500 kg truck traveling north at 20 m/s. They stick together after the collision. What is the magnitude of the velocity of the combined vehicles immediately after impact?
Answer: 15.7 m/s, calculated using conservation of momentum in two dimensions When you encounter collision problems where objects stick together, you're dealing with a perfectly inelastic collision that requires conservation of momentum in vector form. Since the vehicles are traveling in perpendicular directions (east and north), you must analyze momentum components separately. Using conservation of momentum: $\vec{p}_{initial} = \vec{p}_{final}$ For the x-component (east): $p_x = (1200)(25) + (1500)(0) = 30,000 \text{ kg⋅m/s}$ For the y-component (north): $p_y = (1200)(0) + (1500)(20) = 30,000 \text{ kg⋅m/s}$ The total final momentum magnitude is: $p_{final} = \sqrt{30,000^2 + 30,000^2} = 42,426 \text{ kg⋅m/s}$ With combined mass of 2700 kg, the final velocity is: $v = \frac{42,426}{2700} = 15.7 \text{ m/s}$ Option A incorrectly treats this as a one-dimensional problem by simply averaging speeds weighted by mass, ignoring the perpendicular directions entirely. Option B applies conservation of kinetic energy, which doesn't hold for inelastic collisions where energy is lost to deformation and heat. Option C adds the velocity vectors incorrectly, treating them as scalars rather than accounting for their perpendicular relationship. Remember: In two-dimensional collision problems, always break momentum into components, solve each direction separately, then combine vectorially. Don't let perpendicular motion fool you into thinking this is a simple one-dimensional problem—the vector nature of momentum is crucial.
A rocket of initial mass $M_0$ ejects fuel at a constant rate $\frac{dm}{dt} = k$ with exhaust velocity $v_e$ relative to the rocket. If the rocket starts from rest in deep space, what is the rocket's velocity when its mass has decreased to $\frac{M_0}{2}$ ?
Answer: $v_e \ln(2)$ , derived from conservation of momentum for the rocket-fuel system This is a variable mass problem requiring the rocket equation. From conservation of momentum: $M\frac{dv}{dt} = v_e\frac{dm}{dt}$ , where $\frac{dm}{dt} < 0$ (mass decreases). Rearranging: $dv = -v_e\frac{dm}{M}$ . Integrating from initial conditions ( $v = 0, M = M_0$ ) to final state ( $v = v_f, M = M_0/2$ ): $\int_0^{v_f} dv = -v_e \int_{M_0}^{M_0/2} \frac{dm}{M}$ . This gives $v_f = -v_e[\ln(M)]_{M_0}^{M_0/2} = -v_e[\ln(M_0/2) - \ln(M_0)] = v_e\ln(2)$ . Choice B incorrectly uses constant acceleration. Choice C oversimplifies momentum conservation. Choice D uses the rocket equation form but makes an error: $\ln(M_0/(M_0/2)) = \ln(2)$ , not 2.
An explosion occurs in a system initially at rest, breaking it into three pieces. Two pieces have masses 2.0 kg and 3.0 kg and velocities 5.0 m/s east and 4.0 m/s north respectively. If the third piece has mass 1.0 kg, what is its velocity?
Answer: 16 m/s at 50° south of west, based on vector momentum balance When you encounter an explosion problem, you're dealing with conservation of momentum. Since the system starts at rest, the total initial momentum is zero, so the final momentum of all pieces must also sum to zero. Let's set up a coordinate system with east as positive x and north as positive y. The momentum of each piece is mass times velocity: - Piece 1: $\vec{p_1} = (2.0 \text{ kg})(5.0 \text{ m/s east}) = 10.0 \text{ kg⋅m/s}$ in the +x direction - Piece 2: $\vec{p_2} = (3.0 \text{ kg})(4.0 \text{ m/s north}) = 12.0 \text{ kg⋅m/s}$ in the +y direction For momentum conservation: $\vec{p_1} + \vec{p_2} + \vec{p_3} = 0$ Therefore: $\vec{p_3} = -(\vec{p_1} + \vec{p_2}) = (-10.0, -12.0) \text{ kg⋅m/s}$ The magnitude is: $|\vec{p_3}| = \sqrt{(-10.0)^2 + (-12.0)^2} = \sqrt{244} = 15.6 \text{ kg⋅m/s}$ Since the third piece has mass 1.0 kg: $v_3 = \frac{15.6}{1.0} = 15.6 \approx 16 \text{ m/s}$ The direction is: $\theta = \arctan\left(\frac{12.0}{10.0}\right) = 50°$ south of west (negative x and y components). This confirms answer B is correct. Answer A underestimates both speed and angle. Answer C has the wrong angle despite being close in magnitude. Answer D significantly overestimates both the speed and angle. Remember: In explosion problems, always check that your final momentum vectors sum to the initial momentum. Setting up clear coordinate systems and using vector components prevents calculation errors.
A system consists of two objects connected by a light string over a massless pulley. Object A (2 kg) hangs vertically, while object B (3 kg) sits on a horizontal frictionless surface. When the string is cut, object A falls freely. Is momentum conserved for the system of both objects during A's fall?
Answer: No, because gravity acts on object A, making the net external force on the system non-zero When analyzing momentum conservation, you need to identify all external forces acting on your system. Momentum is conserved only when the net external force is zero. Once the string is cut, object A falls freely under gravity's influence. Since you're considering both objects as your system, gravity becomes an external force acting downward on object A. This creates a net external force on the system, so momentum cannot be conserved. Object A accelerates downward, gaining momentum, while object B remains stationary on the frictionless surface. The system's total momentum changes from zero to a non-zero value. Choice A incorrectly focuses on the string tension, which was indeed an internal force. However, cutting the string doesn't eliminate external forces—gravity still acts on object A and becomes the dominant external force affecting the system. Choice B makes a critical error by assuming the total momentum stays constant. While object B does remain at rest, object A gains downward momentum due to gravitational acceleration, so the system's total momentum increases in magnitude. Choice D incorrectly suggests that removing the pulley constraint itself prevents momentum conservation. The mechanical setup change isn't the issue—it's the unbalanced external force (gravity) that matters. Remember this key principle: momentum conservation requires zero net external force on the system. Always identify your system boundaries first, then catalog all forces acting from outside those boundaries. Gravity acting on part of your system counts as an external force and will prevent momentum conservation.
A 2.0 kg object moving at 4.0 m/s collides with a 3.0 kg object initially at rest. After the collision, the 2.0 kg object moves at 1.0 m/s in the same direction. What is the velocity of the 3.0 kg object after the collision?
Answer: 2.0 m/s in the same direction as the initial motion When you encounter collision problems, you're dealing with conservation of momentum. In any collision where no external forces act on the system, the total momentum before collision equals the total momentum after collision. To solve this, apply the momentum conservation equation: $m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$ Before collision: The 2.0 kg object has momentum of (2.0 kg)(4.0 m/s) = 8.0 kg⋅m/s, and the 3.0 kg object is at rest, so its momentum is zero. Total initial momentum = 8.0 kg⋅m/s. After collision: The 2.0 kg object moves at 1.0 m/s, giving it momentum of (2.0 kg)(1.0 m/s) = 2.0 kg⋅m/s. Let the 3.0 kg object's final velocity be $v_f$ . Setting up the equation: $8.0 = 2.0 + 3.0v_f$ Solving: $6.0 = 3.0v_f$ , so $v_f = 2.0$ m/s This confirms answer A is correct. Answer B (1.5 m/s) would give a total final momentum of only 6.5 kg⋅m/s, violating conservation. Answer C (2.5 m/s) would yield 9.5 kg⋅m/s total, which exceeds the initial momentum. Answer D (3.0 m/s) would produce 11.0 kg⋅m/s, an even greater violation. For collision problems, always write out the momentum conservation equation first, then substitute your known values. This systematic approach prevents calculation errors and ensures you don't miss the fundamental physics principle being tested.
A bullet of mass 20 g traveling at 400 m/s embeds in a 5.0 kg wooden block initially at rest. What fraction of the bullet's initial kinetic energy is lost in this perfectly inelastic collision?
Answer: Approximately 99.6% of the kinetic energy is dissipated during collision When you encounter perfectly inelastic collision problems, you need to apply both conservation of momentum and analyze energy changes. In perfectly inelastic collisions, the objects stick together, and kinetic energy is always lost to heat, sound, and deformation. First, find the final velocity using conservation of momentum. Initially: $p_i = m_{bullet} \cdot v_{bullet} = 0.020 \text{ kg} \times 400 \text{ m/s} = 8.0 \text{ kg⋅m/s}$ After collision, both objects move together: $p_f = (m_{bullet} + m_{block}) \cdot v_f = (0.020 + 5.0) \text{ kg} \cdot v_f$ Setting $p_i = p_f$ : $8.0 = 5.02 \cdot v_f$ , so $v_f = 1.59 \text{ m/s}$ Now compare kinetic energies. Initial: $KE_i = \frac{1}{2}(0.020)(400^2) = 1600 \text{ J}$ Final: $KE_f = \frac{1}{2}(5.02)(1.59^2) = 6.35 \text{ J}$ Energy lost: $\frac{1600 - 6.35}{1600} \times 100\% = 99.6\%$ Answer A is correct with 99.6% energy loss. The other options (B, C, D) all significantly underestimate the energy loss. This happens because most of the bullet's momentum transfers to a much more massive block, resulting in a very slow final velocity. The dramatic mass difference (bullet is only 0.4% of the total final mass) means almost all kinetic energy converts to internal energy. Study tip: In perfectly inelastic collisions with large mass ratios, expect energy losses approaching 100%. The smaller object's kinetic energy gets "absorbed" by the larger mass's inertia.

Start this quiz