Question 1

Q1: 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...

Accepted Answer

According to Newton's first law of motion, if a body is at rest or moving with a constant velocity, its acceleration is zero, and therefore the net external force acting on it is zero. (a) The rain drop is falling with a constant speed, meaning its velocity is constant. Therefore, its acceleration i...

Question 2

Q10: A body of mass 0.40 kg moving initially with a constant speed of 10 m s⁻¹ to the north is subject to a constant force of 8.0 N directed towards the so...

Accepted Answer

Given: Mass, $m = 0.40 	ext{ kg}$ Initial velocity (north), $u = +10 	ext{ m/s}$ (Let's take north as the positive direction) Force (south), $F = -8.0 	ext{ N}$ Duration of force, from $t=0$ to $t=30 	ext{ s}$ Position at $t=0$ is $x=0$. Step 1: Analyze motion for t  30 s First, find the velocit...

Question 3

Q11: 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 hig...

Accepted Answer

Given: Initial velocity of truck, $u_{truck} = 0$ Acceleration of truck, $a_{truck} = 2.0 	ext{ m/s}^2$ Height of truck, $h = 6 	ext{ m}$ Time of drop, $t_{drop} = 10 	ext{ s}$ Time of observation, $t_{obs} = 11 	ext{ s}$ Step 1: Find the velocity of the truck at t = 10 s When the stone is dropp...

Question 4

Q12: 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⁻...

Accepted Answer

(a) At one of its extreme positions: At an extreme position of its oscillation, the bob is momentarily at rest. Its velocity is zero. If the string is cut at this instant, the bob has zero initial velocity. The only force acting on it will be gravity. Therefore, the bob will fall vertically downward...

Question 5

Q13: 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⁻¹, (b) downwards with a uniform ac...

Accepted Answer

A weighing scale measures the normal reaction force ($R$) exerted on the person. The reading is this force divided by $g$. Let's call the reading $M_{apparent} = R/g$. Given: Mass of the man, $m = 70 	ext{ kg}$ Acceleration due to gravity, $g = 10 	ext{ m/s}^2$ Weight of the man, $W = mg = 70 	im...

Question 6

Q14: 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  4 s, 0 < t < 4 s? (b) impulse at t...

Accepted Answer

Given: Mass of the particle, $m = 4 	ext{ kg}$ The graph is a position-time graph. (a) Force on the particle Force is given by $F = ma$. Acceleration is the second derivative of position with respect to time, and velocity is the first derivative (slope of the x-t graph).    For t  4 s: The position...

Question 7

Q15: 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 =...

Accepted Answer

Given: Mass A, $m_A = 10 	ext{ kg}$ Mass B, $m_B = 20 	ext{ kg}$ Applied force, $F = 600 	ext{ N}$ Total mass, $M = m_A + m_B = 30 	ext{ kg}$ First, find the acceleration of the system, which is the same in both cases. $$a = \frac{F}{M} = \frac{600 	ext{ N}}{30 	ext{ kg}} = 20 	ext{ m/s}^2$$...

Question 8

Q16: 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...

Accepted Answer

This is an Atwood machine setup. Given: Mass 1, $m_1 = 8 	ext{ kg}$ Mass 2, $m_2 = 12 	ext{ kg}$ Acceleration due to gravity, $g = 10 	ext{ m/s}^2$ Since $m_2 > m_1$, the 12 kg mass will move downwards and the 8 kg mass will move upwards with the same acceleration, $a$. To Find: Acceleration of t...

Question 9

Q17: 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 di...

Accepted Answer

Let the initial nucleus be at rest. Its initial momentum is zero. $$p_{initial} = 0$$ Let the nucleus disintegrate into two smaller nuclei with masses $m_1$ and $m_2$. Let their velocities after disintegration be $v_1$ and $v_2$ respectively. The final momentum of the system is the vector sum of the...

Question 10

Q18: Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s⁻¹ collide and rebound with the same speed. What is the impulse...

Accepted Answer

Given: Mass of each ball, $m = 0.05 	ext{ kg}$ Initial speed of each ball, $u = 6 	ext{ m/s}$ Final speed of each ball, $v = 6 	ext{ m/s}$ Let's consider ball 1. Let its initial direction of motion be positive. Initial velocity of ball 1, $u_1 = +6 	ext{ m/s}$. After collision, it rebounds, so i...

Question 11

Q19: 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⁻¹, what is the recoil speed of the gun ?

Accepted Answer

Given: Mass of the shell, $m_s = 0.020 	ext{ kg}$ Mass of the gun, $m_g = 100 	ext{ kg}$ Muzzle speed of the shell, $v_s = 80 	ext{ m/s}$ Initially, the gun and shell are at rest, so the total initial momentum is zero. To Find: Recoil speed of the gun, $v_g$ Formula: According to the law of conse...

Question 12

Q2: 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,...

Accepted Answer

The problem states to ignore air resistance. The only external force acting on the pebble in all cases is the force of gravity, which is its weight ($W = mg$) acting vertically downwards. Given: Mass of the pebble, $m = 0.05 	ext{ kg}$ Acceleration due to gravity, $g = 10 	ext{ m/s}^2$ Calculation...

Question 13

Q20: A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball ?...

Accepted Answer

Given: Mass of the ball, $m = 0.15 	ext{ kg}$ Initial speed, $u = 54 	ext{ km/h}$ Final speed, $v = 54 	ext{ km/h}$ Angle of deflection, $	heta = 45^\circ$ First, convert speed to m/s: $u = 54 	imes \frac{1000}{3600} = 15 	ext{ m/s}$. So, $|\vec{u}| = |\vec{v}| = 15 	ext{ m/s}$. Let's set up...

Question 14

Q21: 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. Wh...

Accepted Answer

Given: Mass of stone, $m = 0.25 	ext{ kg}$ Radius of circle, $r = 1.5 	ext{ m}$ Angular speed, $\omega = 40 	ext{ rev/min}$ Maximum tension, $T_{max} = 200 	ext{ N}$ Part 1: Tension in the string First, convert angular speed to SI units (rad/s). $$\omega = 40 \frac{	ext{rev}}{	ext{min}} = 40 \...

Question 15

Q22: 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...

Accepted Answer

According to Newton's first law of motion (the law of inertia), a body in motion continues to move in a straight line with constant velocity unless acted upon by an external force.  When the stone is being whirled, the tension in the string provides the centripetal force that constantly changes the...

Question 16

Q23: 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 sudden...

Accepted Answer

(a) A horse cannot pull a cart and run in empty space: To move forward, the horse must push against something. On the ground, the horse pushes the ground backward with its hooves. According to Newton's third law, the ground exerts an equal and opposite force (the reaction force) on the horse in the...

Question 17

Q3: 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...

Accepted Answer

In all cases where the stone is dropped, we neglect air resistance. The only force acting on it is gravity. Given: Mass of the stone, $m = 0.1 	ext{ kg}$ Acceleration due to gravity, $g = 10 	ext{ m/s}^2$ (a) Dropped from a stationary train: The only force acting on the stone is gravity.  Force $F...

Question 18

Q4: 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 mo...

Accepted Answer

The particle is moving in a circle on a smooth horizontal table. The forces acting on the particle are: 1. Its weight, $mg$, acting vertically downwards. 2. The normal reaction from the table, $N$, acting vertically upwards. 3. The tension in the string, $T$, acting horizontally towards the center o...

Question 19

Q5: 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⁻¹. How long does the body take to stop?

Accepted Answer

Given: Retarding force, $F = -50 	ext{ N}$ (negative sign indicates retardation) Mass of the body, $m = 20 	ext{ kg}$ Initial speed, $u = 15 	ext{ m/s}$ Final speed, $v = 0 	ext{ m/s}$ (since the body stops) To Find: Time taken to stop, $t$ Formula: First, we find the acceleration using Newton's...

Question 20

Q6: A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s⁻¹ to 3.5 m s⁻¹ in 25 s. The direction of the motion of the body remain...

Accepted Answer

Given: Mass of the body, $m = 3.0 	ext{ kg}$ Initial speed, $u = 2.0 	ext{ m/s}$ Final speed, $v = 3.5 	ext{ m/s}$ Time interval, $t = 25 	ext{ s}$ To Find: Magnitude and direction of the force, $F$ Formula: First, we find the acceleration using the first equation of motion: $$v = u + at$$ Then,...

NCERT Solutions