Exams › JEE Main › Physics › Laws of Motion
306 questions with worked solutions.
Q1. Which of the following has the same SI unit as impulse?
Answer: Momentum
Impulse equals change in momentum, with SI unit N*s = kg*m/s, which is exactly the unit of momentum. Energy (J), power (W) and velocity (m/s) have different units.
Answer: g/4
For the sliding monkey, the branch tension is T = m(g - a). The branch holds if T <= 0.75 mg, i.e. m(g - a) <= 0.75 mg -> a >= 0.25 g. The least safe acceleration is g/4.
Answer: 6 s
The car's deceleration can be calculated using Newton's second law, where the net force (friction) divided by mass gives the acceleration. With a friction force of 5000 N and a mass of 1000 kg, the deceleration is 5 m/s². Using the formula for time to stop (time = initial velocity / deceleration), we find that it takes 6 seconds for the car to come to rest.
Answer: 2.5
Thrust force = (exhaust speed)*(mass rate) = 50*0.1 = 5 N. Acceleration = thrust/mass = 5/2 = 2.5 m/s^2.
Answer: Mg ≤ F ≤ Mg√(1+μ²)
The floor's contact force combines normal N = Mg and friction f (0 to mu*Mg at impending motion): F = sqrt((Mg)^2 + f^2). Minimum (no pull) is Mg; maximum (impending slip) is sqrt((Mg)^2 + (mu Mg)^2) = Mg*sqrt(1+mu^2). So Mg <= F <= Mg*sqrt(1+mu^2).
Answer: Motion in a straight line while momentum remains constant
With no net force the body has zero acceleration, so its velocity and hence momentum stay constant. Straight-line motion with constant (unchanging) momentum is the only option consistent with no force acting.
Answer: m1m2(1 + μk)g/(m1 + m2)
System acceleration a=(m2 - mu_k m1)g/(m1+m2). For the hanging block, m2 g - T = m2 a, so T = m2(g-a) = m1 m2 (1+mu_k) g/(m1+m2).
Answer: Each balance shows a reading of M kg.
Both balances measure the weight of the block, which is the same for each balance since they are in a direct line of support. Therefore, each balance will show a reading equal to the mass of the block, M kg.
Answer: It is moving upward or downward with constant speed
If tension equals weight, net force is zero, so acceleration is zero. Zero acceleration means the elevator moves with constant velocity (constant speed) up or down, or stays at rest.
Answer: 1: 4
At top v^2=3gr: T_top = mv^2/r - mg = 3mg - mg = 2mg. At bottom v_b^2 = 3gr + 2g(2r) = 7gr: T_bottom = mv_b^2/r + mg = 7mg + mg = 8mg. Ratio = 2mg:8mg = 1:4.
Answer: 1/6 m
Slipping is on the verge when dy/dx = mu. Here dy/dx = x^2/2 = 0.5 gives x = 1, so the maximum height is y = x^3/6 = 1/6 m.
Answer: 27 m s⁻¹
Impulse on wall = 2 m v = 0.54 N.s with m=0.010 kg, so v = 0.54/(2x0.010) = 27 m/s.
Answer: 0.9 Ns
The average impulse is calculated by integrating the force over the time the bullet is in the barrel. Given the force function, the impulse can be found by evaluating the area under the force-time graph, which results in 0.9 Ns.
Answer: 15 m/s downward
Using g = 9.8, after 5 s the body's velocity is 100 - 9.8(5) = 51 m/s upward, so momentum = 50(51) = 2550 kg m/s up. Conservation: 20(150) + 30 v2 = 2550, giving v2 = (2550 - 3000)/30 = -15 m/s, i.e. 15 m/s downward.
Answer: the mass of the cube
Sliding starts when F = mu mg; overturning starts when F h = mg(a/2), i.e. F = mg a/(2h). Which occurs first depends on comparing mu with a/(2h), so it depends on the side length a, the height h of the force, and mu, but the mass m cancels out. Hence it does NOT depend on the mass of the cube.
Answer: F1/m
When the particle is in equilibrium, the net force acting on it is zero, meaning the forces balance each other out. If F1 is removed, the remaining forces F2 and F3 will no longer be balanced, resulting in an acceleration equal to the net force (which is F1) divided by the mass (m), hence the acceleration becomes F1/m.
Answer: all of them have the same acceleration
All objects slide down the incline under the influence of gravity, and since there is no rolling involved, their acceleration is determined solely by the gravitational force acting on them, which is the same for all shapes regardless of their mass distribution.
Answer: 13 N and 5 N
The two forces must satisfy both the condition of their magnitudes summing to 18 N and the Pythagorean theorem, as the resultant is at right angles to the smaller force. The only pair that meets these criteria is 13 N and 5 N, where 13 N is the larger force and 5 N is the smaller force, resulting in a correct resultant of 12 N.
Answer: 9: 7
The acceleration of the system is determined by the difference in weights of the two masses. Using Newton's second law, the ratio of the masses can be derived from the relationship between the gravitational force and the acceleration, leading to the conclusion that m1: m2 = 9: 7.
Answer: 0.06
Deceleration a = v/t = 6/10 = 0.6 m/s^2. Friction provides this: mu = a/g = 0.6/10 = 0.06 (about 0.06 even with g = 9.8).
Answer: PM/(M + m)
The force transmitted by the rope to the block is determined by the system's total mass and the applied force. Since both the block and the rope are in motion, the effective mass that the applied force P acts upon is the sum of the masses M and m, leading to the formula P multiplied by the mass of the block divided by the total mass.
Answer: Each balance shows M kg.
The whole weight Mg is transmitted through both spring balances in the series chain (each is light, so it adds no weight). Tension is Mg everywhere, so each balance reads M kg.
Answer: 7.0 × 10⁵ N
Thrust - mg = ma, so Thrust = m(g+a) = 3.5e4 * (10+10) = 7.0 x 10^5 N.
Answer: 2.0
To find the mass of the block, we can use the formula for static friction, which is the product of the coefficient of static friction and the normal force. The normal force on the slope can be calculated using the weight of the block, which is equal to mass times gravity. Given that the frictional force is 10 N and the coefficient of static friction is 0.8, we can set up the equation 10 N = 0.8 * (mass * g * cos(30°)). Solving this yields a mass of 2.0 kg.
Answer: μₖ = 1 - 1/n²
Smooth: a1 = g sin45. Rough: a2 = g(sin45 - mu cos45). Same distance d=0.5*a*t^2 with t_rough = n*t_smooth gives a2 = a1/n^2, so (sin45 - mu cos45) = sin45/n^2. Since sin45=cos45, 1 - mu = 1/n^2, giving mu_k = 1 - 1/n^2.
Answer: 3
The ratio F₁/F₂ can be derived from the forces acting on the body on the inclined plane. Since F₁ must overcome both the gravitational component pulling the body down and the friction opposing the upward movement, while F₂ must counteract the gravitational pull and the friction acting downwards, the relationship simplifies to F₁ being three times F₂ when tan θ = 2μ.
Answer: 1/6 m
The correct option is 1/6 m because at this height, the gravitational force acting on the block is balanced by the maximum static friction force, which is determined by the coefficient of friction and the normal force. As the height increases, the angle of the surface increases, leading to a greater component of gravitational force parallel to the surface, which can cause the block to slide.
Answer: 1/6 m
The maximum height at which the block can be placed without slipping is determined by balancing the gravitational force and the frictional force. Given the surface's slope and the coefficient of friction, the calculations show that the maximum height is 1/6 m, as this is the point where the frictional force is sufficient to counteract the component of gravitational force acting parallel to the surface.
Answer: 20N
To calculate the external force required to move the body up the incline with the same acceleration, we need to consider both the gravitational force acting down the slope and the force needed to achieve the desired acceleration. The total force required is the sum of the gravitational component along the incline and the force needed for acceleration, which results in 20N.
Answer: 0.6
omega = 3.5 rev/s = 7*pi = 21.99 rad/s. The coin stays put when friction supplies the centripetal force: mu*g >= omega^2*r, so mu = omega^2*r/g = (21.99^2 * 0.0125)/10 = 0.60.
Answer: 1 - 1/n²
The time taken to slide down the rough incline compared to the smooth incline is influenced by the frictional force, which reduces the acceleration. The relationship derived from the equations of motion and the forces acting on the object leads to the conclusion that the coefficient of kinetic friction is expressed as 1 - 1/n², indicating how friction affects the sliding time.
Answer: 2√(p/k)
dp/dt = kt, so integral from 0 to T gives kT^2/2 = 3p - p = 2p. Thus T^2 = 4p/k and T = 2 sqrt(p/k).
Answer: 120°
R^2 = 13F^2 + 12F^2 cos(theta). Doubling Q to 6F: R'^2 = 40F^2 + 24F^2 cos(theta). Requiring R' = 2R gives 40 + 24cos = 52 + 48cos, so cos(theta) = -1/2 and theta = 120 deg.
Answer: 100 N
Equilibrium of the mass-point on the rope: T cos45 = mg and T sin45 = F. Dividing gives F = mg = 10*10 = 100 N.
Answer: 0.20 m s⁻¹
Let man speed = v (backward), son speed = u (forward). Momentum: 20u = 50v -> u = 2.5v. Relative speed u + v = 0.70 -> 3.5v = 0.70 -> v = 0.20 m/s.
Answer: 9.859 × 10⁻⁴ N
The normal force exerted by the side walls of the groove is determined by the centripetal force required to keep the block moving in a circular path. Given the mass, radius, and time for one complete revolution, the calculation shows that the normal force is 9.859 × 10⁻⁴ N, which corresponds to the necessary centripetal force for the block's motion.
Answer: 5
The least force to start motion is applied at angle theta = arctan(mu): F_min = mu m g / sqrt(1+mu^2). With mu = 1/sqrt(3), m = 1, g = 10: F_min = (10/sqrt(3))/sqrt(4/3) = 5 N.
Answer: 3
The relationship between the time of ascent and descent indicates that the forces acting on the body, including friction, must balance in such a way that the acceleration during ascent is half that of descent. Given the angle and the coefficient of friction, solving the equations of motion leads to the conclusion that x must equal 3 to satisfy these conditions.
Answer: 0.2 kg m s⁻¹, 0.05 m s⁻¹
The impulse imparted to the gun is equal to the momentum of the bullet, which is calculated as the product of its mass and velocity (0.004 kg * 50 m/s = 0.2 kg m/s). The recoil velocity of the gun can be found using the conservation of momentum, resulting in a velocity of 0.05 m/s when considering the mass of the gun.
Answer: 25
To keep the block adhered to the wall, the applied horizontal force must counteract the force of gravity acting on the block. The weight of the block is 0.5 kg multiplied by the acceleration due to gravity (10 m/s²), which equals 5 N. The maximum static friction force is the coefficient of static friction (0.2) multiplied by the normal force (the applied horizontal force). Setting the maximum friction equal to the weight gives the equation: 0.2 * F = 5 N, leading to F = 25 N.
Answer: (400 î + 100 ĵ) m
a = F/m = (8,2) m/s^2. From rest, r = (1/2)a t^2 = 50*(8,2) = (400,100) m.
Answer: 60 kg s⁻¹
The correct option is right because it applies the rocket equation, which relates thrust, mass flow rate, and exhaust velocity. To achieve the desired acceleration, the mass flow rate of the fuel must be 60 kg/s, ensuring that the thrust produced is sufficient to overcome both the gravitational force and provide the necessary upward acceleration.
Answer: 500
a_x = 20/2 = 10 m/s^2. Starting from rest, x = (1/2)(10)(10)^2 = 500 m.
Answer: 4F0T/3M
The correct option is derived from integrating the force over the time interval to find the impulse, which gives the change in momentum. The force function is a quadratic function of time, and when integrated from 0 to 2T, it results in a total impulse that leads to a final velocity of 4F0T/3M.
Answer: Statement - I is false but Statement - II is true
Three forces represented by the sides of a triangle taken in order add to zero, giving translatory equilibrium, so Statement-II is true. But such forces act along different lines and are not concurrent, so Statement-I is false. Hence Statement-I is false but Statement-II is true.
Answer: 9.8 m
The block comes to rest due to the work done against friction, which can be calculated using the initial kinetic energy and the frictional force. The distance covered is determined by equating the initial kinetic energy to the work done by friction, leading to a distance of 9.8 m.
Answer: Rope will break while climbing upward
When the monkey climbs upward with an acceleration of 5 m/s², the tension in the rope increases due to both the weight of the monkey and the additional force from the upward acceleration, resulting in a total tension of 700 N, which exceeds the rope's maximum capacity of 350 N, causing it to break.
Answer: 0.036 s
The time of contact can be calculated using the impulse-momentum theorem, which states that the impulse (force multiplied by time) equals the change in momentum. Here, the change in momentum is twice the initial momentum of the ball (since it reverses direction), and dividing this by the force gives the time of contact.
Answer: When the elevator moves downward with constant acceleration
When the elevator moves downward with constant acceleration, the net force acting on the person decreases, resulting in a sensation of reduced weight or weight loss.
Answer: Mg
The contact force is the resultant of the normal reaction and friction. Since the block moves at constant velocity, all forces balance, so the contact force equals and opposes the weight: magnitude = Mg.