CBSE Class 11 Physics Kinematics Advanced Problems

ADVANCED LEVEL ASSIGNMENT
CLASS 12 PHYSICS : KINEMATICS

1. A particle of mass 10^-2 kg is moving along the positive X-axis under the influence of a force F (x) = - \( \frac{k}{2x²} \) where k = 10^-2 Nm². At time t = 0, it is at x = 1.0 m and its velocity is v = 0.
(a) find the velocity when it reaches x = 0.5 m (b) find the time at which it reaches x = 0.25 m

Answer:
1. (a) By the work-energy theorem, (1/2)mv² = ∫F dx from x to 1 = (k/2)(1/x - 1).
2. So v² = (k/m)(1/x - 1), with k/m = 10⁻²/10⁻² = 1 m³/s². At x = 0.5 m, v² = 1 × (2 - 1) = 1, so v = 1 m/s, directed along the negative x-direction.
3. (b) v = -√(1/x - 1) m/s, so the time to go from x = 1 to x = 0.25 is t = ∫ dx/√(1/x - 1) taken from 0.25 to 1 = ∫ √(x/(1 - x)) dx.
4. The integral of √(x/(1 - x)) dx is sin⁻¹√x - √(x(1 - x)).
5. t = [π/2 - 0] - [π/6 - √(0.25 × 0.75)] = 1.5708 - 0.5236 + 0.4330 = 1.48 s.
6. Hence, the velocity is 1 m/s along the negative x-direction and the time is 1.48 s.

Teacher's Note:
a) The force is attractive towards the origin, so the particle speeds up as it moves in.
b) Use the substitution x = sin²φ to evaluate the time integral.

2. A car starts moving rectilinearly, first with acceleration a = 5.0 m/s² (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate a, comes to a stop. The total time of motion equals τ = 25 s. The average velocity during that time is equal to <v> = 72 km per hour. How long does the car move uniformly ?

Answer:
1. Average velocity = 72 km/h = 20 m/s, so the total distance = 20 × 25 = 500 m.
2. Let the car accelerate for time t, move uniformly for time T and decelerate for time t. Then 2t + T = 25.
3. Maximum speed v = at. Distance = (1/2)at² + (at)T + (1/2)at² = at(t + T) = at(25 - t).
4. 5t(25 - t) = 500, so t² - 25t + 100 = 0, which gives t = 5 s or t = 20 s.
5. t = 20 s would make T negative, so t = 5 s and T = 25 - 10 = 15 s.
6. Hence, the car moves uniformly for 15 s.

Teacher's Note:
a) Reject the root that gives a negative time.
b) Check: the speed is 25 m/s and the distance is 62.5 + 375 + 62.5 = 500 m.

3. At the moment t = 0 a particle leaves the origin and moves in the positive direction of the x axis. Its velocity varies with time as v = v₀ (1 - t / τ), where v₀ is the initial velocity vector whose modulus equals v₀ = 10.0 cm/s ; τ = 5.0 s. Find :
(a) the x coordinate of the particle at the moments of time 6.0, 10 and 20 s
(b) the moments of time when the particle is at the distance 10.0 cm from the origin
(c) the distance s covered by the particle during the first 4.0 and 8.0 s

Answer:
1. x(t) = ∫v dt = v₀(t - t²/2τ) = 10(t - t²/10) cm. The velocity is zero at t = τ = 5 s, so the particle turns back after 5 s.
2. (a) x(6) = 10(6 - 3.6) = 24 cm = 0.24 m; x(10) = 10(10 - 10) = 0; x(20) = 10(20 - 40) = -200 cm = -2.0 m.
3. (b) For x = +10 cm: t² - 10t + 10 = 0, so t = 5 - √15 = 1.1 s or t = 5 + √15 = 8.9 s. For x = -10 cm: t² - 10t - 10 = 0, so t = 5 + √35 = 10.9 s.
4. (c) In the first 4 s the particle moves in one direction, so s = x(4) = 10(4 - 1.6) = 24 cm.
5. In 8 s: it reaches x(5) = 25 cm at t = 5 s and then returns to x(8) = 10(8 - 6.4) = 16 cm. So s = 25 + (25 - 16) = 34 cm.
6. Hence, the answers are 0.24 m, 0 and -2.0 m; 1.1 s, 8.9 s and 10.9 s; and 24 cm and 34 cm.

Teacher's Note:
a) The printed answer -4.0 m for x(20) is wrong; the correct value is -2.0 m because x(20) = 10(20 - 40) cm.
b) Distance is the total path length, so add the forward and backward parts separately.
c) Distance 10 cm from the origin includes both sides, x = +10 cm and x = -10 cm.

4. The velocity of a particle moving in the positive direction of the x axis varies as v = α√x, where α is a positive constant. Assuming that at the moment t = 0 the particle was located at the point x = 0, find :
(a) the time dependence of the velocity and the acceleration of the particle
(b) the mean velocity of the particle averaged over the time that the particle takes to cover the first s meters of the path.

Answer:
1. (a) v = dx/dt = α√x, so dx/√x = α dt. Integrating, 2√x = αt (x = 0 at t = 0), so x = α²t²/4.
2. Velocity v = dx/dt = α²t/2.
3. Acceleration a = dv/dt = α²/2, which is constant.
4. (b) The time to cover distance s: 2√s = αt, so t = 2√s/α.
5. Mean velocity = s/t = s × α/(2√s) = α√s/2.

Teacher's Note:
a) The answer shows that the motion is uniformly accelerated.
b) The mean velocity over the time is s divided by the time taken.

5. A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as a = α√v , where α is a positive constant. At the initial moment the velocity of the point is equal to v₀. What distance will it traverse before it stops ? What time will it take to cover that distance ?

Answer:
1. The deceleration gives dv/dt = -α√v and also v dv/dx = -α√v, so dx = -√v dv/α.
2. Distance: s = (1/α)∫√v dv taken from 0 to v₀ = (2/3α) v₀^3/2.
3. Time: dv/√v = -α dt, so ∫dv/√v from v₀ to 0 gives 2√v₀ = αt.
4. Hence, t = 2√v₀/α.

Teacher's Note:
a) Use v dv/dx when the distance is wanted, and dv/dt when the time is wanted.
b) The particle stops in a finite time and distance, even though the deceleration falls as v falls.

6. The current velocity of a river grows in proportion to the distance from its bank and reaches the maximum value v₀ in the middle. Near the banks the velocity is zero. A boat is moving along the river in such a manner that it is always perpendicular to the current and the speed of the boat in still water is u. Find the distance through which the boat crossing the river will be carried away by the current if the width of the river is c. Also determine the trajectory of the boat.

Answer:
1. Take the x axis along the current and the y axis across the river, with the starting bank at y = 0. The boat crosses at speed u, so y = ut.
2. For y less than c/2 the current velocity is v_x = 2v₀y/c, and for y greater than c/2 it is v_x = 2v₀(c - y)/c.
3. First half: dx/dt = 2v₀ut/c, so x = v₀ut²/c = v₀y²/(uc). Hence y² = (uc/v₀)x for the first half.
4. At y = c/2 the drift is x = v₀c/(4u). The second half is the mirror image, so the total drift is 2 × v₀c/(4u).
5. Hence, the drift = \( \frac{c v_0}{2u} \), and the trajectory is the parabola y² = \( \frac{uc}{v_0} \) x for the first half and its mirror image for the second half.

Teacher's Note:
a) Break the crossing into two halves because the current changes its law at the middle.
b) The trajectory is symmetric about the mid-line of the river.

7. A particle moves in the plane xy with velocity v = ai + bxj, where i and j are the unit vectors of the x and y axes, and a and b are constants. At the initial moment of time the particle was located at the point x = y = 0. Find the equation of the particle's trajectory y (x).

Answer:
1. The velocity components are dx/dt = a and dy/dt = bx.
2. From the first, x = at. So dy/dt = bat.
3. Integrating, y = (1/2)abt².
4. Substituting t = x/a: y = (1/2)ab × x²/a².
5. Hence, y = \( \frac{b}{2a} \) x², a parabola.

Teacher's Note:
a) Eliminate t between x(t) and y(t) to get the trajectory.
b) Both x and y are zero at t = 0, so no constants of integration remain.

8. A particle of mass 1 kg which moves along the x-axis is subjected to an accelerating force which increases linearly with time and a retarding force which increases directly with displacement (constant of proportionality being one with proper dimensions is both the cases. At time t = 0, displacement and velocity both are zero. Find the displacement as a function of time t.

Answer:
1. With the constants of proportionality equal to one, the accelerating force is t and the retarding force is x.
2. Newton's second law for m = 1 kg: d²x/dt² = t - x.
3. A particular solution is x = t. The general solution is x = t + C cos t + D sin t.
4. At t = 0, x = 0 gives C = 0. The velocity v = 1 + D cos t is zero at t = 0, so D = -1.
5. Hence, x = t - sin t.

Teacher's Note:
a) This is the equation of simple harmonic motion with a driving term.
b) Check: x'' = sin t and t - x = sin t, so the equation is satisfied.

9. A car starts from rest with an acceleration of 6 m/s² which decreases to zero linearly with time, in 10 second, after which the car continues at a constant speed. Find the time required for the car to travel 400 m from the start.

Answer:
1. The acceleration is a = 6(1 - t/10) for 0 ≤ t ≤ 10 s.
2. Velocity v = 6(t - t²/20). At t = 10 s, v = 6(10 - 5) = 30 m/s.
3. Distance x = 6(t²/2 - t³/60). At t = 10 s, x = 6(50 - 16.67) = 200 m.
4. The remaining 200 m is covered at 30 m/s in 200/30 = 6.67 s.
5. Total time = 10 + 6.67 = 16.67 s.

Teacher's Note:
a) First check whether the distance of 400 m is reached within the acceleration stage; here it is not.
b) Integrate the acceleration twice to get the distance.

10. An elevator car whose floor-to-ceiling distance is equal to 2.7 m starts ascending with constant acceleration 1.2 m/s². 2.0 s after the start a bolt falls from the ceiling of the car. Find :
(a) the bolt's free fall time
(b) the displacement and the distance covered by the bolt during the free fall in the reference frame fixed to the elevator shaft.
(c) Find displacement bolt with respect to elevator.

Answer:
1. Take g = 9.8 m/s². (a) In the frame of the elevator the bolt falls with relative acceleration g + a = 9.8 + 1.2 = 11 m/s², so 2.7 = (1/2)(11)t² and t = 0.70 s.
2. (b) At the moment of release the bolt has the lift's velocity, 1.2 × 2 = 2.4 m/s upward. Displacement in the shaft frame = 2.4 × 0.70 - (1/2)(9.8)(0.70)² = 1.68 - 2.41 = -0.72 m, that is, 0.7 m downward.
3. The bolt first rises v²/2g = (2.4)²/19.6 = 0.29 m and then falls 0.29 + 0.72 = 1.01 m. Distance covered = 0.29 + 1.01 = 1.3 m.
4. (c) The displacement of the bolt with respect to the elevator is the floor-to-ceiling distance, 2.7 m downward.
5. Hence, the answers are 0.7 s; 0.7 m and 1.3 m; and 2.7 m.

Teacher's Note:
a) The bolt starts with the upward velocity of the lift, so it moves up a little before falling.
b) Displacement and distance differ because the bolt changes its direction.

11. Two swimmers leave point A on one bank of the river to reach point B lying right across on the other bank. One of them crosses the river along the straight line AB while the other swims at right angles to the stream and then walks the distance that he has been carried away by the stream to get to point B. What was the velocity u of his walking if both swimmers reach the destination simultaneously ? The stream velocity v = 2.0 km/hour and the velocity u of each swimmer with respect to water equals 2.5 km per hour.

Answer:
1. Let the river width be d. The first swimmer heads upstream at an angle, so his speed across is √(2.5² - 2.0²) = 1.5 km/h, and his time is d/1.5.
2. The second swimmer swims at right angles to the stream. His time across is d/2.5, and the stream carries him 2.0 × d/2.5 = 0.8d.
3. He then walks 0.8d at speed u_w, taking 0.8d/u_w.
4. Equal times: d/1.5 = d/2.5 + 0.8d/u_w, so 0.667 - 0.4 = 0.8/u_w, that is, 0.267 = 0.8/u_w.
5. Hence, u_w = 3.0 km/h.

Teacher's Note:
a) The question uses u for both the swimming speed and the walking speed; keep them apart in the working.
b) Compare the two times to find the walking speed.

12. Two boats, A and B, move away from a buoy anchored at the middle of a river along the mutually perpendicular straight lines ; the boat A along the river, and the boat B across the river. Having moved off an equal distance from the buoy the boats returned. Find the ratio of times of motion of boats τ_A / τ_B if the velocity of each boat with respect to water is η = 1.2 times greater than the stream velocity.

Answer:
1. Let the stream velocity be v and the speed of each boat in water be ηv. Let each boat go a distance l from the buoy.
2. Boat A: downstream speed (η + 1)v and upstream speed (η - 1)v. τ_A = l/((η + 1)v) + l/((η - 1)v) = 2lη/((η² - 1)v).
3. Boat B heads upstream at an angle so as to cross directly, with the speed across √(η² - 1)v. τ_B = 2l/(√(η² - 1)v).
4. τ_A/τ_B = η/√(η² - 1) = 1.2/√(1.44 - 1) = 1.2/0.663.
5. Hence, τ_A/τ_B = 1.8.

Teacher's Note:
a) Boat B must keep its heading tilted against the stream to stay on a line across the river.
b) The boat moving along the stream always takes longer.

13. Point A moves uniformly with velocity v so that the vector v is continually "aimed" at point B which in its turn moves rectilinearly and uniformly with velocity u < v. At the initial moment of time v ⊥ u and the points are separated by a distance l. How soon will the points converge ?

Answer:
1. Let r be the separation of A and B, X the separation measured along the direction of u, and α the angle between the line AB and the direction of u.
2. A moves along AB at speed v and B moves at speed u along its own line, so dr/dt = u cos α - v and dX/dt = u - v cos α.
3. Multiply the first by v and the second by u and add: d(vr + uX)/dt = u² - v², which is a constant.
4. At t = 0, r = l and X = 0 (because AB is perpendicular to u), so vr + uX = vl. When the points meet, r = 0 and X = 0, so vr + uX = 0.
5. Hence, (0 - vl) = (u² - v²)t, so t = vl/(v² - u²).

Teacher's Note:
a) Look for a combination of r and X whose rate of change is constant.
b) Check: if u = 0, t = l/v, which is the time for A to reach a fixed point B.

14. The distance between two moving particles P and Q at any time is 'a'. If v_r be their relative velocity and if u and v be the components of v_r along and perpendicular to PQ, then show that their closest distance is \( \frac{a v}{vr} \) and that the time that elapses before they arrive at their nearest distance is \( \frac{a u}{vr²} \).

Answer:
1. Work in the frame of P. Then Q moves in a straight line with the constant velocity v_r, starting at distance a from P.
2. Let θ be the angle between v_r and the line PQ. Then u = v_r cos θ and v = v_r sin θ.
3. The closest distance is the perpendicular distance from P to the line of motion of Q: d = a sin θ = a v/v_r.
4. Q travels along its path from the starting point to the foot of this perpendicular, a distance a cos θ = a u/v_r.
5. Time = distance/speed = (a u/v_r)/v_r = a u/v_r².
6. Hence proved: the closest distance is a v/v_r and the time taken is a u/v_r².

Teacher's Note:
a) Going to the frame of one particle reduces the problem to a point moving on a straight line.
b) The nearest approach occurs when the relative velocity is perpendicular to the line joining the particles.

15. An aircraft flies with constant airspeed (speed of aircraft in still air) 200 km/hr from position A to position B, which is 100 km north-east of A and then flies back to A. Throughout the whole flight the wind velocity is 60 km/hr from the west. Find total time of flight from A to B and back.

Answer:
1. The wind from the west blows towards the east at 60 km/h. The line AB is at 45° (north-east), and the ground velocity must be along AB in each direction.
2. The component of the wind perpendicular to AB is 60 sin 45° = 42.4 km/h, and the component along AB is 60 cos 45° = 42.4 km/h.
3. The aircraft cancels the perpendicular component: its heading makes an angle with AB where sin φ = 42.4/200 = 0.212, so cos φ = 0.977.
4. Ground speed from A to B = 200 × 0.977 + 42.4 = 237.9 km/h, so the time = 100/237.9 = 0.420 h = 25.2 min.
5. Ground speed from B to A = 200 × 0.977 - 42.4 = 153.0 km/h, so the time = 100/153.0 = 0.654 h = 39.2 min.
6. Total time = 25.2 + 39.2 = 64.4 min.

Teacher's Note:
a) The printed answer is 64.2 min; the correct value from the data is 64.4 min.
b) The perpendicular part of the wind has to be cancelled by the heading, so it does not help or hinder the motion along AB.
c) The wind helps on the way out and slows the return.

16. A river of width 'a' with straight parallel banks flows due north with speed u. The points O and A are on opposite banks and A is due east of O. Coordinate axes Ox and Oy are taken in the east and north directions respectively. A boat, whose speed is v relative to water, starts from O and crosses the river.
If the boat is steered due east and u varies with x as : u = x (a - x) \( \frac{v}{a²} \). Find:
(a) equation of trajectory of the boat
(b) time taken to cross the river
(c) absolute velocity of boatman when he reaches the opposite bank.
(d) the displacement of boatman when he reaches the oppsite bank from the initial position.

Answer:
1. The boat is steered due east, so dx/dt = v and x = vt. The northward velocity is the river speed, dy/dt = u = x(a - x)v/a².
2. (a) dy/dx = (dy/dt)/(dx/dt) = x(a - x)/a². Integrating, y = x²/(2a) - x³/(3a²).
3. (b) The boat reaches x = a at the time t = a/v.
4. (c) At x = a, u = a(a - a)v/a² = 0, so the river speed is zero there. The absolute velocity is v due east.
5. (d) At x = a, y = a/2 - a/3 = a/6. The displacement is a î + (a/6) ĵ, that is, a due east and a/6 due north.

Teacher's Note:
a) The river speed is zero at both banks and maximum at the middle, so the boat drifts north the most near the centre.
b) Use dy/dx to eliminate time in the trajectory.

17. A bird flies in the xy-plane with a velocity vector given by v = (α - βt²) i + γ t j, with α = 2.1 m/s, β = 2.8 m/s² and where the positive y-direction is vertically upward. At t = 0 the bird is at the origin.
(a) Calculate the position and acceleration vectors of the bird as functions of time.
(b) What is the bird's altitude (y-coordinate) as it flies over x = 0 for the first time after t = 0?

Answer:
1. (a) Integrating the velocity with r = 0 at t = 0: r = (αt - βt³/3) i + (γt²/2) j.
2. Differentiating the velocity: a = -2βt i + γ j.
3. (b) The bird is over x = 0 when αt - βt³/3 = 0, that is, t = 0 or t² = 3α/β. The first time after t = 0 is t = √(3α/β) = √(3 × 2.1/2.8) = 1.5 s.
4. At this time y = γt²/2 = γ × 2.25/2 = 1.125γ, that is, y = 3αγ/(2β).
5. The value of γ is not printed in the question, so the altitude is 3αγ/(2β) = 1.125γ metres, with γ in m/s².

Teacher's Note:
a) The question gives numbers for α and β only, so the altitude is left in terms of γ.
b) The first time over x = 0 is the smallest positive root of x(t) = 0.

18. Two towers AB and CD are situated at distance d apart as shown in figure. AB is 20 m high and CD is 30 m high from the ground. An object of mass m is thrown from the top of AB horizontally with a velocity of 10 m/s towards CD. Simultaneously another object of mass 2m is thrown from the top of CD at an angle 60º to the horizontal towards AB with the same magnitude of initial velocity as that of the first object. The two objects move in the same vertical plane, collide in mid-air and stick to each other.
(a) calculate the distance d between the towers
(b) find the position where the objects hit the ground.

Answer:
1. Take g = 10 m/s². Both objects have the same acceleration g, so in the frame of one object the other moves in a straight line with constant relative velocity.
2. Velocity of the first object: (10, 0). Velocity of the second: (-10 cos 60°, -10 sin 60°) = (-5, -8.66) m/s. Relative velocity of the first with respect to the second = (15, 8.66) m/s.
3. The second object starts 10 m higher and d to the right. They meet when the relative displacement is covered: vertically 8.66t = 10, so t = 1.155 s, and horizontally 15t = d.
4. (a) d = 15t = 15 × 1.155 = 17.32 m.
5. (b) At the collision the first object has velocity (10, -gt) and the second (-5, -8.66 - gt). By momentum conservation, the horizontal velocity of the combined mass = (m × 10 + 2m × (-5))/3m = 0, and the vertical velocity = (-11.55 + 2(-8.66 - 11.55))/3 = -17.3 m/s.
6. The collision takes place at a height 20 - (1/2)(10)(1.155)² = 13.3 m, at a horizontal distance 10 × 1.155 = 11.55 m from AB.
7. The combined mass has no horizontal velocity, so it falls vertically and hits the ground at 11.55 m from AB (the time of fall is 0.65 s).

Teacher's Note:
a) Use g = 10 m/s², as the question does not give g.
b) In the frame of a falling object, the other object moves in a straight line, so the problem reduces to a simple meeting condition.
c) The equal and opposite horizontal momenta cancel, so the combined mass falls straight down.

19. A large heavy box is sliding without friction down a smooth plane of inclination θ. From a point P on the bottom of a box, a particle is projected inside the box. The initial speed of the particle with respect to the box is u and the direction of projection makes an angle α with the bottom as shown in figure.
(a) Find the distance along the bottom of the box between the point of projection P and the point Q where the particle lands. (Assume that the particle does not hit any other surface of the box. Neglect air resistance).
(b) If the horizontal displacement of the particle as seen by an observer on the ground is zero, find the speed of the box with respect to the ground at the instant when the particle was projected.

Answer:
1. (a) The box slides down with acceleration g sin θ along the incline. In the frame of the box the effective acceleration of the particle is g - g sin θ (along the incline) = only the component g cos θ perpendicular to the incline, pointing into the bottom of the box.
2. Component of u along the bottom = u cos α, and perpendicular to the bottom = u sin α.
3. Time of flight T = 2u sin α/(g cos θ).
4. Distance PQ = u cos α × T = 2u² sin α cos α/(g cos θ) = u² sin 2α/(g cos θ).
5. (b) In the ground frame there is no horizontal acceleration of the particle, so its horizontal velocity stays constant. Zero horizontal displacement means zero horizontal velocity.
6. Horizontal component of u (relative to the box) = u cos α cos θ - u sin α sin θ = u cos(α + θ). The horizontal component of the box velocity V is V cos θ in the opposite direction.
7. So u cos(α + θ) - V cos θ = 0, that is, V = u cos(α + θ)/cos θ.

Teacher's Note:
a) In the box frame, only the part of gravity perpendicular to the incline remains effective, because the box itself accelerates down the slope with g sin θ.
b) Horizontally there is no force on the particle, so its horizontal velocity cannot change.

20. An object A is kept fixed at the point x = 3 m and y = 1.25 m on a plank P raised above the ground. At time t = 0 the plank starts moving along the +X direction with an acceleration 1.5 m/s². At the same instant a stone is projected from the origin with a velocity u as shown. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 45º to the horizontal. All the motions are in X-Y plane. Find u and the time after which the stone hits the object. Take g = 10 m/s².

Answer:
1. Let the stone be projected with components u_x and u_y. The object moves along x with a = 1.5 m/s², so its position at time t is x = 3 + 0.75t², y = 1.25.
2. For the stone: x = u_xt and y = u_yt - 5t².
3. The stone is moving downward at 45°, so at the hit v_y = -v_x: u_y - 10t = -u_x, so u_y = 10t - u_x.
4. Putting this in y: (10t - u_x)t - 5t² = 1.25, so 5t² - u_xt = 1.25. Also u_xt = 3 + 0.75t².
5. So 5t² - 3 - 0.75t² = 1.25, which gives 4.25t² = 4.25 and t = 1 s.
6. u_x = 3 + 0.75 = 3.75 m/s and u_y = 10 - 3.75 = 6.25 m/s.
7. Hence, u = (3.75 i + 6.25 j) m/s and the stone hits the object after 1 s.

Teacher's Note:
a) The horizontal velocity of the stone stays constant, so the 45° condition gives v_y = -u_x.
b) The position of the object is only a function of time, which simplifies the matching condition.
c) Check: y = 6.25 - 5 = 1.25 m at t = 1 s.

21. A cart is moving along X-direction with a velocity of 4 m/s. A person on the cart throws a stone with a velocity of 6 m/s relative to himself. In the frame of reference of the cart the stone is thrown in Y-Z plane making an angle of 30º with vertical z-axis. At the highest point of its trajectory, the stone hits an object of equal mass hung vertically from branch of a tree by means of a string of length L. A completely inelastic collision occurs in which the stone gets embedded in the object. Determine :
(a) the speed of the combined mass immediately after the collision with respect to an observer on the ground
(b) the length L of the string such that the tension in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass. (g = 9.8 m/s²)

Answer:
1. (a) In the cart frame the stone's velocity is 6 m/s at 30° to the z-axis: components v_y = 6 sin 30° = 3 m/s and v_z = 6 cos 30° = 5.2 m/s. On the ground the stone also has v_x = 4 m/s.
2. At the highest point v_z = 0, so the stone's velocity just before the collision is (4, 3, 0) m/s, of magnitude 5 m/s.
3. The object is at rest and has equal mass, so by momentum conservation the combined velocity = (4, 3, 0)/2 = (2, 1.5, 0) m/s. Speed = √(2² + 1.5²) = 2.5 m/s.
4. (b) The tension is zero at the horizontal position only if the speed there is zero (at the horizontal position the tension supplies the centripetal force mv²/L, since gravity is perpendicular to the string).
5. By energy conservation, (1/2)(2.5)² = gL, so L = 6.25/(2 × 9.8) = 0.32 m.
6. Hence, the speed is 2.5 m/s and L = 0.32 m.

Teacher's Note:
a) The cart's velocity adds to the stone's horizontal velocity when seen from the ground.
b) At the highest point the vertical velocity of the stone is zero, so the collision is purely horizontal.
c) At the horizontal position gravity has no component along the string, so zero tension means zero speed.

22. Two guns situated on the top of a hill of height 10 m fire one shot each with the same speed 5√3 m/s at some interval of time. One gun fires horizontally and other fires upwards at an angle of 60º with the horizontal. The shots collide in air at point P. Find
(a) the time interval between the firings and
(b) the coordinates of the point P. Take origin of the coordinate system at the foot of the hill right below the muzzle and trajectories in x-y plane. (g = 10 m/s²)

Answer:
1. Let the horizontal shot be in the air for time t₁ and the other shot for time t₂ before they meet. For the horizontal shot: x = 5√3 t₁, y = 10 - 5t₁².
2. For the second shot the speed components are 5√3 cos 60° = 5√3/2 m/s and 5√3 sin 60° = 7.5 m/s: x = (5√3/2)t₂, y = 10 + 7.5t₂ - 5t₂².
3. Equal x: 5√3 t₁ = (5√3/2)t₂, so t₂ = 2t₁.
4. Equal y: -5t₁² = 7.5(2t₁) - 5(4t₁²), so 15t₁² = 15t₁ and t₁ = 1 s, t₂ = 2 s.
5. (a) The time interval between the firings = t₂ - t₁ = 1 s.
6. (b) x = 5√3 × 1 = 5√3 m and y = 10 - 5 = 5 m, so P is at (5√3 m, 5 m).

Teacher's Note:
a) The shot fired upwards must be fired first because it stays in the air longer.
b) Both shots must be at the same place at the same moment, so equate x and y together.

23. On a frictionless horizontal surface, assumed to be the x - y plane, a small trolley A is moving along a straight line parallel to the y-axis (see figure) with a constant velocity of (√3-1) m/s. At a particular instant, when the line OA makes an angle of 45º with the x-axis, a ball is thrown along the surface from the origin O. Its velocity makes an angle φ with the x-axis and it hits the trolley.
(a) The motion of the ball is observed from the frame of the trolley. Calculate the angle θ made by the velocity vector of the ball with the x-axis in this frame.
(b) Find the speed of the ball with respect to the surface if φ = 4θ/3.

Answer:
1. (a) In the frame of the trolley, the trolley is at rest, and the ball must move in a straight line to hit it. At the instant of throwing the trolley lies along OA, so the ball's relative velocity must be directed along OA.
2. OA makes 45° with the x-axis, so θ = 45°.
3. (b) φ = 4θ/3 = 60°. Let the speed of the ball with respect to the surface be v and the trolley velocity V = (√3 - 1) m/s along +y.
4. Relative velocity of the ball = (v cos 60°, v sin 60° - V). It must be at 45° to the x-axis, so v cos 60° = v sin 60° - V.
5. V = v(sin 60° - cos 60°) = v(√3 - 1)/2, so v = 2V/(√3 - 1) = 2 m/s.
6. Hence, θ = 45° and the speed of the ball is 2 m/s.

Teacher's Note:
a) Two bodies in uniform motion meet only if the relative velocity points along the line joining them.
b) The velocity components of the ball are measured relative to the surface; subtract the trolley velocity to go to the trolley frame.

24. A table with smooth horizontal surface is fixed in a cabin that rotates with uniform angular velocity ω in a circular path of radius R (as shown in the figure). A smooth groove AB of length L (<<R) is made on the surface of table. The groove makes an angle θ with radius OA of the circle in which the cabin rotates. A small particle is kept at the point A in the groove and is released to move along AB. Find the time taken by the particle to reach the point B.

Answer:
1. In the frame of the cabin, the particle experiences the centrifugal force mω²R directed away from O. Since L is much less than R, this force is almost constant along the groove.
2. The component of this force along the groove AB = mω²R cos θ, since the groove makes an angle θ with the radius OA.
3. The groove is smooth, so the acceleration along the groove is a = ω²R cos θ.
4. Starting from rest, L = (1/2)at² = (1/2)ω²R cos θ × t².
5. Hence, t = \( \sqrt{\frac{2L}{\omega^2 R \cos\theta}} \).

Teacher's Note:
a) Always add the pseudo force when you use a rotating or accelerating frame.
b) Only the component of the pseudo force along the groove accelerates the particle; the smooth groove cancels the rest.

25. A ball is placed on a smooth chute inclined at an angle α = 30° to the horizontal and rotating at frequency n = 30 rmp about vertical axis passing through its lower end. At what distance from the end of the chute does the ball rest.

Answer:
1. Work in the frame of the chute. At distance L along the chute from the lower end O, the ball is at a distance r = L cos α from the vertical axis.
2. The centrifugal force is mω²r = mω²L cos α, directed horizontally outward. Its component along the chute, upward, is mω²L cos²α.
3. The component of gravity along the chute, downward, is mg sin α.
4. For rest, mω²L cos²α = mg sin α, so L = g sin α/(ω² cos²α).
5. ω = 2πn = 2π × 30/60 = π rad/s. L = 9.8 × 0.5/(π² × 0.75) = 0.66 m, which is about 0.67 m if g = π² is taken.
6. Hence, the ball rests at about 0.67 m from the lower end of the chute.

Teacher's Note:
a) The chute is smooth, so only the components of the forces along the chute decide equilibrium.
b) The ball is at a horizontal distance L cos α from the axis, not L.
c) The printed value 0.67 m comes from using g = 10 m/s² and π² = 10; with g = 9.8 m/s² the result is 0.66 m.