RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Physics · 10 questions · 24 marks
Kinematics tells you how a body moves; Newton's laws tell you why. This chapter introduces inertia, the momentum form of the second law, and the action-reaction pair that students most often mishandle, then puts them to work on free-body diagrams for pulleys, lifts and blocks in contact. Friction and banked roads round it off, showing how everyday constraints appear as forces in the equations.
A cricket player draws his hands backwards while catching a fast ball. This action mainly:
Answer
Increases the time of contact and so reduces the average force on his hands — the ball must be brought to rest, so the change in momentum Δp is fixed. Since F = Δp/Δt, stretching Δt by moving the hands backwards lowers the average force, which is what saves the fielder's palms.
A 5 kg block rests on a horizontal floor with μ_s = 0.4. A horizontal push of 15 N is applied but the block does not move (g = 10 m/s²). What is the frictional force acting on it?
Answer
15 N — the limiting static friction is μ_s N = 0.4 × 5 × 10 = 20 N, which exceeds the applied 15 N, so the block stays at rest. Static friction is self-adjusting and takes exactly the value needed for equilibrium, which here is 15 N opposing the push, not its maximum value of 20 N.
A person of mass 60 kg stands on a weighing machine inside a lift accelerating downward at 2 m/s² (g = 10 m/s²). What reading does the machine show?
Answer
480 N — the machine reads the normal reaction N. For downward acceleration, mg − N = ma, so N = m(g − a) = 60(10 − 2) = 60 × 8 = 480 N. The person feels lighter than the true weight of 600 N, though their mass is unchanged.
A force of 12 N acts for 0.25 s on a body initially at rest, giving it a momentum of 3 kg m/s. If the same force acts for 0.5 s instead, the final momentum will be:
Answer
6 kg m/s — impulse equals change in momentum, J = FΔt. Doubling the time from 0.25 s to 0.5 s doubles the impulse from 12 × 0.25 = 3 N s to 12 × 0.5 = 6 N s, and since the body starts from rest the final momentum equals the impulse, i.e. 6 kg m/s.
Assertion (A): A horse pulling a cart can make it move even though the cart pulls back on the horse with an equal and opposite force. Reason (R): The action and reaction of Newton's third law act on two different bodies and therefore never cancel each other.
Answer
Both A and R are true and R is the correct explanation of A — the horse's pull acts on the cart and the cart's pull acts on the horse, so when you draw a free-body diagram of the cart alone only the forward pull and friction appear. The cart accelerates whenever that pull exceeds the friction on it, and the reaction force, acting on a different body, cannot cancel it.
A bullet of mass 20 g is fired from a 4 kg rifle with a muzzle speed of 400 m/s. Find the recoil speed of the rifle.
Answer
Momentum is conserved because no external horizontal force acts during firing, and the system is initially at rest. 0 = m_bullet v_bullet + m_rifle v_rifle 0 = (0.020)(400) + (4)v_rifle 0 = 8 + 4v_rifle, so v_rifle = −2 m/s. The rifle recoils at 2 m/s, the negative sign showing it moves opposite to the bullet.
Blocks of mass 3 kg and 2 kg lie in contact on a frictionless horizontal surface. A horizontal force of 25 N is applied to the 3 kg block, pushing both. Find the common acceleration and the contact force between the blocks.
Answer
Treating the two blocks as one system of mass 3 + 2 = 5 kg: a = F/m_total = 25/5 = 5 m/s². For the 2 kg block alone, the only horizontal force is the contact push N from the 3 kg block: N = m₂a = 2 × 5 = 10 N. Check with the 3 kg block: 25 − N = 3 × 5 = 15, so N = 10 N, which agrees. The contact force is 10 N.
Two masses m₁ = 6 kg and m₂ = 4 kg are connected by a light inextensible string passing over a frictionless pulley. Derive expressions for the acceleration of the system and the tension in the string, and evaluate both for these masses with g = 10 m/s².
Answer
Let m₁ > m₂, so m₁ descends and m₂ rises with the same magnitude of acceleration a, because the string is inextensible. The string is light and the pulley frictionless, so the tension T is the same throughout. For m₁ (taking downward as positive for it): m₁g − T = m₁a ... (i) For m₂ (taking upward as positive for it): T − m₂g = m₂a ... (ii) Adding (i) and (ii) eliminates T: (m₁ − m₂)g = (m₁ + m₂)a, so a = (m₁ − m₂)g/(m₁ + m₂). Substituting this back into (ii): T = m₂(g + a) = m₂g + m₂(m₁ − m₂)g/(m₁ + m₂) = g[m₂(m₁ + m₂) + m₂(m₁ − m₂)]/(m₁ + m₂), which simplifies to T = 2m₁m₂g/(m₁ + m₂). Numerically: a = (6 − 4)(10)/(6 + 4) = 20/10 = 2 m/s². T = 2(6)(4)(10)/10 = 480/10 = 48 N. Sanity check: T = 48 N lies between m₂g = 40 N and m₁g = 60 N, as it must, since the lighter block accelerates upward and the heavier one downward.
(a) Define coefficient of friction and angle of repose, and derive the relation between them. (b) A 10 kg crate is pushed along a rough floor with μ_k = 0.3 by a horizontal force of 50 N (g = 10 m/s²). Find the acceleration of the crate and the distance it travels in 4 s from rest.
Answer
(a) The coefficient of friction μ is the ratio of the limiting frictional force to the normal reaction between two surfaces: μ = f/N. It is a dimensionless number depending on the nature and roughness of the surfaces, not on the area of contact. The angle of repose is the maximum angle of an inclined plane at which a body placed on it just remains on the point of sliding down. Derivation: consider a body of mass m resting on an incline of angle θ. Resolving the weight, the component along the incline is mg sin θ and perpendicular to it is mg cos θ. The normal reaction is N = mg cos θ. At the angle of repose the body is on the verge of sliding, so friction has reached its limiting value: f = μ_s N = μ_s mg cos θ, and it exactly balances the component down the plane: mg sin θ = μ_s mg cos θ Cancelling mg: tan θ = μ_s. So the tangent of the angle of repose equals the coefficient of static friction. (b) Normal reaction N = mg = 10 × 10 = 100 N (the applied force is horizontal, so it does not change N). Kinetic friction f_k = μ_k N = 0.3 × 100 = 30 N, opposing the motion. Net force = 50 − 30 = 20 N, so a = 20/10 = 2 m/s². Distance in 4 s from rest: s = ut + ½at² = 0 + ½(2)(16) = 16 m.
Read the following and answer the questions that follow: A highway engineer is designing a curved exit ramp of radius 80 m. To help vehicles turn safely she banks the road at an angle θ. The design speed for the ramp is 20 m/s, and the tyres on a wet surface give a coefficient of friction of only 0.2 (take g = 10 m/s²). (a) If the ramp were flat rather than banked, what maximum speed could a vehicle negotiate on the wet surface? (b) Find the banking angle at which a vehicle could take the curve at exactly 20 m/s even with no friction at all. (c) Explain physically what supplies the centripetal force on a frictionless banked road. (d) Why does the safe banking angle not depend on the mass of the vehicle?
Answer
(a) On a flat road, friction alone supplies the centripetal force: μmg = mv²/r, so v_max = √(μrg) = √(0.2 × 80 × 10) = √160 ≈ 12.6 m/s. This is well short of the 20 m/s design speed, which is precisely why banking is needed. (b) For a frictionless banked road, v = √(rg tan θ), so tan θ = v²/(rg) = (20)²/(80 × 10) = 400/800 = 0.5. Hence θ = tan⁻¹(0.5) ≈ 26.6°. (c) On a frictionless banked road the only forces are the vehicle's weight mg vertically down and the normal reaction N perpendicular to the road surface. Because the surface is tilted, N has a horizontal component N sin θ pointing towards the centre of the curve, and it is this component that provides the centripetal force. The vertical component N cos θ simultaneously balances the weight mg. (d) Writing the two equations, N sin θ = mv²/r and N cos θ = mg, and dividing the first by the second gives tan θ = v²/(rg). The mass m appears on both sides and cancels, so the required banking angle depends only on the design speed and the radius of the curve. A loaded truck and a small car therefore need the same bank angle, which is what makes a single road design workable for all traffic.
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