Exercises — Laws of Motion

23 questions✓ Free · step-by-step

4.13 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.1
Give the magnitude and direction of the net force acting on (a) a drop of rain falling down with a constant speed, (b) a cork of mass 10 g floating on water, (c) a kite skillfully held stationary in the sky, (d) a car moving with a constant velocity of 30 km/h on a rough road, (e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.
Hint. By Newton's second law, zero acceleration always means zero net force — check each scenario for whether velocity is actually changing, not for how many individual forces seem to be involved.
Step 1 — What all five scenarios share. In every case the object described has constant velocity: (a) constant speed, (b) and (c) stationary (zero velocity, unchanging), (d) constant velocity, (e) presumably moving at constant velocity with literally no forces present at all.

Step 2 — Applying Newton's second law. Since acceleration is zero in every one of these five situations, the net force must also be zero in every case — even though (a)-(d) each have several individual forces in play (gravity, air resistance, buoyancy, tension, friction, driving force), those forces exactly balance out to a net of zero.

✦ Answer: The net force is zero in all five cases (a)-(e), since each describes an object moving at constant velocity (including being at rest).
Where students slip. Assuming a moving object (like the raindrop or the car) must have some nonzero net force just because it's in motion — constant velocity, no matter how fast, always means zero net force; force is tied to changing velocity, not to motion itself.
4.24 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.2
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (a) during its upward motion, (b) during its downward motion, (c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45 degrees with the horizontal direction? Ignore air resistance.
Hint. With air resistance ignored, gravity is the only force acting on the pebble at every single instant of its flight, no matter which way it happens to be moving at that instant.
Step 1 — The only force in play. Since air resistance is ignored, gravity (weight = mg) is the sole force acting on the pebble throughout its entire flight, regardless of whether it is moving up, down, or momentarily at rest.

Step 2 — Computing its magnitude. mg = 0.05 × 10 = 0.5 N.

Step 3 — (a), (b), (c) All the same. During upward motion, during downward motion, and even at the highest point where velocity is momentarily zero, the net force is unchanged: 0.5 N, directed vertically downward — the highest point in particular shows that zero velocity does not mean zero force.

Step 4 — Does the 45° launch angle change anything? No. Gravity acts the same way regardless of the initial launch direction, since it does not depend on the pebble's own velocity at all — only its mass and the local gravitational field.

✦ Answer: The net force is 0.5 N downward throughout the flight, in all three cases, and this does not change if the pebble is launched at 45° instead of straight up.
Where students slip. Assuming the net force must be zero at the highest point since velocity is zero there — velocity and force are independent quantities; the pebble is still being pulled down by gravity at that exact instant, which is exactly why it doesn't simply hover.
4.34 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.3
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg, (a) just after it is dropped from the window of a stationary train, (b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h, (c) just after it is dropped from the window of a train accelerating with 1 m/s^2, (d) lying on the floor of a train which is accelerating with 1 m/s^2, the stone being at rest relative to the train. Neglect air resistance throughout.
Hint. The moment the stone leaves the train's window, it is no longer in contact with the train at all — whatever the train does afterward (accelerate, keep constant speed) cannot exert any further force on a stone that isn't touching it.
Step 1 — (a), (b), (c) Once dropped, only gravity remains. In all three of these cases, the stone has left the train and is no longer in contact with it, so the train's own state of motion (stationary, constant velocity, or accelerating) is irrelevant from that moment on — with air resistance neglected, the only force acting on the falling stone is its weight, mg = 0.1 × 10 = 1 N, directed downward.

Step 2 — (d) A different case — the stone stays on the train. Here the stone is not dropped; it remains on the accelerating train's floor, moving along with it. To accelerate at the same 1 m/s² as the train, the stone needs a net horizontal force, supplied by friction between the stone and the floor: F = ma = 0.1 × 1 = 0.1 N, directed horizontally, in the same direction as the train's acceleration.

✦ Answer: (a), (b), (c) 1 N downward in every case, since only gravity acts once the stone is falling freely. (d) 0.1 N horizontal, in the direction of the train's acceleration, since the stone is still moving with the train.
Where students slip. Assuming (c)'s answer must involve the train's 1 m/s² acceleration somehow — once released, the stone carries no memory of the train's acceleration; it simply falls under gravity like any dropped object, unlike part (d) where the stone is still attached to (moving with) the accelerating train.
4.42 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.4
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is: (i) T, (ii) T - mv^2/l, (iii) T + mv^2/l, (iv) 0. T is the tension in the string. [Choose the correct alternative].
Hint. On a smooth (frictionless) horizontal table, ask what forces actually act on the particle in the horizontal plane where the circular motion happens.
Step 1 — Identify the horizontal forces. The table is smooth, so there is no friction. The only horizontal force acting on the particle is the string's tension, which pulls it directly toward the peg at the centre of the circle.

Step 2 — Apply Newton's second law for circular motion. The net force toward the centre must supply the centripetal force needed for circular motion, and here that entire role is played by the tension alone — there is no other horizontal force to add or subtract.

✦ Answer: (i) T — the tension alone is the net centripetal force on the particle.
Where students slip. Trying to subtract mv²/l from T as in option (ii) — mv²/l is not a separate, independently-existing force to subtract; it is simply the required centripetal force, which tension supplies in full.
4.52 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.5
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m/s. How long does the body take to stop?
Hint. Find the deceleration from F = ma first, then use v = u - at with the final speed set to zero.
Step 1 — Find the deceleration. a = F/m = 50/20 = 2.5 m/s².

Step 2 — Find the time to stop. Using v = u − at with v = 0, u = 15 m/s: 0 = 15 − 2.5t, so t = 15/2.5 = 6 s.

✦ Answer: The body takes 6 seconds to stop.
Where students slip. Forgetting that a 'retarding' force acts opposite to the direction of motion — this is why it decelerates the body toward a stop rather than speeding it up, and why it must be treated as a negative acceleration in the kinematic equation.
4.62 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.6
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m/s to 3.5 m/s in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
Hint. Find the acceleration from the change in speed over the given time, then apply F = ma.
Step 1 — Find the acceleration. a = (v − u)/t = (3.5 − 2.0)/25 = 1.5/25 = 0.06 m/s².

Step 2 — Find the force. F = ma = 3.0 × 0.06 = 0.18 N.

✦ Answer: The force has magnitude 0.18 N, directed the same way as the body's motion, since the body speeds up without changing direction.
Where students slip. Forgetting to state the direction — since the body speeds up (rather than slows down) while its direction of motion stays fixed, the force must point the same way as the velocity, not opposite to it.
4.73 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.7
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
Hint. Since the two forces are perpendicular, their resultant follows directly from the Pythagorean theorem.
Step 1 — Find the resultant force. Since the forces are perpendicular, resultant F = √(8² + 6²) = √(64+36) = √100 = 10 N.

Step 2 — Find the acceleration. a = F/m = 10/5 = 2 m/s².

Step 3 — Find the direction. The resultant makes an angle θ = tan⁻¹(6/8) ≈ 37° with the direction of the 8 N force.

✦ Answer: The acceleration is 2 m/s², directed at about 37° from the 8 N force (toward the 6 N force).
Where students slip. Simply adding 8 N and 6 N to get 14 N — since the two forces act at right angles to each other, they must be combined using the Pythagorean rule (vector addition), not plain arithmetic addition.
4.83 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.8
The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
Hint. The retarding force must decelerate the whole system — driver plus vehicle together, not just the vehicle's own mass.
Step 1 — Convert speed and find the deceleration. 36 km/h = 10 m/s. a = (0 − 10)/4.0 = −2.5 m/s² (a deceleration of 2.5 m/s²).

Step 2 — Use the total mass of driver plus vehicle. Total mass = 400 + 65 = 465 kg.

Step 3 — Find the force. F = ma = 465 × 2.5 = 1162.5 N, since it is this combined mass that must be decelerated together.

✦ Answer: The average retarding force is about 1162.5 N, directed opposite to the vehicle's motion.
Where students slip. Using only the three-wheeler's own mass (400 kg) and leaving out the driver — the driver decelerates along with the vehicle too, so the retarding force must act on their combined mass of 465 kg.
4.92 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.9
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m/s^2. Calculate the initial thrust (force) of the blast.
Hint. The upward thrust has to do two jobs at once: overcome the rocket's own weight, and still leave enough net force to produce the stated acceleration.
Step 1 — Set up the force balance. Net upward force = Thrust − Weight, and this net force must equal ma. So Thrust = ma + mg = m(a + g).

Step 2 — Substitute the values. Thrust = 20,000 × (5.0 + 10) = 20,000 × 15 = 300,000 N.

✦ Answer: The initial thrust is 3.0 × 10⁵ N (300,000 N).
Where students slip. Computing only ma = 20,000 × 5.0 = 100,000 N and stopping there — that figure is just the net force producing the acceleration; the thrust itself must be larger still, since part of it is spent simply holding the rocket's weight up before any net acceleration can happen.
4.105 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.10
A body of mass 0.40 kg moving initially with a constant speed of 10 m/s to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = -5 s, 25 s, 100 s.
Hint. Split the timeline into three distinct regimes: before the force starts (constant velocity), while the force acts (constant deceleration, even through a possible reversal), and after the force stops (constant velocity again, at whatever speed it had left).
Step 1 — Set up directions and the acceleration during 0 to 30 s. Take north as positive. a = F/m = −8.0/0.40 = −20 m/s² for the whole 30 s the force acts.

Step 2 — Position at t = -5 s (before the force starts). For t < 0 the body simply moves at its original constant 10 m/s: x(−5) = 0 − (10)(5) = −50 m.

Step 3 — Position at t = 25 s (still within the 30 s force window). Using x = ut + ½at² with u = 10, a = −20, t = 25: x(25) = 10(25) + ½(−20)(25²) = 250 − 6250 = −6000 m. (The body actually stops and reverses direction at t = 0.5 s, but the same constant-acceleration formula still applies straight through that reversal.)

Step 4 — Position at t = 100 s (force has already stopped at t = 30 s). First find the state at t = 30 s: v(30) = 10 − 20(30) = −590 m/s; x(30) = 10(30) + ½(−20)(30²) = 300 − 9000 = −8700 m. After t = 30 s the force is gone, so the body moves at this constant −590 m/s for the remaining 70 s: x(100) = −8700 + (−590)(70) = −8700 − 41300 = −50000 m.

✦ Answer: x(−5 s) = −50 m; x(25 s) = −6000 m; x(100 s) = −50000 m.
Where students slip. Using the same constant-acceleration formula all the way out to t = 100 s — the force only acts for 30 s; after that the body moves at whatever constant velocity it had at t = 30 s, since no force remains to change it further.
4.114 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.11
A truck starts from rest and accelerates uniformly at 2.0 m/s^2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance.)
Hint. The instant the stone is dropped, it keeps whatever horizontal velocity the truck had at that moment, but stops sharing the truck's acceleration from then on.
Step 1 — Truck's (and stone's initial) velocity at the moment of dropping. v = at = 2.0 × 10 = 20 m/s horizontal, which becomes the stone's horizontal velocity the instant it's released.

Step 2 — What happens after release. Once separated from the truck, the stone is simply a free-falling object: its horizontal velocity stays at 20 m/s (no horizontal force acts on it anymore), while it gains vertical speed under gravity alone.

Step 3 — (a) Velocity at t = 11 s (1 s after being dropped). Vertical component: v_y = g × 1 = 10 m/s downward. Combined speed: √(20² + 10²) = √500 ≈ 22.4 m/s, directed about tan⁻¹(10/20) ≈ 26.6° below the horizontal.

Step 4 — (b) Acceleration at t = 11 s. Once falling freely, the only force on the stone is gravity, so its acceleration is simply g = 10 m/s² downward — not the truck's 2.0 m/s², since the stone is no longer in contact with (or influenced by) the truck.

✦ Answer: Velocity ≈ 22.4 m/s, about 26.6° below the horizontal; acceleration = 10 m/s² vertically downward.
Where students slip. Carrying the truck's 2.0 m/s² acceleration into the stone's motion after it's dropped — the stone only keeps the truck's velocity at the instant of release, not its ongoing acceleration, since the two are no longer physically connected.
4.123 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.12
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m/s. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position?
Hint. The bob's velocity at the exact instant the string is cut is what determines everything afterward — check what that velocity actually is at each of the two points.
Step 1 — (a) Cutting at an extreme position. At an extreme position of the swing, the bob is momentarily at rest (this is a turning point of the oscillation, where velocity is zero). With no horizontal velocity to carry it anywhere, cutting the string here leaves only gravity acting, so the bob simply falls straight down in a vertical line.

Step 2 — (b) Cutting at the mean position. At the mean position, the bob has its maximum speed (1 m/s), directed horizontally (tangent to its circular arc at the lowest point of the swing). Cutting the string here launches the bob as a horizontal projectile, so it follows a parabolic path under gravity, just like a ball thrown horizontally.

✦ Answer: (a) A vertical straight-line path (free fall from rest). (b) A parabolic path (horizontal projectile motion, starting at 1 m/s).
Where students slip. Assuming the bob keeps moving along its circular arc for a moment after the string is cut — the instant the string is cut, the centripetal force vanishes completely, and the bob immediately follows whatever velocity it had at that instant, in a straight line modified only by gravity.
4.135 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.13
A man of mass 70 kg stands on a weighing scale in a lift which is moving (a) upwards with a uniform speed of 10 m/s, (b) downwards with a uniform acceleration of 5 m/s^2, (c) upwards with a uniform acceleration of 5 m/s^2. What would be the readings on the scale in each case? (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
Hint. The scale reads the normal force pushing up on the man, which only equals his true weight mg when there's no acceleration — going through R = m(g+a) or m(g-a) for the accelerating cases keeps the sign bookkeeping straight.
Step 1 — (a) Uniform speed (zero acceleration). With no acceleration, the scale reading equals the man's actual weight: R = mg = 70 × 10 = 700 N.

Step 2 — (b) Accelerating downward at 5 m/s². When accelerating downward, the apparent weight decreases: R = m(g − a) = 70 × (10 − 5) = 70 × 5 = 350 N.

Step 3 — (c) Accelerating upward at 5 m/s². When accelerating upward, the apparent weight increases: R = m(g + a) = 70 × (10 + 5) = 70 × 15 = 1050 N.

Step 4 — (d) Free fall. If the lift falls freely, its acceleration equals g itself, so R = m(g − g) = 0 — the scale reads zero, the classic weightlessness case.

✦ Answer: (a) 700 N (b) 350 N (c) 1050 N (d) 0 N.
Where students slip. Reading (b) and (c) backwards — accelerating downward makes you feel lighter (reading drops below mg), while accelerating upward makes you feel heavier (reading rises above mg); mixing these up is the single most common mistake on this question.
4.144 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.14
Figure 4.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only).
Hint. Read the shape of each segment of the graph carefully: a perfectly straight line (even a sloped one) means constant velocity, not acceleration — only a sudden kink in the graph signals a force acting.
Step 1 — Reading the three segments. For t < 0, x stays flat at 0 (particle at rest). From 0 to 4 s, x rises in a straight line from 0 to 3 m — a constant slope, meaning constant velocity of 3/4 = 0.75 m/s, not changing speed. For t > 4 s, x is flat again at 3 m (particle at rest once more).

Step 2 — (a) Force in each continuous segment. Since each of these three segments has constant velocity (zero in the first and third, a steady 0.75 m/s in the middle one), the acceleration — and hence the force — is exactly zero throughout all three: for t < 0, for 0 < t < 4 s, and for t > 4 s.

Step 3 — (b) Where the force actually acts. The velocity itself jumps abruptly at the two kinks in the graph — from 0 to 0.75 m/s right at t = 0, and back from 0.75 m/s to 0 right at t = 4 s — meaning brief impulsive forces act at exactly these two instants, not as sustained forces over any interval.

Step 4 — Computing the two impulses. Impulse at t = 0: m·Δv = 4 × (0.75 − 0) = 3 kg m/s. Impulse at t = 4 s: m·Δv = 4 × (0 − 0.75) = −3 kg m/s.

✦ Answer: Force is zero in all three continuous intervals (t<0, 0<t<4s, t>4s). Impulse = +3 kg m/s at t = 0, and −3 kg m/s at t = 4 s.
Where students slip. Assuming the sloped middle segment (0 to 4 s) must mean a nonzero force is acting, just because the particle is moving there — a straight sloped line on an x-t graph still means constant velocity, and force only shows up as a kink (sudden change of slope), which happens only at the two endpoints.
4.153 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.15
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
Hint. The whole 30 kg system accelerates the same way regardless of which end is pulled — the tension only needs to accelerate whichever single mass isn't the one being pulled directly.
Step 1 — Acceleration of the whole system. Since the surface is smooth (frictionless) and the string is light, the two masses move together: a = F/(m_A + m_B) = 600/30 = 20 m/s², the same in both cases.

Step 2 — (i) Force applied to A (10 kg). Here the string must pull B (20 kg) along by itself: T = m_B × a = 20 × 20 = 400 N.

Step 3 — (ii) Force applied to B (20 kg). Now the string must pull A (10 kg) along instead: T = m_A × a = 10 × 20 = 200 N.

✦ Answer: (i) T = 400 N (ii) T = 200 N.
Where students slip. Assuming the tension must be the same in both cases since it's the same force F and the same two masses — the tension only has to accelerate whichever mass is NOT the one F is directly applied to, so pulling from the heavier end (B) leaves a lighter mass (A) for the string to drag, giving a smaller tension.
4.163 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.16
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
Hint. This is the standard two-mass-over-a-pulley setup — the heavier mass falls, the lighter one rises, and both share the same magnitude of acceleration since the string is inextensible.
Step 1 — Find the system's acceleration. a = (m2 − m1)g / (m1 + m2) = (12 − 8)(10) / (8 + 12) = 40/20 = 2 m/s².

Step 2 — Find the tension using the lighter mass (which accelerates upward). T − m1g = m1a, so T = m1(g + a) = 8 × (10 + 2) = 96 N.

Step 3 — Check using the heavier mass (which accelerates downward). m2g − T = m2a, so T = m2(g − a) = 12 × (10 − 2) = 96 N — matches, confirming the result.

✦ Answer: Acceleration = 2 m/s²; tension = 96 N.
Where students slip. Using the same sign convention (g+a or g−a) for both masses — the lighter mass accelerates upward against gravity (so g+a) while the heavier one accelerates downward with gravity assisting (so g−a); mixing these up gives two different, inconsistent tensions instead of the matching 96 N.
4.173 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.17
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Hint. Momentum conservation applies to the whole disintegration event — total momentum right before it happens must equal total momentum right after.
Step 1 — Momentum before disintegration. The original nucleus is at rest, so the total momentum of the system is exactly zero before it splits.

Step 2 — Momentum after disintegration. By conservation of momentum, the total momentum immediately after must still be zero: p1 + p2 = 0, where p1 and p2 are the momenta of the two product nuclei.

Step 3 — What this forces on their directions. Rearranging gives p1 = −p2, meaning m1v1 = −m2v2. Since mass is always a positive quantity, this equation can only hold if v1 and v2 point in exactly opposite directions — there is no other way for two positive masses' momenta to be negatives of each other.

✦ Answer: Since the total momentum must remain zero, the two product nuclei's momenta must be equal in magnitude and opposite in sign, which forces their velocities to point in exactly opposite directions.
Where students slip. Concluding only that the two product nuclei have equal and opposite momentum, without following through to velocity direction — since mass can't be negative, equal-and-opposite momentum specifically requires equal-and-opposite direction of travel, which is the actual claim being asked to show.
4.183 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.18
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m/s collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
Hint. Impulse equals the change in momentum — take initial and final velocity as signed quantities along one axis, since the ball reverses direction entirely.
Step 1 — Set up signed velocities for one ball. Take its initial velocity as +6 m/s; after rebounding it moves at −6 m/s (same speed, opposite direction).

Step 2 — Compute the impulse on that ball. Impulse = m(v_f − v_i) = 0.05 × (−6 − 6) = 0.05 × (−12) = −0.6 kg m/s, since reversing direction entirely doubles the velocity change compared to just stopping — this comes out to 0.6 N s directed opposite to its original motion.

Step 3 — The other ball, by symmetry. The second ball's velocity flips from −6 m/s to +6 m/s, giving impulse = 0.05 × (6 − (−6)) = +0.6 kg m/s — equal in magnitude and opposite in direction to the first ball's impulse, consistent with Newton's third law during the collision.

✦ Answer: Each ball receives an impulse of magnitude 0.6 N s (kg m/s), directed opposite to its own original velocity.
Where students slip. Computing the change in speed (6 to 6, i.e. zero) instead of the change in velocity — the ball's speed is unchanged, but its direction fully reverses, and impulse depends on the vector change in velocity, not on the scalar speed.
4.193 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.19
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m/s, what is the recoil speed of the gun?
Hint. The gun-and-shell system starts at rest, so conservation of momentum requires their momenta afterward to cancel out exactly.
Step 1 — Set up conservation of momentum. Initial total momentum = 0 (both at rest before firing). So m_shell × v_shell + m_gun × v_gun = 0.

Step 2 — Solve for the gun's recoil speed. v_gun = −(m_shell × v_shell)/m_gun = −(0.020 × 80)/100 = −1.6/100 = −0.016 m/s.

✦ Answer: The gun recoils at 0.016 m/s, in the direction opposite to the shell's motion.
Where students slip. Using the shell's speed directly as an estimate for the gun's recoil — the gun is vastly more massive (100 kg vs 0.020 kg), so by momentum conservation its recoil speed must be correspondingly tiny, not comparable to the shell's 80 m/s.
4.204 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.20
A batsman deflects a ball by an angle of 45 degrees without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
Hint. Since the ball's speed is unchanged and only its direction is deflected by 45 degrees, the change-in-velocity vector's size can be found from the isosceles triangle formed by the initial and final velocity vectors, using |Δv| = 2v sin(θ/2).
Step 1 — Convert the speed. 54 km/h = 15 m/s.

Step 2 — Find the magnitude of the velocity change. Since initial and final speeds are equal (v = 15 m/s) with a 45° angle between their directions, |Δv| = 2v sin(45°/2) = 2 × 15 × sin(22.5°) ≈ 30 × 0.3827 ≈ 11.48 m/s.

Step 3 — Compute the impulse. Impulse = m|Δv| = 0.15 × 11.48 ≈ 1.72 kg m/s (N s).

✦ Answer: The impulse imparted to the ball is about 1.72 N s, directed along the bisector of the angle between the ball's initial and final directions of motion.
Where students slip. Computing the impulse as simply m times the speed change (which is zero, since speed is unchanged) — the ball's direction changes even though its speed doesn't, and impulse depends on the vector change in velocity, which is nonzero here because of the 45° deflection.
4.214 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.21
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Hint. Convert the given rate of revolution into a linear speed first, then apply the centripetal-force formula T = mv squared over r.
Step 1 — Convert 40 rev/min to a linear speed. Frequency = 40/60 rev/s. v = 2πr × frequency = 2π × 1.5 × (40/60) = 2π × 1 = 2π ≈ 6.28 m/s.

Step 2 — Find the tension at this speed. T = mv²/r = 0.25 × (2π)² / 1.5 = 0.25 × 39.48 / 1.5 ≈ 6.58 N, since the string supplies exactly this much centripetal force.

Step 3 — Find the maximum speed for T = 200 N. Rearranging T = mv²/r: v_max = √(T×r/m) = √(200 × 1.5 / 0.25) = √1200 ≈ 34.6 m/s.

✦ Answer: Tension ≈ 6.58 N at 40 rev/min; the maximum whirling speed the string can survive is about 34.6 m/s.
Where students slip. Using 40 (rev/min) directly as if it were already a speed in m/s — revolutions per minute must first be converted into an actual linear speed (via the circle's circumference) before it can go into the centripetal-force formula.
4.222 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.22
If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks: (a) the stone moves radially outwards, (b) the stone flies off tangentially from the instant the string breaks, (c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle?
Hint. The instant the string breaks, the centripetal force disappears completely — whatever velocity the stone has at that exact instant is the only thing left to determine where it goes next.
Step 1 — What velocity direction is at the moment of breaking. At every point of circular motion, velocity is directed along the tangent to the circle at that point, never radially.

Step 2 — What happens once tension vanishes. With the string broken, there is no more centripetal force to keep curving the stone's path, so it simply continues moving in a straight line along whatever direction its velocity was pointing at that instant — the tangent direction.

✦ Answer: (b) — the stone flies off tangentially from the instant the string breaks.
Where students slip. Picking (a), radially outward — the stone's velocity was never directed radially to begin with (it was always tangential), so once the constraining force disappears, it continues along the tangent, not suddenly outward along the radius.
4.234 marksNCERT Cl-11 Physics Part I, Ch4 Exercises, Q4.23
Explain why (a) a horse cannot pull a cart and run in empty space, (b) passengers are thrown forward from their seats when a speeding bus stops suddenly, (c) it is easier to pull a lawn mower than to push it, (d) a cricketer moves his hands backwards while holding a catch.
Hint. For (a), think about what the horse actually pushes against to move forward at all — for (c), think about how the angle of the applied force changes the normal force (and hence friction) from the ground.
Step 1 — (a) Horse and cart in empty space. A horse moves forward by pushing backward against the ground, which by Newton's third law pushes the horse forward — this external reaction force from the ground is what actually propels the horse-cart system. In empty space, with nothing to push against, the horse's pull on the cart and the cart's equal pull back on the horse are purely internal forces within the horse-cart system, and internal forces alone cannot accelerate a system's overall centre of mass.

Step 2 — (b) Passengers thrown forward. By Newton's first law (inertia), a passenger's body tends to keep moving at its original speed even as the bus decelerates rapidly, since nothing directly forces the passenger's body to slow down at the same rate as the bus itself — so relative to the now-slower bus, the passenger continues forward.

Step 3 — (c) Pulling vs pushing a lawn mower. Pushing at a downward angle adds a downward component to the applied force, increasing the normal force from the ground (and hence the friction opposing motion, since friction = μ × normal force). Pulling at an upward angle instead reduces the normal force, so there's less friction to overcome, making it easier.

Step 4 — (d) Cricketer moving hands back. By the impulse-momentum relation (F = Δp/Δt), the same change in the ball's momentum spread over a longer time produces a smaller average force. Moving the hands back while catching extends the stopping time, which is exactly why it reduces the force (and sting) felt on the hands.

✦ Answer: (a) the horse needs an external reaction force from the ground, which is absent in empty space (b) inertia keeps the passenger's body moving at the original speed (c) pulling reduces the normal force and hence friction, unlike pushing which increases it (d) drawing the hands back increases the stopping time, reducing the average force for the same momentum change.
Where students slip. In (a), describing the horse's pull and the cart's reaction pull as somehow 'cancelling out the horse's effort' — the real point is narrower: those two forces are internal to the horse-cart system and simply cannot produce any net external acceleration of that system by themselves, no matter how hard the horse pulls.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Physics Part I textbook, Reprint 2026-27 (keph104.pdf) — one end-of-chapter Exercises set (23 questions, 4.1-4.23); the chapter's own note to 'take g = 10 m/s^2 for simplicity in numerical calculations' was followed throughout, a deliberate departure from g = 9.8 used in Ch1-3; Fig 4.16's exact position-time shape was rendered from the PDF and read directly rather than assumed. Questions are referenced from the NCERT textbook for identification.

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