NCERT Solutions · Class 11 Physics Laws of Motion

Once the stone leaves the person's hand, the truck can no longer push it — only gravity acts on it. Everything about this problem comes from separating "what the stone was doing while riding on the truck" from "what happens to it in the air."Step $\displaystyle 1$: The stone's velocity at the instant of release.The truck starts from rest \(\displaystyle (u = 0) \) and accelerates uniformly at \(\displaystyle a = 2.0 \text{ m s}^{-2} \). Using Newton's first equation of motion, \[v = u + at \] at \(\displaystyle t = 10\text{ s} \): \[v = 0 + (2.0 \text{ m s}^{-2})(10 \text{ s}) = 20 \text{ m s}^{-1} \]So at the moment the stone is dropped, it is moving horizontally at $\displaystyle 20$ m s⁻¹ — the same as the truck, because it has been riding along with it.Step $\displaystyle 2$: What happens after release — horizontal motion.The instant the stone leaves the person's hand, it is no longer in contact with the truck. With air resistance neglected, the only force on the stone is gravity, which is purely vertical. By Newton's first law, with no horizontal force acting, the stone's horizontal velocity stays exactly as it was at release: \(\displaystyle v_x = 20 \text{ m s}^{-1} \), unchanged for as long as it falls. This is the point that trips people up — the truck keeps accelerating after the stone is dropped, but that acceleration belongs to the truck, not to the stone anymore.Step $\displaystyle 3$: What happens after release — vertical motion.Vertically, the stone starts from rest (it had no vertical velocity while sitting on the truck) and falls freely under gravity, \(\displaystyle g = 10 \text{ m s}^{-2} \) (taken as given for this problem). One second after release — that is, at \(\displaystyle t = 11\text{ s} \), since the stone was dropped at \(\displaystyle t = 10\text{ s} \) — its downward speed is \[v_y = g t' = (10 \text{ m s}^{-2})(1 \text{ s}) = 10 \text{ m s}^{-1} \] (A quick check that the stone is still airborne at this instant: falling $\displaystyle 6$ m from rest under \(\displaystyle g = 10 \text{ m s}^{-2} \) takes \(\displaystyle t' = \sqrt{2h/g} = \sqrt{1.2} \approx 1.095 \text{ s} \), which is later than \(\displaystyle t' = 1\text{ s} \), so the stone has not yet hit the ground.)(a) Velocity at t = $\displaystyle 11$ sThe stone's velocity is the vector sum of its unchanged horizontal part and its vertical free-fall part — this is just projectile motion, with the "launch" being horizontal instead of the more usual case of launching from the ground. \[v = \sqrt{v_x^2 + v_y^2} = \sqrt{(20 \text{ m s}^{-1})^2 + (10 \text{ m s}^{-1})^2} = \sqrt{500} \text{ m s}^{-1} \approx 22.4 \text{ m s}^{-1} \]Speed alone is not the full answer here — velocity is a vector, so the direction matters. Taking the angle \(\displaystyle \theta \) below the horizontal (measured from the direction the truck is moving): \[\tan\theta = \frac{v_y}{v_x} = \frac{10}{20} = \frac{1}{2} \quad\Rightarrow\quad \theta = \tan^{-1}\!\left(\tfrac{1}{2}\right) \approx 26.6^\circ \]So the stone moves at about $\displaystyle 22.4$ m s⁻¹, directed $\displaystyle 26.6$° below the horizontal, in the direction the truck is travelling.(b) Acceleration at t = $\displaystyle 11$ sThis is the part where it's easy to guess wrong: since the truck was accelerating and the stone was on the truck, it's tempting to think the stone's acceleration should somehow still involve the truck's \(\displaystyle 2.0 \text{ m s}^{-2} \). It does not. Acceleration comes only from the net force acting right now, and once the stone is in the air, the truck exerts no force on it at all — only gravity does. So the stone's acceleration is simply \[a_{\text{stone}} = g = 10 \text{ m s}^{-2}, \text{ directed vertically downward.} \] This is true throughout the fall, not just at \(\displaystyle t = 11\text{ s} \) — the horizontal velocity is carried along unchanged (zero horizontal acceleration), while gravity alone accelerates the stone downward.Answer: (a) velocity \(\displaystyle \approx 22.4 \text{ m s}^{-1} \), directed at \(\displaystyle \tan^{-1}(1/2) \approx 26.6^\circ \) below the horizontal, in the direction of the truck's motion; (b) acceleration \(\displaystyle = 10 \text{ m s}^{-2} \), vertically downward (equal to \(\displaystyle g \), unaffected by the truck).

At the extreme position the bob is momentarily at rest — a point of zero speed, not maximum speed — so cutting the string there leaves nothing to carry it sideways, and it simply falls straight down.In an oscillation, the extreme position is where the bob has swung as far as it can and is about to turn back: all its kinetic energy has converted to gravitational potential energy, so its speed there is zero. This is the detail people get backwards — it's the mean position, not the extreme one, where the speed is largest.With the bob momentarily at rest, cutting the string removes the tension, and the only force left on it is its own weight, \(\displaystyle mg \), acting vertically downward. Newton's second law with zero initial velocity and a purely vertical force gives purely vertical motion — there is no sideways force to bend the path.The bob falls straight down along a vertical line, exactly like a ball released from rest.At the mean position the string is vertical, so the bob's velocity is horizontal — cutting the string there launches it as a horizontal projectile, so it traces a parabola, not a straight line.At the mean position the velocity of a pendulum bob is always perpendicular to the string (tangential to the circular arc), and since the string is vertical there, the velocity must be horizontal. This is the point of maximum speed, given as \(\displaystyle v = 1 \text{ m s}^{-1} \).The instant the string is cut, the tension — which had been supplying the centripetal force toward the pivot — vanishes. The only force remaining is gravity, \(\displaystyle mg \), pointing straight down. This is precisely the setup of a horizontally launched projectile: constant horizontal velocity, constant downward acceleration \(\displaystyle g \).Take the cut point as the origin, with \(\displaystyle x \) horizontal (along the initial velocity) and \(\displaystyle y \) measured vertically downward.Horizontal direction has no force acting on the bob, so the horizontal velocity stays constant: \[x = vt \]Vertical direction starts from rest and accelerates under gravity, by \(\displaystyle y = \frac{1}{2}at^2 \) with \(\displaystyle a = g \): \[y = \frac{1}{2}gt^2 \]Eliminating \(\displaystyle t = x/v \) from the first equation: \[y = \frac{1}{2}g\left(\frac{x}{v}\right)^2 = \frac{g}{2v^{2}}x^{2} \]Substituting \(\displaystyle g = 10 \text{ m s}^{-2} \) (as given) and \(\displaystyle v = 1 \text{ m s}^{-1} \): \[y = \frac{10}{2(1)^{2}}x^{2} = 5x^{2} \]with \(\displaystyle x \) and \(\displaystyle y \) in metres. This is the equation of a parabola, curving downward and outward from the point where the string was cut — the standard shape of horizontal-projectile motion.The mass ($\displaystyle 0.1$ kg) and string length ($\displaystyle 2$ m) given in the problem do not enter this calculation at all — they would matter for finding the tension in the string or the time to reach the floor, but the shape of the path after release depends only on the direction and size of the velocity at the moment of cutting, together with \(\displaystyle g \).It is also worth noting that once the string is cut, the bob does not keep moving along the circular arc even briefly — with the tension gone there is no centripetal force, so from that instant its path is governed only by its instantaneous velocity and gravity.Answer: (a) A vertical straight line — the bob falls straight down, since its speed is zero at the extreme position. (b) A parabola, \(\displaystyle y = 5x^{2} \) (with \(\displaystyle x, y\) in metres) — since the bob leaves the mean position with a horizontal speed of \(\displaystyle 1 \text{ m s}^{-1}\) and then undergoes projectile motion under gravity.

A weighing scale never measures gravity directly — it measures the normal force pushing back on the man's feet, and that force changes whenever the lift accelerates.The man's weight \(\displaystyle mg \) is constant throughout — mass doesn't change and neither does \(\displaystyle g \). What changes is the normal reaction \(\displaystyle N \) that the scale exerts upward on him. By Newton's third law, the man presses down on the scale with a force equal in magnitude to \(\displaystyle N \), and that is exactly what the scale's pointer shows — displayed as if it were a mass, \(\displaystyle N/g \). So the "reading" is an apparent weight, not the true weight.Take the upward direction as positive, and let \(\displaystyle a \) be the lift's acceleration (positive if the acceleration points up, negative if it points down). Two forces act on the man: weight \(\displaystyle mg \) downward and normal force \(\displaystyle N \) upward. Newton's second law along the vertical gives\[N - mg = ma \implies N = m(g+a) \]Here \(\displaystyle m = 70 \text{ kg} \) and \(\displaystyle g = 10 \text{ m s}^{-2} \). The scale's reading, expressed as a mass, is \(\displaystyle R = N/g \).(a) Upward with uniform speed \(\displaystyle 10\text{ m s}^{-1} \)A uniform speed means zero acceleration — this is the detail that trips people up, since the lift is moving fast but not speeding up. So \(\displaystyle a = 0 \): \[N = m(g+a) = 70(10+0) = 700 \text{ N} \] \[R = \frac{700}{10} = 70 \text{ kg} \] The scale reads exactly the man's true weight, unchanged from when the lift is at rest — constant velocity never affects the reading, no matter how fast.(b) Downward with uniform acceleration \(\displaystyle 5 \text{ m s}^{-2} \)The acceleration points downward, so with "up" as positive, \(\displaystyle a = -5 \text{ m s}^{-2} \): \[N = 70(10-5) = 70(5) = 350 \text{ N} \] \[R = \frac{350}{10} = 35 \text{ kg} \] The scale under-reads the man's weight — this is the sensation of feeling lighter as a lift starts descending.(c) Upward with uniform acceleration \(\displaystyle 5 \text{ m s}^{-2} \)Now the acceleration points upward, so \(\displaystyle a = +5 \text{ m s}^{-2} \): \[N = 70(10+5) = 70(15) = 1050 \text{ N} \] \[R = \frac{1050}{10} = 105 \text{ kg} \] The scale over-reads — the extra push needed to accelerate the man upward is felt as extra weight.(d) Lift mechanism fails, falling freely under gravityFree fall means the lift's acceleration equals \(\displaystyle g \) itself, directed downward: \(\displaystyle a = -g = -10 \text{ m s}^{-2} \). \[N = m(g-g) = m(0) = 0 \text{ N} \] \[R = \frac{0}{10} = 0 \text{ kg} \] With no normal force at all, the man and the scale are both in free fall together, so he exerts no force on it — this is the state of apparent weightlessness, identical to what an astronaut in orbit feels, even though true gravity has not vanished.Answer: (a) $\displaystyle 700$ N (scale reads $\displaystyle 70$ kg — unchanged); (b) $\displaystyle 350$ N (scale reads $\displaystyle 35$ kg); (c) $\displaystyle 1050$ N (scale reads $\displaystyle 105$ kg); (d) $\displaystyle 0$ N (scale reads $\displaystyle 0$ kg — apparent weightlessness).

A straight-line position-time graph means constant velocity — so wherever the graph is a straight segment, the force is zero. Force only shows up at the instant the graph bends, because that is where the velocity itself jumps.Figure $\displaystyle 4.16$ is a graph of position \(\displaystyle x\) against time \(\displaystyle t\) made of three straight pieces: Mass of the particle: \(\displaystyle m = 4\ \text{kg}\).(a) Force in each intervalVelocity is the slope of the \(\displaystyle x\)-\(\displaystyle t\) graph, \(\displaystyle v=\dfrac{dx}{dt}\).For \(\displaystyle t<0\): \(\displaystyle x\) is constant, so \[v = 0 . \]For \(\displaystyle 0<t<4\ \text{s}\): the segment is a straight line (constant slope), so the velocity is constant at \[v = \frac{\Delta x}{\Delta t} = \frac{3\ \text{m} - 0}{4\ \text{s} - 0} = 0.75\ \text{m s}^{-1}. \]For \(\displaystyle t>4\ \text{s}\): \(\displaystyle x\) is constant again, so \[v = 0 . \]In each of these three intervals the velocity does not change with time, so the acceleration is \[a = \frac{dv}{dt} = 0 . \]By Newton's second law, \(\displaystyle F = ma\), so \[F = 0 \quad \text{for } t<0,\ \ 0<t<4\ \text{s},\ \text{and}\ t>4\ \text{s}. \]It is tempting to think a force must act during \(\displaystyle 0<t<4\ \text{s}\) because \(\displaystyle x\) is changing there — but changing position is not the same as changing velocity. A straight segment on an \(\displaystyle x\)-\(\displaystyle t\) graph is uniform motion, and uniform motion needs zero net force, exactly like the particle being at rest.(b) Impulse at \(\displaystyle t=0\) and \(\displaystyle t=4\ \text{s}\)At the two instants \(\displaystyle t=0\) and \(\displaystyle t=4\ \text{s}\) the graph has a sharp kink: the slope changes abruptly, so the velocity itself jumps in essentially zero time. That is not an ordinary finite force — it is an impulsive force, and what we can calculate is its impulse, \(\displaystyle J=\Delta p = m(v_{\text{after}}-v_{\text{before}})\).At \(\displaystyle t=0\): velocity jumps from \(\displaystyle v_{\text{before}}=0\) (the \(\displaystyle t<0\) value) to \(\displaystyle v_{\text{after}}=0.75\ \text{m s}^{-1}\) (the \(\displaystyle 0<t<4\ \text{s}\) value). \[J = m(v_{\text{after}}-v_{\text{before}}) = 4\ \text{kg} \times (0.75 - 0)\ \text{m s}^{-1} = 3\ \text{kg m s}^{-1}. \] This is positive, so the impulse points in the direction the particle starts moving in (the positive \(\displaystyle x\)-direction).At \(\displaystyle t=4\ \text{s}\): velocity jumps from \(\displaystyle v_{\text{before}}=0.75\ \text{m s}^{-1}\) back to \(\displaystyle v_{\text{after}}=0\). \[J = m(v_{\text{after}}-v_{\text{before}}) = 4\ \text{kg} \times (0 - 0.75)\ \text{m s}^{-1} = -3\ \text{kg m s}^{-1}. \] The negative sign means this impulse points opposite to the direction of motion — it is the "braking" push that brings the particle to rest.Both values come out to exactly \(\displaystyle 3\ \text{kg m s}^{-1}\) in magnitude because the graph's grid values (\(\displaystyle 3\ \text{m}\), \(\displaystyle 4\ \text{s}\), \(\displaystyle 4\ \text{kg}\)) are exact readings, so no further rounding is needed.Answer: F = $\displaystyle 0$ for t < $\displaystyle 0$, $\displaystyle 0$ < t < $\displaystyle 4$ s, and t > $\displaystyle 4$ s (the particle moves at constant velocity, including zero, throughout each interval, so there is no net force in any of them). Impulse at t = $\displaystyle 0$ is +$\displaystyle 3$ kg m s⁻¹ (in the direction of motion, starting the particle moving); impulse at t = $\displaystyle 4$ s is −$\displaystyle 3$ kg m s⁻¹ (opposite to the direction of motion, stopping the particle).

The string can only pull — so the tension is exactly the force needed to drag whichever body is not touched directly by F.Both bodies sit on a smooth (frictionless) horizontal surface, so once F is applied, the string stays taut and the two masses move together as one system with a common acceleration \(\displaystyle a\). Split the problem into two steps for each case: first find the common acceleration from the whole system, then isolate the body that is pulled only by the string to find the tension in it.Let \(\displaystyle m_A = 10\ \text{kg}\), \(\displaystyle m_B = 20\ \text{kg}\), and \(\displaystyle F = 600\ \text{N}\).Step $\displaystyle 1$ — acceleration of the system (same in both cases).By Newton's second law applied to the two bodies together (mass \(\displaystyle m_A+m_B\), since the string is light and internal tension cancels out for the system as a whole): \[F = (m_A+m_B)\,a \] \[a = \frac{F}{m_A+m_B} = \frac{600\ \text{N}}{10\ \text{kg}+20\ \text{kg}} = \frac{600\ \text{N}}{30\ \text{kg}} = 20\ \text{m s}^{-2} \]This acceleration is the same in case (i) and case (ii) — the total mass being dragged is $\displaystyle 30$ kg either way. What changes between the two cases is which body the tension has to move.Case (i): F applied to A.Here A is pulled directly by F, and it drags B along through the string. Look at B alone: the only horizontal force acting on it is the tension T pulling it forward (there is no other horizontal push on B — the surface is smooth). By Newton's second law on B: \[T = m_B\,a = 20\ \text{kg} \times 20\ \text{m s}^{-2} = 400\ \text{N} \]Case (ii): F applied to B.Now B is pulled directly by F, and it drags A along through the string. Look at A alone: the only horizontal force on it is the tension T. By Newton's second law on A: \[T = m_A\,a = 10\ \text{kg} \times 20\ \text{m s}^{-2} = 200\ \text{N} \]A common slip here is to use the pulled body's mass, or the total mass, to find T — but the string only ever has to accelerate the body on the far end of it from F, since that trailing body has no other horizontal force acting on it. That is why the two answers are different even though F and the total mass are the same in both cases.Both masses were given to $\displaystyle 2$ significant figures, so both accelerations and tensions are quoted to $\displaystyle 2$ significant figures.Answer: The acceleration of the system is \(\displaystyle 20\ \text{m s}^{-2}\) in both cases. (i) With F applied to the $\displaystyle 10$ kg body, the tension in the string is $\displaystyle 400$ N. (ii) With F applied to the $\displaystyle 20$ kg body, the tension in the string is $\displaystyle 200$ N.

When two masses hang from a string over a pulley, the heavier one falls and the lighter one rises with the same magnitude of acceleration — treat the whole string-and-masses system as one, using Newton's second law separately for each mass and eliminating the tension.Let \(\displaystyle m_1 = 8 \) kg (the lighter mass, which rises) and \(\displaystyle m_2 = 12 \) kg (the heavier mass, which falls), joined by a light inextensible string over a frictionless pulley. Take \(\displaystyle g = 10 \) m s\(\displaystyle ^{-2}\) as instructed, and let \(\displaystyle a \) be the common magnitude of acceleration and \(\displaystyle T \) the tension in the string (the same throughout, since the string is light and the pulley frictionless).Setting up the two equations of motionFor \(\displaystyle m_2 \), which accelerates downward, Newton's second law (\(\displaystyle F_{\text{net}} = ma \)) along the direction of motion gives\[m_2 g - T = m_2 a \quad \text{...(i)} \]For \(\displaystyle m_1 \), which accelerates upward, the tension exceeds the weight, so\[T - m_1 g = m_1 a \quad \text{...(ii)} \]A common slip here is to write \(\displaystyle m_1 g - T \) for the rising mass out of habit — but the net force must point the way the object actually accelerates, which for \(\displaystyle m_1 \) is upward, so it is \(\displaystyle T - m_1 g \).Solving for the accelerationAdding (i) and (ii) eliminates \(\displaystyle T \):\[m_2 g - m_1 g = m_2 a + m_1 a \]\[a = \frac{(m_2 - m_1)g}{m_1 + m_2} \]Substituting the values:\[a = \frac{(12 - 8) \text{ kg} \times 10 \text{ m s}^{-2}}{(8 + 12) \text{ kg}} = \frac{4 \times 10}{20} \text{ m s}^{-2} = \frac{40}{20} \text{ m s}^{-2} = 2 \text{ m s}^{-2} \]The $\displaystyle 12$ kg mass accelerates downward and the $\displaystyle 8$ kg mass accelerates upward, each at $\displaystyle 2$ m s\(\displaystyle ^{-2}\).Solving for the tensionSubstitute \(\displaystyle a = 2 \) m s\(\displaystyle ^{-2}\) back into equation (ii):\[T = m_1(g + a) = 8 \text{ kg} \times (10 + 2) \text{ m s}^{-2} = 8 \times 12 \text{ N} = 96 \text{ N} \]As a check, equation (i) should give the same tension:\[T = m_2(g - a) = 12 \text{ kg} \times (10 - 2) \text{ m s}^{-2} = 12 \times 8 \text{ N} = 96 \text{ N} \]Both equations agree, confirming the result. Since \(\displaystyle g \) is given as an exact value ($\displaystyle 10$ m s\(\displaystyle ^{-2}\)) for this problem and the masses are whole numbers, the results are stated to the precision of the given data, with no further rounding needed.Answer: The acceleration is $\displaystyle 2$ m s\(\displaystyle ^{-2}\), with the $\displaystyle 12$ kg mass moving down and the $\displaystyle 8$ kg mass moving up; the tension in the string is $\displaystyle 96$ N.

Momentum, not force, decides the directions here — and momentum was zero before the split, so it must be zero after.The law of conservation of momentum says: if no external force acts on a system, its total momentum stays the same before and after any internal event. A nucleus disintegrating into fragments is exactly this kind of system — the forces that tear it apart are internal (nuclear) forces between its own parts, so no external force acts on the system as a whole, and momentum must be conserved.Before disintegration. The nucleus of mass \(\displaystyle M\) is at rest in the lab frame, so its momentum is zero: \[\vec{p}_i = M\vec{v} = M\times 0 = 0 \]After disintegration. Let the nucleus split into two smaller nuclei of masses \(\displaystyle m_1\) and \(\displaystyle m_2\), moving with velocities \(\displaystyle \vec{v}_1\) and \(\displaystyle \vec{v}_2\). The total momentum of the products is \[\vec{p}_f = m_1\vec{v}_1 + m_2\vec{v}_2 \]Applying conservation of momentum, \(\displaystyle \vec{p}_f = \vec{p}_i\): \[m_1\vec{v}_1 + m_2\vec{v}_2 = 0 \] \[\Rightarrow\ m_1\vec{v}_1 = -\,m_2\vec{v}_2 \]This single vector equation is the whole argument. Read it carefully: it does not merely say the two momenta are equal in size — it says one momentum vector is the negative of the other. A negative of a vector is not "some other direction," it is the same line, reversed: same magnitude, exactly opposite sense.Since mass is a positive scalar, dividing through by \(\displaystyle m_1\) doesn't flip any sign: \[\vec{v}_1 = -\frac{m_2}{m_1}\vec{v}_2 \] so \(\displaystyle \vec{v}_1\) is a negative scalar multiple of \(\displaystyle \vec{v}_2\) — the two velocity vectors are anti-parallel, lying along one common line but pointing away from each other.One more check: could both fragments simply stay at rest, making the equation trivially true without any real "direction" to speak of? No — a disintegration converts some rest energy (mass defect) into kinetic energy of the products, so at least one of \(\displaystyle \vec{v}_1,\vec{v}_2\) must be nonzero. But the relation \(\displaystyle m_1\vec{v}_1=-m_2\vec{v}_2\) forces the other to be nonzero too (if \(\displaystyle \vec{v}_1=0\) then \(\displaystyle \vec{v}_2\) would have to be zero as well). So both fragments genuinely move, and by the equation above they move in exactly opposite directions along the same straight line — which is why, in a cloud-chamber photograph of such a decay, the two recoiling tracks always lie back-to-back on one line through the point where the original nucleus sat.Answer: By conservation of momentum, \(\displaystyle m_1\vec{v}_1 = -m_2\vec{v}_2\); since mass is positive, this forces \(\displaystyle \vec{v}_1\) and \(\displaystyle \vec{v}_2\) to point in exactly opposite directions along the same line (equal and opposite momenta), so the two product nuclei must recoil away from each other, back-to-back.

Impulse is the change in momentum, \(\displaystyle \vec{J}=\Delta \vec p = m\vec v_{f}-m\vec v_{i}\) — and "rebounds with the same speed" does NOT mean the impulse is zero. The speed is unchanged, but the velocity reverses, and impulse depends on velocity (a vector), not speed.Take the direction of the first ball's initial motion as positive. Call the balls A and B, each of mass \[m = 0.05\ \text{kg}. \]Before the collision, since they move in opposite directions with the same speed \(\displaystyle 6\ \text{m s}^{-1}\): \[u_A = +6\ \text{m s}^{-1}, \qquad u_B = -6\ \text{m s}^{-1}. \]They "collide and rebound with the same speed" — each ball leaves the collision moving at \(\displaystyle 6\ \text{m s}^{-1}\) but in the direction it came from, i.e. each velocity reverses sign: \[v_A = -6\ \text{m s}^{-1}, \qquad v_B = +6\ \text{m s}^{-1}. \]Impulse on ball A (due to ball B): \[J_A = m(v_A-u_A) = (0.05\ \text{kg})\big[(-6)-(+6)\big]\ \text{m s}^{-1} = (0.05\ \text{kg})(-12\ \text{m s}^{-1}) \] \[J_A = -0.6\ \text{kg m s}^{-1}. \]The minus sign says the impulse on A points opposite to A's original direction of motion — i.e. along the direction B was originally travelling.Impulse on ball B (due to ball A): \[J_B = m(v_B-u_B) = (0.05\ \text{kg})\big[(+6)-(-6)\big]\ \text{m s}^{-1} = (0.05\ \text{kg})(+12\ \text{m s}^{-1}) \] \[J_B = +0.6\ \text{kg m s}^{-1}. \]This impulse on B points opposite to B's original direction of motion — i.e. along the direction A was originally travelling.Notice \(\displaystyle J_A = -J_B\): the impulse ball B gives ball A is equal in magnitude and exactly opposite in direction to the impulse ball A gives ball B. This is Newton's third law at work — the two balls push on each other with equal and opposite forces during the collision, and since they act over the same short contact time, the impulses (force × time) are also equal and opposite.Both results carry the same magnitude, \[|J| = 2mv = 2(0.05\ \text{kg})(6\ \text{m s}^{-1}) = 0.6\ \text{kg m s}^{-1}, \] kept to one significant figure to match the precision of the given speed \(\displaystyle 6\ \text{m s}^{-1}\).Answer: Each ball receives an impulse of magnitude \(\displaystyle 0.6\ \text{kg m s}^{-1}\) (equivalently \(\displaystyle 0.6\ \text{N s}\)), directed opposite to that ball's own initial velocity — i.e. each impulse points along the direction the other ball was originally moving in. The two impulses are equal in magnitude and opposite in direction to each other.

Momentum is conserved because no external horizontal force acts on the gun–shell system during firing.Before the shot, the gun and shell sit together, motionless — total momentum zero. The explosive push of the propellant is an internal force between the gun and the shell, so it cannot change the total momentum of the system; it only redistributes it between the two parts. This is the law of conservation of momentum applied to an isolated system.Setting up the equationLet \(\displaystyle m = 0.020 \) kg be the mass of the shell, \(\displaystyle M = 100 \) kg be the mass of the gun, \(\displaystyle v \) be the velocity of the shell just after firing, \(\displaystyle V \) be the recoil velocity of the gun.Take the direction in which the shell is fired as positive, so \(\displaystyle v = 80 \) m \(\displaystyle \text{s}^{-1} \).Initial momentum of the system (both at rest): \[p_i = 0 \]Final momentum, shell and gun moving in opposite senses: \[p_f = mv + MV \]Conservation of momentum, \(\displaystyle p_i = p_f \), gives \[mv + MV = 0 \]Solving for the recoil velocity\[V = -\frac{mv}{M} \]Substituting the values: \[V = -\frac{(0.020\ \text{kg})(80\ \text{m s}^{-1})}{100\ \text{kg}} \]\[V = -\frac{1.6\ \text{kg m s}^{-1}}{100\ \text{kg}} = -0.016\ \text{m s}^{-1} \]Reading the signThe minus sign is not a mistake to drop — it says the gun's velocity is opposite in direction to the shell's velocity. If the shell flies forward out of the barrel, the gun kicks backward into the shoulder of whoever is holding it. This is exactly the "recoil" a gun is known for.Significant figuresThe mass of the shell (\(\displaystyle 0.020\) kg) and the muzzle speed (\(\displaystyle 80\) m \(\displaystyle \text{s}^{-1}\)) are each given to two significant figures, so the result is stated to two significant figures: \(\displaystyle 0.016\) m \(\displaystyle \text{s}^{-1}\). There is no intermediate rounding — the division above was carried out in full before rounding.Answer: The gun recoils with a speed of $\displaystyle 0.016$ m s⁻¹ ($\displaystyle 1.6$ cm s⁻¹), directed opposite to the direction in which the shell is fired.

Impulse is a change in momentum, and momentum is a vector — so a ball whose speed never changes can still be given a large impulse, purely because its direction changed.The law. The impulse–momentum theorem says\[\vec{J} \;=\; \int \vec{F}\,dt \;=\; \Delta \vec{p} \;=\; m\vec{v}_f - m\vec{v}_i \]where \(\displaystyle m\) is the mass of the ball, \(\displaystyle \vec{v}_i\) its velocity just before the bat hits it, and \(\displaystyle \vec{v}_f\) its velocity just after. \(\displaystyle \vec J\) has the units of momentum, \(\displaystyle \text{kg m s}^{-1}\), which is the same thing as \(\displaystyle \text{N s}\).The step people get wrong: the speed is unchanged, so it is tempting to write \(\displaystyle m(v_f - v_i) = m(15 - 15) = 0\). That subtracts speeds. The theorem subtracts velocities, and these two velocities point different ways, so \(\displaystyle \Delta\vec p\) is far from zero.The data. Convert the speed before anything else — never substitute km/h into an SI formula:\[v \;=\; 54\ \text{km h}^{-1} \;=\; 54 \times \frac{1000\ \text{m}}{3600\ \text{s}} \;=\; 15\ \text{m s}^{-1} \]and \(\displaystyle m = 0.15\ \text{kg}\). The speed is the same before and after, so \(\displaystyle |\vec v_i| = |\vec v_f| = 15\ \text{m s}^{-1}\). (The value of \(\displaystyle g\) is not needed — no weight enters an impulse over so short a contact.)The geometry, stated carefully. Put the bat at the point \(\displaystyle O\). Draw the two paths as rays from \(\displaystyle O\): ray \(\displaystyle OA\) toward the bowler, the side the ball came from, and ray \(\displaystyle OB\), the side the ball goes to. The deflection quoted is the angle between these two paths at the bat:\[\angle AOB = 45^\circ \]This is the trap in the problem. The angle between the two paths is not the same as the angle through which the velocity vector turns. The ball travels into \(\displaystyle O\) along \(\displaystyle OA\), so \(\displaystyle \vec v_i\) points along \(\displaystyle -\widehat{OA}\); it then travels out along \(\displaystyle OB\). The velocity vector therefore swings through \(\displaystyle 180^\circ - 45^\circ = 135^\circ\), even though the two paths make only \(\displaystyle 45^\circ\) with each other. Getting this backwards is what turns a cosine into a sine below.Set up axes. Let the \(\displaystyle x\)-axis lie along the bisector of \(\displaystyle \angle AOB\), pointing outward from \(\displaystyle O\) between the two rays; the \(\displaystyle y\)-axis is perpendicular to it. Each ray then makes \(\displaystyle 22.5^\circ\) with the \(\displaystyle x\)-axis:\[\widehat{OA} = (\cos 22.5^\circ,\; +\sin 22.5^\circ), \qquad \widehat{OB} = (\cos 22.5^\circ,\; -\sin 22.5^\circ) \]Hence\[\vec v_i = -v\,(\cos 22.5^\circ,\; \sin 22.5^\circ), \qquad \vec v_f = +v\,(\cos 22.5^\circ,\; -\sin 22.5^\circ) \]Subtract the velocities.\[\vec v_f - \vec v_i = v\big(\cos 22.5^\circ + \cos 22.5^\circ,\; -\sin 22.5^\circ + \sin 22.5^\circ\big) = \big(2v\cos 22.5^\circ,\; 0\big) \]The \(\displaystyle y\)-components cancel exactly. What survives lies entirely along the \(\displaystyle x\)-axis — that is, along the bisector of the two paths, pointing away from the bat, back into the region the ball came from. That is physically what you expect: the bat pushed the ball back the way it came.Substitute.\[|\vec J| = m\,\big|\vec v_f - \vec v_i\big| = 2mv\cos 22.5^\circ = 2 \times 0.15\ \text{kg} \times 15\ \text{m s}^{-1} \times \cos 22.5^\circ \]\[|\vec J| = 4.5\ \text{kg m s}^{-1} \times 0.9238795 = 4.1575\ \text{kg m s}^{-1} \]Round once, at the end. The mass \(\displaystyle 0.15\ \text{kg}\) and the speed \(\displaystyle 54\ \text{km h}^{-1}\) each carry two significant figures, so the answer is quoted to two:\[|\vec J| \approx 4.2\ \text{kg m s}^{-1} \;=\; 4.2\ \text{N s} \]The direction is half the answer. Impulse is a vector, so a magnitude alone is incomplete: \(\displaystyle \vec J\) points along the bisector of the incoming and outgoing paths — at \(\displaystyle 22.5^\circ\) to each of them — directed outward from the bat, i.e. back toward the side the ball arrived from. By Newton's third law the ball pushes on the bat with an equal impulse in exactly the opposite direction.A check on the sine-versus-cosine choice: if instead you read the \(\displaystyle 45^\circ\) as the angle through which the velocity vector turns, you would get \(\displaystyle 2mv\sin 22.5^\circ = 1.7\ \text{N s}\), directed perpendicular to the bisector. The bisector direction found above is the signature of the correct reading — for a ball turned back on itself through \(\displaystyle 135^\circ\), \(\displaystyle \Delta\vec p\) must lie along the bisector of the two paths, and only \(\displaystyle 2mv\cos 22.5^\circ\) does.Answer: \(\displaystyle |\vec{J}| = 2mv\cos 22.5^\circ = 4.2\ \text{kg m s}^{-1}\) ($\displaystyle 4.2$ N s), directed along the bisector of the ball's incoming and outgoing paths, at \(\displaystyle 22.5^\circ\) to each, pointing back toward the side from which the ball came.

The string is the only thing pulling the stone toward the centre — its tension has to supply the entire centripetal force needed to keep the stone moving on the circle.Start by turning the rotation rate into an angular speed, since a count of "rev/min" cannot be substituted directly into the circular-motion formulas — this conversion is the step most often skipped.\[\nu = 40\ \text{rev/min} = \frac{40}{60}\ \text{rev/s} = \frac{2}{3}\ \text{s}^{-1} \]\[\omega = 2\pi\nu = 2\pi \times \frac{2}{3}\ \text{rad/s} = \frac{4\pi}{3}\ \text{rad/s} \approx 4.19\ \text{rad/s} \]By Newton's second law applied to circular motion, a mass \(\displaystyle m\) moving on a circle of radius \(\displaystyle r\) at angular speed \(\displaystyle \omega\) needs a net force \(\displaystyle F = m\omega^2 r\), directed toward the centre, to keep it on that path. Here the string tension \(\displaystyle T\) is the only horizontal pull on the stone, so \(\displaystyle T\) itself must equal this centripetal force — the problem sets the circle in a horizontal plane, so the tension is not also being asked to hold up the stone's weight.\[T = m\omega^2 r \]Substitute \(\displaystyle m = 0.25\ \text{kg}\), \(\displaystyle r = 1.5\ \text{m}\), \(\displaystyle \omega = \dfrac{4\pi}{3}\ \text{rad/s}\), keeping the exact fraction through the arithmetic rather than rounding early:\[T = 0.25\ \text{kg} \times \left(\frac{4\pi}{3}\ \text{rad/s}\right)^{2} \times 1.5\ \text{m} = 0.25 \times 17.55 \times 1.5\ \text{N} \approx 6.58\ \text{N} \]Every given quantity here ($\displaystyle 0.25$ kg, $\displaystyle 1.5$ m, $\displaystyle 40$ rev/min) carries two significant figures, so the tension rounds to \(\displaystyle T \approx 6.6\ \text{N}\).For the second part, ask instead: at what speed does the required centripetal force reach the string's breaking tension? Written in terms of the linear speed \(\displaystyle v\), the same centripetal force is \(\displaystyle F = \dfrac{mv^{2}}{r}\). Setting this equal to the maximum tension the string can bear, \(\displaystyle T_{\max} = 200\ \text{N}\), and solving for \(\displaystyle v\):\[T_{\max} = \frac{m v_{\max}^{2}}{r} \quad\implies\quad v_{\max} = \sqrt{\frac{T_{\max}\, r}{m}} \]\[v_{\max} = \sqrt{\frac{200\ \text{N} \times 1.5\ \text{m}}{0.25\ \text{kg}}} = \sqrt{1200\ \text{m}^2/\text{s}^2} = 20\sqrt{3}\ \text{m/s} \approx 34.64\ \text{m/s} \]This is the fastest the stone can travel around the $\displaystyle 1.5$ m circle: any faster, and the centripetal force the string would have to supply exceeds $\displaystyle 200$ N and it snaps. Reported to three significant figures, \(\displaystyle v_{\max} \approx 34.6\ \text{m/s}\).Answer: The tension in the string is about $\displaystyle 6.6$ N; the stone can be whirled at a maximum speed of about $\displaystyle 34.6$ m/s (\(\displaystyle =20\sqrt{3}\ \text{m/s}\)) before the string breaks.

Once the string is gone, no force acts on the stone — so it keeps moving exactly the way it was moving the instant before, in a straight line. That direction is tangential, not radial.Think about what the string was doing while it was intact. As the stone moves along the circle, its velocity vector \(\displaystyle \vec{v} \) always points along the tangent to the circle at the stone's current position — that is simply the definition of circular motion, true at every speed, not just near breaking point. The string's tension supplies the centripetal force, \[F_c = \frac{mv^2}{r}, \] directed radially inward, from the stone toward the center. This force does not speed the stone up or slow it down (it acts perpendicular to \(\displaystyle \vec{v} \) at every instant); its only job is to continuously bend the path into a circle by pulling the velocity vector toward the center.Now apply Newton's first law of motion: a body continues in its state of rest or of uniform motion in a straight line unless acted upon by an external force. The instant the string snaps, the tension — the only force that was keeping the stone on a curved path — disappears. There is no other horizontal force present (this is a horizontal circle, so gravity is balanced by a vertical support/normal force and plays no role in the horizontal motion). With zero net force, the stone cannot keep curving. It must move in a straight line with whatever velocity it had at the breaking instant.Since that velocity was tangential all along, the stone flies off tangentially, along the straight line that touches the circle at the point where the string broke.Why the other two options fail: So the trajectory immediately after the break is a straight line along the tangent at the point of release, continuing in that direction (subject only to gravity afterward, which curves it into a projectile path in the vertical plane — but that curving is downward due to gravity, not "radially outward" from the original circle).Answer: (b) — the stone flies off tangentially from the instant the string breaks, because with the tension gone there is no force left to bend its path, and Newton's first law says it must continue in the straight line given by its velocity at that instant, which is always directed along the tangent to the circle.

(a) A "reaction force" needs two bodies — the horse and the ground.A horse walks (and pulls a cart) by pushing backward against the ground with its feet. By Newton's third law of motion — for every action there is an equal and opposite reaction — the ground pushes back on the horse's feet with an equal and opposite force, directed forward. It is this ground reaction, not the horse's muscular effort by itself, that actually accelerates the horse–cart system forward.In empty space there is no ground, no floor, no medium of any kind to push against. A force needs a second body to act on before that body can push back; with nothing there, the horse's feet can exert no action force on anything, so there is no reaction force to propel it forward. With no external force on the horse–cart system, Newton's first law says it simply stays as it is (at rest, or drifting at whatever constant velocity it already had) — it cannot accelerate. A horse cannot pull a cart and run in empty space because the forward push comes from the ground, and there is no ground to push against.(b) The lower body stops with the bus; the upper body keeps going by inertia.By Newton's first law of motion, a body continues in its state of rest or of uniform motion unless an external unbalanced force acts on it — this tendency is inertia. While the bus moves at a steady speed \(\displaystyle v \), the passenger's whole body, including the parts not touching any seat or rail, is also moving at \(\displaystyle v \).When the brakes are applied suddenly, a large retarding force acts on the bus through its wheels and brings the bus body to rest (or to a much lower speed) in a very short time. This retarding force reaches the passenger only through contact — the feet on the floor, the seat against the back — so the lower part of the body is dragged to a stop along with the bus. The upper part of the body is not rigidly fixed to the bus, and no such backward force acts on it directly, so by inertia it continues moving forward at close to the original speed \(\displaystyle v \) for a moment after the lower body has stopped. The mismatch between a suddenly-halted lower body and a still-moving upper body is what throws the passenger forward.(c) Pulling tilts part of the mower's weight off the ground; pushing adds to it.Let the mower have weight \(\displaystyle mg \), let \(\displaystyle \mu \) be the coefficient of friction between the mower and the ground, and let the force \(\displaystyle F \) applied along the handle make an angle \(\displaystyle \theta \) with the horizontal. The mower moves only horizontally, so in either case the vertical forces must balance.When pulling, the handle rises from the mower up to the hand, so \(\displaystyle F \) has an upward component \(\displaystyle F\sin\theta \) besides its forward component \(\displaystyle F\cos\theta \). Balancing vertically, \[N_{\text{pull}} = mg - F\sin\theta. \] The upward pull is partly carrying the mower's weight, so the normal reaction — and with it the limiting friction \(\displaystyle f_{\text{pull}} = \mu N_{\text{pull}} \) that opposes motion — is reduced.When pushing, the handle slopes down from the hand to the mower, so the same component \(\displaystyle F\sin\theta \) now points downward, adding to the weight: \[N_{\text{push}} = mg + F\sin\theta. \] The push presses the mower harder into the ground, raising the normal reaction and the limiting friction \(\displaystyle f_{\text{push}} = \mu N_{\text{push}} \).Since \(\displaystyle N_{\text{pull}} < mg < N_{\text{push}} \), it follows that \(\displaystyle f_{\text{pull}} < f_{\text{push}} \) for the same applied force and angle. Less friction to overcome means less force is needed to keep the mower moving at a given speed, which is why it is easier to pull a lawn mower than to push it.(d) Withdrawing the hands stretches out the stopping time and so lowers the peak force.Newton's second law, written in terms of momentum, is \[F = \frac{\Delta p}{\Delta t}, \] where \(\displaystyle \Delta p \) is the change in the ball's momentum during the catch (from its momentum just before reaching the hands down to zero, once it is held) and \(\displaystyle \Delta t \) is the time taken to bring it to rest.\(\displaystyle \Delta p \) is fixed once the ball's mass and incoming speed are fixed — the ball must end up at rest in the fielder's hand regardless of how the catch is taken. What the fielder controls is \(\displaystyle \Delta t \): by drawing the hands backward as the ball arrives, its speed is brought to zero more gradually, over a longer interval \(\displaystyle \Delta t \). Because \(\displaystyle \Delta p \) stays the same, a larger \(\displaystyle \Delta t \) means a smaller average force \(\displaystyle F \) acts on the hands — and, by Newton's third law, an equally smaller reaction force acts on the ball from the hands. Keeping the hands still instead would stop the ball in a much shorter \(\displaystyle \Delta t \), producing a much larger force and a painful (or injurious) impact. This is why a cricketer moves the hands backward while holding a catch.Answer: (a) no ground to push against means no reaction force on the horse, so it cannot generate a forward force in empty space (Newton's third law needs two bodies). (b) the lower body stops with the bus by contact forces, while the upper body keeps moving forward by inertia (Newton's first law), throwing the passenger forward. (c) pulling lifts part of the mower's weight off the ground, lowering the normal reaction and hence the friction \(\displaystyle f_{\text{pull}}=\mu(mg-F\sin\theta) \); pushing adds to the weight instead, giving \(\displaystyle f_{\text{push}}=\mu(mg+F\sin\theta) > f_{\text{pull}} \), so pulling takes less force. (d) withdrawing the hands increases the time of impact \(\displaystyle \Delta t \), which lowers the average force \(\displaystyle F=\Delta p/\Delta t \) needed to stop the ball for the same change in its momentum.