Explain Force Magnitude and Direction - Grade 11 Physics

1. Which statement correctly describes the horizontal contact forces between the blocks?
   Answer: A pushes B with $12\ \text{N}$ right, while B pushes A with $12\ \text{N}$ left. Whenever you see two objects pushed together as a system, you need to apply Newton's Second Law twice — once to the whole system, and once to an individual block — and remember that Newton's Third Law guarantees contact forces are always equal and opposite. Start with the entire system. The total mass is $2.0 + 3.0 = 5.0\ \text{kg}$, and the applied force is $20\ \text{N}$, so the system's acceleration is $a = \frac{20}{5.0} = 4.0\ \text{m/s}^2$. Now isolate block B alone. The only horizontal force acting on B is the contact force from A pushing it right. Using Newton's Second Law for B: $F_{A \text{ on } B} = m_B \cdot a = 3.0 \times 4.0 = 12\ \text{N}\ \text{(right)}$. By Newton's Third Law, B must push back on A with exactly $12\ \text{N}\ \text{(left)}$. That confirms answer D. Choice A uses block A's mass instead of block B's mass to find the contact force, giving $2.0 \times 4.0 = 8.0\ \text{N}$ — a classic mix-up of which block you're analyzing. Choice B correctly finds $12\ \text{N}$ for A pushing B but then incorrectly assigns a different value ($8.0\ \text{N}$) for B pushing A, violating Newton's Third Law, which requires these forces to be equal in magnitude. Choice C assumes the full $20\ \text{N}$ is transmitted to B, ignoring that some force is "used up" accelerating block A itself. Your strategy: always isolate the block that the contact force is directly accelerating, apply $F = ma$ to it, then use Newton's Third Law to get the reaction force — same magnitude, opposite direction, every time.
2. Which statement correctly compares the net forces on the passenger at the two instants?
   Answer: The first is $120\ \text{N}$ upward, and the second is $120\ \text{N}$ downward. Whenever you see an elevator-and-scale problem, your key tool is Newton's Second Law applied to the passenger: the net force equals the scale reading (normal force, $N$) minus the passenger's weight ($W$), with direction mattering. Start by finding the weight: $W = mg = 60 \times 10 = 600\ \text{N}$ downward. First instant (moving down, scale reads $720\ \text{N}$): $F_{\text{net}} = N - W = 720 - 600 = +120\ \text{N}\ \text{(upward)}$ A scale reading greater than weight means the net force points upward — the elevator is decelerating (slowing its downward motion). Second instant (moving up, scale reads $480\ \text{N}$): $F_{\text{net}} = N - W = 480 - 600 = -120\ \text{N}\ \text{(downward)}$ A reading less than weight means net force points downward — the elevator is decelerating (slowing its upward motion). This confirms answer C. Answer A reverses the directions — it incorrectly assigns downward net force to the first instant and upward to the second, swapping the results of the subtraction. Answer B confuses the scale reading itself with the net force. The scale reads the normal force, not the net force; you must always subtract weight first. Answer D mistakes weight ($600\ \text{N}$) for net force. Weight is just one of the two forces acting; the net force is their difference. Study tip: Always remember — scale reading = normal force, not net force. Subtract weight from the normal force to find $F_{\text{net}}$, and let the sign tell you the direction.
3. What are the magnitudes and directions of the net forces during the first and second intervals?
   Answer: First: $3.0\ \text{N}$ forward; second: $5.0\ \text{N}$ forward. When a question gives you speed changes over a distance (not time), your go-to tool is the work-energy theorem: $W_{net} = \Delta KE$, which means $F_{net} \cdot d = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$. For the first interval, the sled accelerates from $2.0\ \text{m/s}$ to $4.0\ \text{m/s}$ over $4.0\ \text{m}$: $\Delta KE = \frac{1}{2}(2.0)(4.0^2 - 2.0^2) = \frac{1}{2}(2.0)(12) = 12\ \text{J}$ $F_1 = \frac{12\ \text{J}}{4.0\ \text{m}} = 3.0\ \text{N forward}$ For the second interval, from $4.0\ \text{m/s}$ to $6.0\ \text{m/s}$ over $4.0\ \text{m}$: $\Delta KE = \frac{1}{2}(2.0)(6.0^2 - 4.0^2) = \frac{1}{2}(2.0)(20) = 20\ \text{J}$ $F_2 = \frac{20\ \text{J}}{4.0\ \text{m}} = 5.0\ \text{N forward}$ This confirms answer A. B is wrong because it assumes both intervals produce the same force — but the speed change squared matters, not just the speed change. $6^2 - 4^2 \neq 4^2 - 2^2$ even though both intervals increase speed by $2\ \text{m/s}$. C doubles the correct answers, likely from forgetting the $\frac{1}{2}$ in the kinetic energy formula. D reverses the magnitudes and incorrectly assigns a backward direction to the second force — the sled is still speeding up, so the net force must be forward. Study tip: When you see distance (not time) given with speed changes, reach for work-energy immediately. Don't try to find acceleration using kinematics — it's possible but slower and more error-prone.
4. How do the forces applied by the supports to the string change?
   Answer: Each force becomes $2.25$ times as large, and the forces point in opposite directions. Whenever you see a wave speed problem involving a stretched string, your anchor equation is $v = \sqrt{T/\mu}$, where $T$ is the tension and $\mu$ is the linear mass density. This immediately connects wave speed to the tension the supports apply. Since $\mu$ stays constant, squaring both sides gives $v^2 \propto T$. When the wave speed increases by a factor of 1.5, the tension must increase by a factor of $(1.5)^2 = 2.25$. So each support must now pull with a force 2.25 times the original tension. And because the string is horizontal and in equilibrium (at rest overall), the two supports must pull in opposite directions — each one pulling outward, away from the other — creating balanced tension throughout the string. That confirms answer B. Looking at the wrong choices: A makes the mistake of applying the speed ratio directly to the force ratio, forgetting that tension scales with the square of the wave speed, not linearly. C gets the magnitude right (2.25×) but claims the forces point in the same direction — if both supports pulled the same way, there would be a net force and the string would accelerate, violating the "at rest" condition. D gets the direction correct (opposite) but inverts the relationship entirely, as if increasing the wave speed reduces tension, which contradicts $v = \sqrt{T/\mu}$. Study tip: Anytime wave speed on a string changes, immediately square the speed ratio to find the tension ratio. The equation $T \propto v^2$ (at constant $\mu$) is the core tool for this entire class of problems.
5. What is the magnitude and direction of the average force exerted by the wall on the ball, neglecting other impulses during contact?
   Answer: $750\ \text{N}$ west, because the total momentum change includes both the incoming and outgoing momenta. Whenever you see a collision or rebound problem, your first instinct should be the impulse-momentum theorem: the average force equals the change in momentum divided by the contact time, or $F = \Delta p / \Delta t$. The critical step is calculating total momentum change, which means accounting for both the initial and final velocities — including their directions. Set east as positive. The ball's initial momentum is $p_i = (0.15)(+20) = +3\ \text{N}\cdot\text{s}$, and its final momentum is $p_f = (0.15)(-30) = -4.5\ \text{N}\cdot\text{s}$ (west is negative). The change in momentum is: $\Delta p = p_f - p_i = -4.5 - 3 = -7.5\ \text{N}\cdot\text{s}$ Dividing by contact time: $F = -7.5 / 0.010 = -750\ \text{N}$, meaning 750 N directed west. That confirms B is correct. Choice A fails because it only uses the incoming momentum $(0.15 \times 20)/0.010 = 300\ \text{N}$, ignoring the outgoing momentum entirely — you must use the change, not just one side. Choice C gets the magnitude right but assigns the wrong direction. Since the ball rebounds west, the wall pushes it west; a greater outgoing speed does not flip the direction of force eastward. Choice D forgets to multiply by mass — speed change alone ($20 + 30 = 50\ \text{m/s}$) divided by time gives units of acceleration, not force. A reliable strategy: always define a positive direction first, assign signs to both velocities, and compute $\Delta p = mv_f - mv_i$ before dividing by time. Sign errors are the most common trap in impulse problems.
6. Compared with the original electric force on $q$, what is the new electric force on $q$?
   Answer: It is half as large and points away from $Q$. Whenever you see a question like this, recognize it's testing two things at once: how Coulomb's Law scales mathematically, and how charge signs determine force direction. Keep both in mind as you work through each change systematically. Start with Coulomb's Law: $F = k\frac{|q||Q|}{r^2}$. The original force has some magnitude $F_0 = k\frac{qQ}{r^2}$, and since $q$ is positive and $Q$ is negative, the force on $q$ is attractive — it points toward $Q$. Now apply each change. Doubling $q$ multiplies the force by 2. Increasing separation to $2r$ means $r^2$ becomes $(2r)^2 = 4r^2$, dividing the force by 4. Combined, the new magnitude is $F_0 \times 2 \times \frac{1}{4} = \frac{F_0}{2}$ — half the original. Then consider direction: reversing $Q$'s sign makes both charges positive, so the force becomes repulsive — pointing away from $Q$. That confirms answer A. Answer B is wrong because it correctly identifies the halved magnitude but gets the direction wrong — it ignores that flipping $Q$ from negative to positive turns attraction into repulsion. Answer C incorrectly doubles the magnitude; it accounts for doubling $q$ but forgets that the distance increase reduces the force by a factor of 4, giving a net halving. Answer D makes both errors simultaneously — it claims the force doubles and remains attractive. A useful strategy: treat magnitude and direction as two completely separate calculations. Change them one at a time so you don't mix up the math with the sign logic.
7. How does the average net force during interval II compare with the average net force during interval I?
   Answer: It has twice the magnitude and points west. Whenever you see a question comparing forces across two time intervals, your go-to tool is Newton's Second Law combined with the definition of acceleration: $F_{net} = ma = m\frac{\Delta v}{\Delta t}$. Start by finding the acceleration in each interval. During interval I, the velocity changes from 2.0 m/s east to 6.0 m/s east, so $\Delta v = +4.0\ \text{m/s}$ (east) over 2.0 s, giving $a_I = +2.0\ \text{m/s}^2$. During interval II, the velocity changes from 6.0 m/s east to 2.0 m/s east, so $\Delta v = -4.0\ \text{m/s}$ (west) over only 1.0 s, giving $a_{II} = -4.0\ \text{m/s}^2$. Since mass is constant, the force in interval II is $m(-4.0)$, which is twice the magnitude of interval I's force $m(+2.0)$, and points west (negative direction). That confirms answer D. Answer A makes two errors: it assumes both forces are equal in magnitude and both point east, ignoring that the puck is decelerating (not accelerating) in interval II. Answer B gets the direction right (west) but misses that the shorter time interval doubles the magnitude — a common trap when students only look at $\Delta v$ and forget $\Delta t$. Answer C correctly identifies the factor-of-two magnitude difference but assigns the wrong direction; since the puck is slowing down, the net force must oppose motion, pointing west. Your takeaway: always compute $\frac{\Delta v}{\Delta t}$ fully — both magnitude and sign. A smaller time interval with the same speed change means a larger force, and the direction of force follows the sign of $\Delta v$, not the direction of motion.
8. How does the air-resistance force at terminal speed compare with the air-resistance force at the earlier instant?
   Answer: It is twice as large, and both air-resistance forces point upward. Whenever you see a problem involving falling objects and air resistance, your first tool is Newton's second law: the net force equals mass times acceleration, and you must account for the direction of every force. Air resistance always opposes motion. Since the drop falls downward, air resistance points upward in both situations — this immediately eliminates answers A and D, which claim otherwise. Now apply $F_{net} = ma$ at each instant. Taking downward as positive: At the earlier instant (acceleration = $g/2$): $mg - F_1 = m\left(\frac{g}{2}\right) \implies F_1 = mg - \frac{mg}{2} = \frac{mg}{2}$ At terminal speed (acceleration = 0): $mg - F_2 = 0 \implies F_2 = mg$ Comparing: $F_2 = mg = 2 \times \frac{mg}{2} = 2F_1$ The air-resistance force at terminal speed is twice as large, and both forces point upward — confirming C. Looking at the wrong answers: A correctly identifies the factor of two but wrongly says the forces point downward — air resistance never acts in the same direction as motion. B gets the direction right (upward) but inverts the ratio; students who subtract incorrectly or confuse which instant has larger drag fall into this trap. D suggests the direction flips, which would only happen if the drop reversed direction — it never does here. Study tip: At terminal velocity, net force is zero, so drag equals weight — this is a reliable anchor point. Always draw a free-body diagram and assign directions before doing any algebra.
9. How does the net force on the cart at point A compare with the net force at point B?
   Answer: At A it is twice as large and points left; at B it points right. When a spring-mass system oscillates, the restoring force follows Hooke's Law: $F = -kx$, where $x$ is the displacement from equilibrium. The force always points toward equilibrium, and its magnitude is directly proportional to displacement. Keep both of those facts in your mind simultaneously — direction and magnitude depend entirely on position. At point A, the cart is $0.10\ \text{m}$ to the right, so the spring pulls it leftward with force $F_A = k(0.10)$. At point B, the cart is $0.050\ \text{m}$ to the left, so the spring pushes it rightward with force $F_B = k(0.050)$. The ratio is $F_A / F_B = 0.10/0.050 = 2$, meaning A's force is twice as large. Direction at A is left (toward equilibrium); direction at B is right (toward equilibrium). That perfectly describes answer A, confirming it is correct. Answer B gets the magnitudes right but flips both directions — a classic sign error. Remember, the force always points toward equilibrium, not away from it. Answer C gets the directions right but inverts the magnitude ratio, perhaps from confusing which displacement is larger. Answer D claims both forces point left, ignoring that at B the cart is already left of equilibrium, so the force must point right to restore it. A reliable study tip: whenever you see a spring problem, immediately ask two questions — "which side of equilibrium is the object on?" (that gives direction) and "how far is it?" (that gives relative magnitude). Handling these separately prevents the mixed-up combinations that traps like B, C, and D are designed to catch.
10. How does the net force on car B compare with the net force on car A?
    Answer: It is twice as large, and both forces point toward the center. Whenever you see a question comparing forces on objects moving in circles, your starting point should always be the centripetal force formula. For any object moving in a circular path, the net force points toward the center and equals: $F_{net} = \frac{mv^2}{r}$ where m is mass, v is speed, and r is the radius of the circle (the same for both cars here). Now set up the comparison. Let car A have mass m and speed v, so its net force is $F_A = \frac{mv^2}{r}$. Car B has mass $\frac{m}{2}$ and speed $2v$, giving: $F_B = \frac{\frac{m}{2}(2v)^2}{r} = \frac{\frac{m}{2} \cdot 4v^2}{r} = \frac{2mv^2}{r} = 2F_A$ Car B's net force is twice as large, and because it's centripetal, it points toward the center — confirming answer B. Choice A is tempting because mass is halved, but it ignores the squared effect of doubling the speed. Halving mass cuts the force by 2, but quadrupling $v^2$ multiplies by 4, giving a net factor of 2, not ½. Choice C gets the magnitude right but incorrectly says the force points along the motion — that would describe a tangential force causing acceleration along the path, not circular motion. Choice D correctly applies the $v^2$ factor but forgets to account for the halved mass, arriving at 4× instead of 2×; it also wrongly points the force outward (that's the common "centrifugal" misconception). Study tip: Always remember that speed is squared in the centripetal formula — it has a disproportionately large impact compared to mass, which appears to the first power only.

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