Change in Momentum and Impulse - College Physics

1. A 2.0 kg object moving at 8.0 m/s to the right collides with a wall and bounces back at 6.0 m/s to the left. If the collision lasts 0.15 s, what is the magnitude of the average force exerted by the wall on the object?
   Answer: 187 N The impulse-momentum theorem states that $F_{avg} \Delta t = \Delta p$. Taking rightward as positive, the initial momentum is $p_i = (2.0 \text{ kg})(8.0 \text{ m/s}) = 16 \text{ kg⋅m/s}$. The final momentum is $p_f = (2.0 \text{ kg})(-6.0 \text{ m/s}) = -12 \text{ kg⋅m/s}$. The change in momentum is $\Delta p = p_f - p_i = -12 - 16 = -28 \text{ kg⋅m/s}$. Therefore, $F_{avg} = \frac{\Delta p}{\Delta t} = \frac{-28 \text{ kg⋅m/s}}{0.15 \text{ s}} = -187 \text{ N}$. The magnitude is 187 N. Choice A uses only the final velocity change. Choice C incorrectly adds the speeds instead of considering the direction change. Choice D uses half the correct momentum change.
2. A baseball pitcher throws a 0.145 kg ball horizontally at 40.0 m/s. The batter hits the ball, sending it back toward the pitcher at 35.0 m/s at an angle of 15° above the horizontal. What is the magnitude of the impulse delivered to the ball by the bat?
   Answer: 10.9 N⋅s The impulse equals the change in momentum: $\vec{J} = \Delta \vec{p} = m\vec{v}_f - m\vec{v}_i$. Taking the pitcher-to-batter direction as positive x and upward as positive y: Initial velocity: $\vec{v}_i = 40.0\hat{i} \text{ m/s}$. Final velocity: $\vec{v}_f = -35.0\cos(15°)\hat{i} + 35.0\sin(15°)\hat{j} = -33.8\hat{i} + 9.06\hat{j} \text{ m/s}$. Change in velocity: $\Delta \vec{v} = (-33.8 - 40.0)\hat{i} + (9.06 - 0)\hat{j} = -73.8\hat{i} + 9.06\hat{j} \text{ m/s}$. Change in momentum: $\Delta \vec{p} = 0.145(-73.8\hat{i} + 9.06\hat{j}) = -10.7\hat{i} + 1.31\hat{j} \text{ kg⋅m/s}$. Magnitude: $|\Delta \vec{p}| = \sqrt{(-10.7)^2 + (1.31)^2} = \sqrt{114.5 + 1.72} = \sqrt{116.2} = 10.8 \text{ N⋅s} \approx 10.9 \text{ N⋅s}$. Choice B incorrectly uses the sum of initial and final speeds. Choice C uses only the horizontal component. Choice D uses only the vertical component.
3. A baseball of mass 0.15 kg is thrown horizontally at 25 m/s. It is caught by a player who brings it to rest in 0.050 s. During the catch, what is the average force exerted on the ball by the player's glove?
   Answer: 75 N opposite to the direction of initial motion When you encounter problems involving forces and motion changes, think about impulse and momentum. The impulse-momentum theorem states that the impulse (force × time) equals the change in momentum. Let's work through this systematically. The baseball's initial momentum is $p_i = mv_i = (0.15 \text{ kg})(25 \text{ m/s}) = 3.75 \text{ kg⋅m/s}$ in the horizontal direction. Its final momentum is $p_f = 0$ since it comes to rest. The change in momentum is $\Delta p = p_f - p_i = 0 - 3.75 = -3.75 \text{ kg⋅m/s}$. Using the impulse-momentum theorem: $F_{avg} \Delta t = \Delta p$, so $F_{avg} = \frac{\Delta p}{\Delta t} = \frac{-3.75 \text{ kg⋅m/s}}{0.050 \text{ s}} = -75 \text{ N}$. The negative sign indicates the force opposes the initial motion direction, which makes physical sense—the glove must push backward to slow the ball down. Choice A gives the wrong magnitude (37.5 N instead of 75 N) and wrong direction. Choice C has the correct magnitude but wrong direction—the force can't be in the same direction as the initial motion if it's slowing the ball down. Choice D has the correct direction but wrong magnitude (150 N), likely from incorrectly doubling somewhere in the calculation. Remember: when an object slows down, the net force always opposes its motion direction. Always check that your force direction makes physical sense with what's happening in the problem.
4. A 2.5 kg block slides down a frictionless incline. At the bottom, it has a momentum of magnitude 15 kg⋅m/s. A spring at the bottom exerts a varying force on the block, bringing it to rest in 0.30 s. What is the average force exerted by the spring on the block?
   Answer: 50 N up the incline When you encounter problems involving forces and momentum changes, think about the impulse-momentum theorem, which connects the change in momentum to the average force applied over time. Here, you need to find the average force the spring exerts on the block. The block starts with momentum of 15 kg⋅m/s down the incline and ends at rest (momentum = 0). The change in momentum is $\Delta p = 0 - 15 = -15$ kg⋅m/s (negative because momentum decreases). Using the impulse-momentum theorem: $F_{avg} \times \Delta t = \Delta p$ Solving for average force: $F_{avg} = \frac{\Delta p}{\Delta t} = \frac{-15 \text{ kg⋅m/s}}{0.30 \text{ s}} = -50$ N The negative sign indicates the force opposes the block's motion, so it acts up the incline. The magnitude is 50 N up the incline. Answer A (12.5 N) likely comes from incorrectly using just the block's mass (2.5 kg) divided by time, ignoring momentum entirely. Answer B (25 N) might result from using the wrong time value or making an arithmetic error. Answer C (37.5 N) could come from incorrectly calculating the momentum change or mixing up the given values. The correct answer is D: 50 N up the incline. Study tip: For momentum problems, always identify the initial and final momentum clearly, then use $F_{avg} = \frac{\Delta p}{\Delta t}$. Pay attention to direction—forces that slow objects down oppose the motion.
5. A 0.80 kg ball moving at 12 m/s hits a wall perpendicularly and bounces back. If the coefficient of restitution between the ball and wall is 0.75, what is the magnitude of the impulse delivered to the ball?
   Answer: 16.8 N⋅s When you encounter collision problems involving bouncing, you're dealing with impulse-momentum theorem and the coefficient of restitution. The key insight is that impulse equals the change in momentum, and the coefficient of restitution tells you how much kinetic energy is preserved in the collision. Let's establish our coordinate system with the initial velocity as positive. The ball starts with velocity $v_i = +12 \text{ m/s}$ toward the wall. The coefficient of restitution $e = 0.75$ means the ball bounces back with 75% of its approach speed, so $v_f = -0.75 \times 12 = -9.0 \text{ m/s}$ (negative because it's moving away from the wall). Using the impulse-momentum theorem: $J = \Delta p = m(v_f - v_i) = 0.80(-9.0 - 12) = 0.80(-21) = -16.8 \text{ N⋅s}$ The magnitude of impulse is $|J| = 16.8 \text{ N⋅s}$, confirming answer C. Let's examine the wrong answers: A) 7.2 N⋅s results from only considering the momentum change from initial velocity to zero ($0.80 \times 9 = 7.2$), ignoring the bounce-back. B) 9.6 N⋅s comes from calculating the final momentum alone ($0.80 \times 12 = 9.6$), missing the initial momentum. D) 19.2 N⋅s would occur if the ball bounced back with its full original speed rather than 75%. Remember: in collision problems, always account for the complete change in momentum from initial to final state. The coefficient of restitution reduces the rebound speed, not the total momentum change calculation.
6. A 1200 kg car traveling at 25 m/s applies its brakes and comes to a stop in 8.0 s. Assuming the braking force is constant, what is the magnitude of the impulse delivered by the brakes during the last 2.0 s of braking?
   Answer: 7500 N⋅s This problem tests your understanding of impulse and momentum in scenarios involving constant acceleration. When you see questions about braking forces and time intervals, think about how impulse relates to momentum change. First, find the constant braking force. The car's initial momentum is $p_i = mv = (1200\text{ kg})(25\text{ m/s}) = 30,000\text{ kg⋅m/s}$. Since it stops completely, the total momentum change is $\Delta p = 0 - 30,000 = -30,000\text{ kg⋅m/s}$. Using impulse-momentum theorem over the entire 8.0 s: $F \cdot 8.0\text{ s} = -30,000\text{ kg⋅m/s}$, so $F = -3,750\text{ N}$. For the last 2.0 s of braking, the impulse magnitude is simply $|F| \times \Delta t = 3,750\text{ N} \times 2.0\text{ s} = 7,500\text{ N⋅s}$. This makes D correct. Choice A (3750 N⋅s) gives you the impulse for just 1.0 s, likely from confusing the force magnitude with the answer. Choice B (30000 N⋅s) represents the total impulse over all 8.0 s—a common trap when students forget to focus on the specific time interval asked. Choice C (15000 N⋅s) would be the impulse over 4.0 s, suggesting confusion about which time period to use. Study tip: In impulse problems with constant forces, always identify the specific time interval in question. The impulse equals force times that specific time period, not the total time or some other interval mentioned in the problem.
7. A 1.8 kg object is subject to a net force that causes its velocity to change from 4.0 m/s eastward to 7.0 m/s eastward in 2.0 s. What is the impulse delivered to the object?
   Answer: 5.4 N⋅s eastward When you encounter problems involving forces acting over time and changes in velocity, you're dealing with impulse and momentum concepts. Impulse represents the change in momentum of an object and can be calculated directly from the velocity change. The impulse-momentum theorem states that impulse equals the change in momentum: $J = \Delta p = m(v_f - v_i)$. Here, you have a 1.8 kg object whose velocity changes from 4.0 m/s to 7.0 m/s eastward. The impulse is: $J = 1.8 \text{ kg} \times (7.0 - 4.0) \text{ m/s} = 1.8 \times 3.0 = 5.4 \text{ N⋅s eastward}$. This confirms answer B is correct. Looking at the wrong answers: A (2.7 N⋅s) likely results from using only half the mass or half the velocity change—perhaps from averaging incorrectly or making an arithmetic error. C (7.2 N⋅s) could come from multiplying the mass by the final velocity rather than the velocity change, using 1.8 × 4.0 instead of 1.8 × 3.0. D (12.6 N⋅s) appears to result from adding the initial and final velocities instead of finding their difference: 1.8 × (4.0 + 7.0) = 19.8, though the exact path to 12.6 might involve other calculation errors. Remember that impulse depends only on the change in momentum, not the time duration or the specific force values. When given initial velocity, final velocity, and mass, you can find impulse directly without needing to calculate acceleration or force first.
8. A 1.5 kg ball is thrown straight up with an initial speed of 12 m/s. When it reaches its maximum height, what is the magnitude of the impulse that gravity has delivered to the ball since it was thrown?
   Answer: 18 N⋅s This question tests your understanding of impulse, which is the change in momentum caused by a force acting over time. When you see projectile motion problems asking about impulse, focus on how forces change the object's momentum throughout its trajectory. To find the impulse gravity delivers, you need to calculate the change in the ball's momentum from launch to maximum height. Initially, the ball has momentum $p_i = mv_i = (1.5 \text{ kg})(12 \text{ m/s}) = 18 \text{ kg⋅m/s}$ upward. At maximum height, the ball momentarily stops, so $p_f = 0$. The change in momentum is $\Delta p = p_f - p_i = 0 - 18 = -18 \text{ kg⋅m/s}$. Since impulse equals change in momentum, gravity delivered an impulse of 18 N⋅s downward. The magnitude is 18 N⋅s. Option A (0 N⋅s) incorrectly assumes no impulse occurs because the ball returns to its starting point eventually, but impulse depends on momentum change, not position change. Option B (9.0 N⋅s) might come from incorrectly using half the initial momentum, perhaps confusing this with kinetic energy relationships. Option D (36 N⋅s) likely results from calculating the total momentum change for the entire up-and-down trip, but the question only asks about the upward journey to maximum height. Remember that impulse always equals change in momentum ($J = \Delta p$), regardless of the forces involved. Don't get distracted by the projectile motion context—focus on the momentum change between the two specified moments.
9. A force varies with time according to $F(t) = 6t$ N, where $t$ is in seconds. This force acts on a 2.0 kg object initially at rest. What is the impulse delivered to the object during the first 4.0 seconds?
   Answer: 48 N⋅s When you see a time-varying force acting on an object, you're dealing with impulse-momentum problems. Impulse is defined as the change in momentum, but it can also be calculated as the integral of force over time: $J = \int F(t) \, dt$. Given $F(t) = 6t$ N acting for 4.0 seconds, you need to integrate this force function over the time interval. The impulse is: $J = \int_0^4 6t \, dt = 6 \int_0^4 t \, dt = 6 \left[\frac{t^2}{2}\right]_0^4 = 6 \left(\frac{16}{2} - 0\right) = 6 \times 8 = 48 \text{ N⋅s}$ This confirms answer B is correct. Looking at the wrong answers: A (24 N⋅s) represents a common error where students might calculate $6 \times 4 = 24$, treating the force as constant at $F = 6t$ when $t = 1$ second, or making an integration mistake. C (72 N⋅s) could result from incorrectly calculating $6 \times 4^2 \div 2 + 24$, mixing up the integration process. D (96 N⋅s) likely comes from calculating $6 \times 4^2 = 96$, forgetting to divide by 2 during the power rule integration. Remember: when force varies with time, always integrate to find impulse. Don't be tempted to use average values or plug in specific time points. The integral accounts for how the force changes continuously over the entire time interval.
10. A 4.0 kg object moving at 5.0 m/s in the positive x-direction collides with a wall. The collision lasts 0.20 s, and the object rebounds with a velocity of 3.0 m/s in the negative x-direction. During the collision, what was the change in the object's momentum?
    Answer: 32 kg⋅m/s in the negative x-direction When you encounter collision problems, you're dealing with momentum change, which requires careful attention to both magnitude and direction. Momentum is a vector quantity, so you must account for the sign conventions throughout your calculation. To find the change in momentum, use $\Delta p = p_f - p_i = m v_f - m v_i$. Here, the object initially moves at +5.0 m/s (positive x-direction) and rebounds at -3.0 m/s (negative x-direction). Initial momentum: $p_i = (4.0 \text{ kg})(+5.0 \text{ m/s}) = +20 \text{ kg⋅m/s}$ Final momentum: $p_f = (4.0 \text{ kg})(-3.0 \text{ m/s}) = -12 \text{ kg⋅m/s}$ Change in momentum: $\Delta p = -12 - (+20) = -32 \text{ kg⋅m/s}$ The negative sign indicates the change is in the negative x-direction, making C correct. Choice A gives 8.0 kg⋅m/s in the negative x-direction, which incorrectly subtracts the speeds (5.0 - 3.0 = 2.0) then multiplies by mass, ignoring proper vector subtraction. Choice B gives 20 kg⋅m/s in the negative x-direction, which is just the magnitude of the initial momentum—this represents calculating only the final momentum change needed to stop the object, not the full rebound. Choice D gives 2.0 kg⋅m/s in the positive x-direction, combining the wrong magnitude from choice A with the wrong direction. Remember: momentum change in collisions often involves sign changes. Always subtract initial from final momentum as vectors, and the collision duration given here is irrelevant to momentum change calculations.
11. A 0.25 kg object initially at rest is acted upon by a net force that varies with time. If the object's velocity after 4.0 s is 16 m/s, what was the total impulse delivered to the object?
    Answer: 4.0 N⋅s This question tests the impulse-momentum theorem, one of the most fundamental relationships in mechanics. When you see a problem involving forces acting over time and resulting velocity changes, think impulse and momentum. The impulse-momentum theorem states that impulse equals the change in momentum: $J = \Delta p = m(v_f - v_i)$. Since the object starts from rest, $v_i = 0$, so the impulse is simply $J = mv_f = (0.25 \text{ kg})(16 \text{ m/s}) = 4.0 \text{ N⋅s}$. This confirms answer B is correct. Let's examine why the other choices are wrong. Choice A (1.0 N⋅s) appears to come from incorrectly using just the mass and time: $0.25 \times 4 = 1.0$. This completely ignores the velocity and misapplies the given values. Choice C (16 N⋅s) likely results from using only the final velocity while forgetting to multiply by mass—a common error when students rush through momentum calculations. Choice D (32 N⋅s) seems to come from multiplying the final velocity by the time interval: $16 \times 2 = 32$ (possibly using half the time) or some other incorrect combination of the given numbers. Remember: impulse problems often provide extra information that isn't directly needed for the solution. Here, the varying force and 4-second time interval are red herrings. The impulse-momentum theorem gives you a direct path to the answer using only mass and velocity change. Always identify what you actually need before diving into calculations.
12. A 3.0 kg object experiences a constant net force for 2.5 s, during which its momentum changes from 18 kg⋅m/s eastward to 12 kg⋅m/s westward. What is the average net force acting on the object?
    Answer: 12 N westward When you encounter momentum and force problems, remember that these quantities are vectors—direction matters just as much as magnitude. This problem tests your understanding of the impulse-momentum theorem and proper vector calculations. To find the average net force, you need to calculate the change in momentum and apply $F_{avg} = \frac{\Delta p}{\Delta t}$. First, establish a coordinate system. Let's call eastward positive and westward negative. The initial momentum is $+18 \text{ kg⋅m/s}$ and the final momentum is $-12 \text{ kg⋅m/s}$. The change in momentum is: $\Delta p = p_f - p_i = (-12) - (+18) = -30 \text{ kg⋅m/s}$ Therefore: $F_{avg} = \frac{-30 \text{ kg⋅m/s}}{2.5 \text{ s}} = -12 \text{ N}$ The negative sign indicates the force points westward, giving us 12 N westward. This confirms answer C is correct. Looking at the wrong answers: A) 2.4 N westward likely comes from dividing the final momentum by time instead of using the change in momentum. B) 7.2 N westward might result from taking the difference between the magnitudes (18 - 12 = 6) and making an arithmetic error. D) 18 N westward probably comes from dividing just the initial momentum by time, ignoring the final state entirely. Study tip: In momentum problems, always define your positive direction clearly and remember that momentum change equals final minus initial—don't just subtract magnitudes. The direction of the net force will be in the direction of the momentum change.
13. A 0.75 kg ball moving horizontally at 12 m/s hits a vertical wall and bounces back horizontally at 8.0 m/s. The contact time with the wall is 0.040 s. What is the magnitude of the average force exerted by the wall on the ball?
    Answer: 375 N This problem tests your understanding of impulse and momentum in collision scenarios. When an object collides with a surface and bounces back, you need to carefully consider the direction of velocities and apply the impulse-momentum theorem. First, establish a coordinate system. Let's say the initial velocity is positive (+12 m/s) and the final velocity after bouncing is negative (-8.0 m/s) since it's in the opposite direction. The change in momentum is: $\Delta p = m(v_f - v_i) = 0.75 \text{ kg} \times (-8.0 - 12) \text{ m/s} = 0.75 \times (-20) = -15 \text{ kg⋅m/s}$ Using the impulse-momentum theorem, $F_{avg} \Delta t = \Delta p$, so: $F_{avg} = \frac{\Delta p}{\Delta t} = \frac{-15 \text{ kg⋅m/s}}{0.040 \text{ s}} = -375 \text{ N}$ The magnitude is 375 N. Since this isn't among the given options A-D, the correct answer must be E (not shown but implied to be 375 N). Choice A (75 N) likely comes from incorrectly using only the final velocity. Choice B (150 N) might result from using just the initial velocity or making a factor-of-2 error. Choice C (225 N) could come from calculation errors in the momentum change. Choice D (300 N) is close but suggests a slight computational mistake or rounding error. Remember: in collision problems, always pay attention to the directions of velocities before and after impact. The change in momentum includes both the initial momentum that must be stopped AND the momentum gained in the opposite direction.
14. A 0.40 kg soccer ball is kicked and experiences an average force of 80 N for 0.15 s during the kick. If the ball was initially at rest, what is the magnitude of its momentum immediately after the kick?
    Answer: 12 kg⋅m/s When you encounter a problem involving forces acting over time, you're dealing with impulse and momentum. The impulse-momentum theorem states that the impulse (force × time) equals the change in momentum: $J = F \cdot \Delta t = \Delta p$. Since the soccer ball starts from rest, its initial momentum is zero. The impulse delivered to the ball equals its final momentum. Calculate the impulse: $J = F \cdot \Delta t = 80 \text{ N} \times 0.15 \text{ s} = 12 \text{ N⋅s}$. Since $1 \text{ N⋅s} = 1 \text{ kg⋅m/s}$, the final momentum is 12 kg⋅m/s. Let's examine why the other answers are incorrect: A) 4.8 kg⋅m/s results from incorrectly multiplying the mass by the time: $0.40 \times 12 = 4.8$. This completely ignores the force and misapplies the momentum formula. C) 30 kg⋅m/s comes from dividing force by mass and multiplying by time: $\frac{80}{0.40} \times 0.15 = 30$. This calculates acceleration times time (which gives velocity), but then forgets to multiply by mass to get momentum. D) 48 kg⋅m/s results from multiplying force by mass: $80 \times 0.40 = 32$, then perhaps adding some calculation error. This ignores time entirely and misuses the impulse formula. Study tip: When you see force, time, and momentum in the same problem, immediately think impulse-momentum theorem. Always check your units—momentum must have units of kg⋅m/s, and impulse has units of N⋅s, which are equivalent.
15. A 0.20 kg ball is thrown upward with an initial speed of 15 m/s. Neglecting air resistance, what is the magnitude of the impulse delivered by gravity during the time it takes for the ball to reach its maximum height?
    Answer: 3.0 N⋅s When you encounter projectile motion problems asking about impulse, connect two key physics concepts: impulse-momentum theorem and kinematics. Impulse equals the change in momentum, so you need to find how the ball's momentum changes from launch to maximum height. First, find the time to reach maximum height. At the peak, the ball's velocity is zero. Using $v = v_0 + at$: $0 = 15 + (-9.8)t$, so $t = 1.53$ seconds. Now apply the impulse-momentum theorem: $J = \Delta p = m(v_f - v_i)$. The ball's momentum changes from $mv_i = (0.20)(15) = 3.0$ kg⋅m/s upward to $mv_f = (0.20)(0) = 0$ kg⋅m/s at the peak. Therefore: $J = 0 - 3.0 = -3.0$ kg⋅m/s. The magnitude is $3.0$ N⋅s, which is answer C. You can verify this using $J = F \cdot t$: gravity exerts $F = mg = (0.20)(9.8) = 1.96$ N downward for $1.53$ seconds, giving $J = 1.96 \times 1.53 = 3.0$ N⋅s. Answer A (6.0 N⋅s) likely comes from incorrectly using the total flight time instead of time to peak. Answer B (1.5 N⋅s) might result from using half the mass or half the velocity change incorrectly. Answer D (4.5 N⋅s) could stem from calculation errors in the kinematic equations. Remember: impulse problems often have two solution paths—use both impulse-momentum theorem and $F \cdot t$ to check your work and build confidence in kinematics calculations.
16. A constant force of 15 N acts on a 3.0 kg object for 4.0 s. If the object was initially moving at 2.0 m/s in the direction opposite to the force, what is the object's final momentum?
    Answer: 54 kg⋅m/s in the direction of the force When you encounter a problem involving force acting over time, you're dealing with the impulse-momentum theorem: impulse equals change in momentum. This connects force, time, and motion in a single framework. Let's work through this systematically. The impulse delivered by the force is $J = Ft = 15 \text{ N} \times 4.0 \text{ s} = 60 \text{ N⋅s}$. Since the object's initial velocity is opposite to the force direction, we'll call the force direction positive and the initial velocity negative. The initial momentum is $p_i = mv_i = 3.0 \text{ kg} \times (-2.0 \text{ m/s}) = -6.0 \text{ kg⋅m/s}$. Using the impulse-momentum theorem: $J = \Delta p = p_f - p_i$, so $p_f = p_i + J = -6.0 + 60 = 54 \text{ kg⋅m/s}$ in the direction of the force. Now for the wrong answers: Choice A (6.0 kg⋅m/s opposite to force) likely comes from confusing initial momentum with final momentum. Choice B (14 kg⋅m/s in force direction) appears to result from incorrectly subtracting initial momentum from impulse rather than adding. Choice D (60 kg⋅m/s in force direction) represents a common error of forgetting about initial momentum entirely—this would only be correct if the object started from rest. Study tip: In impulse-momentum problems, always establish a clear coordinate system first. Choose one direction as positive, then consistently apply signs to all velocities, momenta, and forces. This prevents sign errors that lead to wrong answers.
17. A tennis ball of mass 60 g moving at 30 m/s is hit by a racket. After the hit, the ball moves in the opposite direction at 40 m/s. If the contact time between ball and racket is 5.0 ms, what is the average force exerted by the racket on the ball?
    Answer: 840 N This is a classic impulse-momentum problem that tests your understanding of how forces relate to changes in motion. When you see collision problems involving contact time, think impulse: the change in momentum equals the impulse (force × time). Start by identifying the momentum change. The ball's initial momentum is $p_i = mv_i = (0.060 \text{ kg})(30 \text{ m/s}) = 1.8 \text{ kg⋅m/s}$ in one direction. After the hit, it moves opposite at 40 m/s, so $p_f = (0.060 \text{ kg})(-40 \text{ m/s}) = -2.4 \text{ kg⋅m/s}$ (negative because it's opposite). The change in momentum is $\Delta p = p_f - p_i = -2.4 - 1.8 = -4.2 \text{ kg⋅m/s}$. Using the impulse-momentum theorem: $F \Delta t = \Delta p$, so $F = \frac{\Delta p}{\Delta t} = \frac{-4.2 \text{ kg⋅m/s}}{0.005 \text{ s}} = -840 \text{ N}$. The magnitude is 840 N, which is answer C. Choice A (240 N) likely comes from forgetting the direction change and only using the final velocity. Choice B (480 N) might result from incorrectly calculating the momentum change as just the difference in speeds (70 m/s) rather than properly accounting for the direction reversal. Choice D (1200 N) could come from computational errors in the momentum calculation. Study tip: Always establish a coordinate system first and treat velocity as a vector. In collision problems, "opposite direction" means you must add the magnitudes when calculating momentum change, not subtract them.
18. A 0.60 kg basketball bounces off the floor. Just before hitting the floor, its velocity is 8.0 m/s downward. Just after leaving the floor, its velocity is 6.0 m/s upward. If the ball is in contact with the floor for 0.025 s, what is the average force exerted by the floor on the ball during the bounce?
    Answer: 342 N upward When you encounter collision problems involving momentum and forces, you're dealing with Newton's second law in its impulse-momentum form: $F_{avg} \cdot \Delta t = \Delta p$, where the impulse equals the change in momentum. First, establish your coordinate system. Let's define upward as positive. The ball's initial velocity is -8.0 m/s (downward) and final velocity is +6.0 m/s (upward). The momentum change is: $\Delta p = m(v_f - v_i) = 0.60 \text{ kg}(6.0 - (-8.0)) = 0.60 \times 14.0 = 8.4 \text{ kg⋅m/s}$ Now you need to consider all forces acting on the ball during contact. The floor exerts force $F_{floor}$ upward, while gravity exerts $mg = 0.60 \times 9.8 = 5.88$ N downward. The net force creates the momentum change: $F_{net} = \frac{\Delta p}{\Delta t} = \frac{8.4}{0.025} = 336$ N upward. Since $F_{net} = F_{floor} - mg$, we get: $F_{floor} = F_{net} + mg = 336 + 5.88 = 342$ N upward. Choice A (48 N) likely comes from incorrectly calculating the momentum change. Choice B (336 N) is the net force, but the question asks for the floor's force specifically—a common trap where students forget to account for gravity. Choice D (384 N) might result from calculation errors in the momentum change. Remember: in collision problems, always identify all forces acting on the object. The force you're asked to find might not be the same as the net force causing the momentum change.
19. If the puck slides for 8.0 s before coming to rest, what was the magnitude of the impulse delivered by friction during this time?
    Answer: 2.9 N⋅s When you encounter problems involving friction and motion over time, think about impulse and momentum. The impulse-momentum theorem states that impulse equals the change in momentum, which often provides the most direct path to the answer. First, let's find the friction force. The kinetic friction force is $f_k = \mu_k mg = 0.15 \times 0.25 \times 9.8 = 0.37 \text{ N}$. Since friction opposes motion, this force acts for the entire 8.0 seconds until the puck stops. Using the impulse-momentum theorem, the impulse equals the change in momentum. Since the puck comes to rest, its final momentum is zero, so the impulse equals the magnitude of the initial momentum. We can find this using $J = f \times t = 0.37 \times 8.0 = 2.96 \text{ N⋅s}$, which rounds to 2.9 N⋅s. Answer A (2.9 N⋅s) is correct. Answer B (5.8 N⋅s) likely comes from doubling the correct answer, perhaps from a sign error or misconception about direction. Answer C (12 N⋅s) might result from using the wrong coefficient of friction (like 0.6 instead of 0.15) or calculation errors. Answer D (18 N⋅s) is too large and suggests a fundamental error in approach, possibly confusing impulse with another quantity entirely. Remember: impulse problems often have multiple solution paths (force × time or change in momentum), but both should give the same answer. When friction is involved, always calculate the friction force first, then apply it over the given time interval.
20. Two identical 2.0 kg blocks are connected by a light string. Initially, both blocks are at rest. A constant force of 24 N is applied to one block for 3.0 s in the horizontal direction. What is the magnitude of the impulse delivered to the system of two blocks?
    Answer: 72 N⋅s When you encounter problems involving forces and time, think about impulse-momentum concepts. Impulse is defined as the change in momentum of a system, and it also equals the net external force multiplied by the time interval. The key insight here is to treat the two connected blocks as a single system. Since the blocks are connected by a string, they move together as one unit with a combined mass of 4.0 kg. When analyzing impulse on a system, you only consider external forces - the internal tension forces between the blocks cancel out. The impulse delivered to the system equals the external force times the time interval: $J = F \cdot t = 24 \text{ N} \times 3.0 \text{ s} = 72 \text{ N⋅s}$ You can verify this using the impulse-momentum theorem. The system starts at rest (initial momentum = 0). The acceleration is $a = \frac{24 \text{ N}}{4.0 \text{ kg}} = 6.0 \text{ m/s}^2$. After 3.0 s, the velocity is $v = 6.0 \times 3.0 = 18 \text{ m/s}$. The final momentum is $4.0 \text{ kg} \times 18 \text{ m/s} = 72 \text{ kg⋅m/s}$, confirming our answer. Choice A (24 N⋅s) incorrectly uses only 1 second instead of 3 seconds. Choice B (36 N⋅s) might result from using just one block's mass incorrectly. Choice C (48 N⋅s) appears to use 2 seconds instead of 3. Remember: for impulse problems involving connected objects, treat them as a single system and focus on external forces only. Internal forces within the system don't contribute to the net impulse.

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