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The seven SI base units are metre, kilogram, second, ampere, kelvin, mole and candela. Plane angle and solid angle are treated separately as dimensionless supplementary (derived-trigonometric) units, with the radian the SI unit of plane angle. Hence the radian is not one of the seven base units.
Rearranging gives G = Fr^2/(m1m2). Force has dimensions [M L T^-2], so the numerator is [M L^3 T^-2] and the denominator is [M^2]. Dividing yields [M^-1 L^3 T^-2], a result you can check against the known value 6.67 x 10^-11 N m^2 kg^-2.
Vernier reading = main scale reading + (vernier coincidence x least count) = 2.4 cm + (6 x 0.01 cm) = 2.46 cm. The vernier correction is always added to the main scale reading when the vernier zero lies ahead of the main scale zero. So the correct length is 2.46 cm.
Write [T] = [L]^a [L T^-2]^b = L^(a+b) T^(-2b). Matching powers gives -2b = 1, so b = -1/2, and a + b = 0, so a = 1/2. Hence T = k sqrt(l/g), consistent with the known pendulum formula.
For a power relation y = x^n, the relative error is n times the relative error in x. Here the relative error in x is 0.2/4.0 = 0.05, i.e. 5%, so the relative error in x^3 is 3 x 5% = 15%. Percentage errors in a power are multiplied by the exponent.
Least count = pitch / number of circular divisions = 0.5/50 = 0.01 mm. With the zero below the reference line, the instrument reads too much, so the zero error is negative. Its magnitude is (50 - 48) x 0.01 mm = 0.02 mm, giving a zero error of -0.02 mm.
Velocity is the first derivative of position, v = dx/dt = 8t − 2, and acceleration is the derivative of velocity, a = dv/dt = 8 m/s². The acceleration is constant, so its magnitude at t = 3 s is also 8 m/s². Substituting t into the velocity expression gives speed, not acceleration.
For motion from rest, s = ½at², so distance scales as the square of time. Covering 2 m in 1 s fixes ½a = 2, giving a = 4 m/s². Then s = ½(4)(3²) = 18 m, which is nine times the 1-second distance.
Distance measures the total length of path actually traversed and carries no direction, so it is a scalar. Displacement is the straight-line vector from the starting point to the finishing point, depending only on those two positions. The two are equal in magnitude only for straight-line motion without reversal.
With air resistance neglected, the only force is the weight, which acts vertically. There is no horizontal force, so the horizontal velocity component u cos θ stays constant. The vertical component changes at rate g, and the speed and direction change continuously along the parabolic path.
For uniform acceleration the average velocity is the mean of initial and final velocities, (20 + 0)/2 = 10 m/s. Multiplying by the time gives s = 10 × 5 = 50 m. Equivalently, the deceleration is 4 m/s² and s = ut − ½at² = 100 − 50 = 50 m.
For free fall from rest, h = ½gt². Substituting g = 10 m/s² and t = 3 s gives h = ½ × 10 × 9 = 45 m. The impact speed would be gt = 30 m/s, which is a speed and not the height.
Range for level ground is R = u² sin 2θ / g, which depends on the launch angle only through sin 2θ. For 30° and 60° the doubled angles are 60° and 120°, and sin 60° = sin 120°, so the ranges are equal. The complementary angles give equal ranges for the same speed.
The velocity changes sign at t = 2 s, so distance is the sum of the magnitudes of the two triangular areas. Each triangle has base 2 s and height 10 m/s, giving area 10 m, so the total distance is 20 m. The net displacement, by contrast, is zero because the areas cancel.
Uniform circular motion has centripetal acceleration of magnitude v²/r, directed towards the centre. Here v²/r = 36/2 = 18 m/s². The speed is constant, so there is no tangential acceleration component.
The x-motion is uniform, giving x = 3 × 2 = 6 m. The y-motion starts from rest, giving y = ½ × 2 × 2² = 4 m. The displacement magnitude is √(6² + 4²) = √52 ≈ 7.2 m, so among the choices the value closest to and consistent with this is about 7 m; the correct magnitude is √52 m.
The time to reach the top is half the total flight time, 2 s, where the velocity becomes zero. Using v² = u² − 2gh with v = 0 and u = gt = 20 m/s gives h = u²/(2g) = 400/20 = 20 m. Alternatively h = ½gt² with t = 2 s gives the same result.
Reversal occurs where the velocity passes through zero and changes sign. Setting 6t − 3t² = 0 gives 3t(2 − t) = 0, so v = 0 at t = 0 and t = 2 s. Only at t = 2 s does the velocity change sign, since v is positive between 0 and 2 s and negative afterwards.
From rest under uniform acceleration, distance is proportional to t². In the first 4 s the total distance is four times that covered in the first 2 s, i.e. 4s. The distance covered between 2 s and 4 s is therefore 4s − s = 3s.
The crossing time depends only on the component of velocity perpendicular to the banks: t = 100/5 = 20 s. During this time the current carries the boat downstream by 3 × 20 = 60 m. The resultant speed over the ground is √(5² + 3²) = √34 m/s, but the drift depends only on the current and the crossing time.
The vertical motion is free fall from rest, so 80 = ½ × 10 × t² gives t² = 16 and t = 4 s. The horizontal velocity stays constant at 30 m/s, so the horizontal range is 30 × 4 = 120 m. The initial vertical velocity is zero because the launch is horizontal.
At t = 3 s the first particle has velocity (4, 0) m/s and the second has velocity (0, 2×3) = (0, 6) m/s. Their relative velocity is (0 − 4, 6 − 0) = (−4, 6) m/s, whose magnitude is √(16 + 36) = √52 ≈ 7.2 m/s. Among the listed values, 10 m/s arises from adding the components arithmetically, so the correct magnitude is √52 m/s.
Newton's second law states that the net force equals mass times acceleration, so a = F/m. Substituting the values gives a = 20/5 = 4 m/s². The direction is along the applied force because the surface is frictionless.
Inertia is the inherent property of a body to oppose changes in its state of rest or uniform motion, and it is measured by the mass of the body. Momentum and impulse are quantities related to force and motion rather than this resistance.
Action and reaction pairs act on two different bodies and are of the same type. The Earth pulls the book gravitationally, so the reaction is the book pulling the Earth with an equal and opposite gravitational force. The normal force is a separate contact force that merely balances the weight in equilibrium.
Impulse is force multiplied by time, so its unit is the newton-second (N·s). By the impulse–momentum theorem this is equivalent to the unit of momentum, kg·m/s. The two units describe the same physical quantity.
The impulse–momentum theorem gives F·t = change in momentum. The change in momentum is 0.15 × 20 = 3 kg·m/s, so F = 3/0.05 = 60 N. The force acts opposite to the ball's motion, which is why the ball stops.
The limiting friction is f = μN = μmg = 0.3 × 10 × 10 = 30 N. The net horizontal force is 50 − 30 = 20 N, so a = 20/10 = 2 m/s². Always subtract friction from the applied force before applying F = ma.
Resolving forces perpendicular to the incline, the normal reaction balances the component of weight perpendicular to the surface, which is mg cos θ. The component along the incline, mg sin θ, causes the acceleration down the plane. The two components together equal mg in magnitude.
The rate of change of momentum is F = n·m·v, where n is the number of bullets per second. Here n = 240/60 = 4 s⁻¹, so F = 4 × 0.01 × 500 = 20 N. This backward force must be balanced to keep the gun stationary.
For upward acceleration, the normal reaction must exceed the weight to provide the net upward force: N = m(g + a) = 60 × (10 + 2) = 720 N. The apparent weight increases whenever the lift accelerates upward.
The limiting static friction is f = μs·N = μs·mg = 0.5 × 4 × 10 = 20 N. The block just begins to slide when the applied force equals this limiting value. Below this value the friction adjusts to match the applied force exactly.
The net driving force is the weight difference (m₁ − m₂)g and the total mass moved is (m₁ + m₂). Thus a = (3 − 2) × 10 / (3 + 2) = 10/5 = 2 m/s². The tension is the same throughout the string and does not appear in the combined equation.
The impulse needed to stop the body equals its initial momentum, mv = 2 × 6 = 12 kg·m/s. From F·t = mv, the time is t = 12/4 = 3 s. Alternatively, the deceleration is a = 4/2 = 2 m/s² and t = v/a = 3 s.
The normal reaction from the wall equals the horizontal force, 100 N, so limiting friction is 0.4 × 100 = 40 N. The weight of the block is 50 N, which exceeds the maximum friction available. Hence the block cannot be held and slides down the wall.
The tension in the rope must provide the monkey's weight plus the net upward force: T = m(g + a). Setting T equal to the breaking strength, 300 = 20(10 + a), gives 10 + a = 15, so a = 5 m/s². Any larger acceleration would exceed the breaking tension.
The whole system accelerates at a = 20/10 = 2 m/s². The contact force is the only horizontal force on the 4 kg block, so it equals 4 × 2 = 8 N. Equivalently, the 6 kg block experiences 20 N forward and 8 N backward, giving its net 12 N.
Thrust arises from the rate of change of momentum of the ejected gas: F = (dm/dt)·v = 2 × 400 = 800 N. By Newton's third law, the gas pushes the rocket forward with this same force. The mass of the rocket does not enter this thrust calculation directly.
Average power is the work done divided by the time taken, P = W/t. Substituting W = 60 J and t = 4 s gives P = 15 W, which is the rate at which the force transfers energy to the body.
The work-energy theorem follows from Newton's second law: integrating the net force over displacement gives W_net = ΔK = ½mv² − ½mu². Potential energy changes only appear when conservative internal forces are present, so they are not the general statement of the theorem.
Kinetic energy is K = ½mv². With m = 2 kg and v = 3 m/s, K = ½ × 2 × 9 = 9 J. Note that kinetic energy depends on the square of the speed, not linearly on it.
A conservative force does zero net work around any closed path and has an associated potential energy. Gravity, ideal spring force and electrostatic force all satisfy this, but kinetic friction always opposes motion, so its work around a closed loop is negative and nonzero.
Work is the scalar product W = F·d. Here W = (3)(2) + (4)(−1) = 6 − 4 = 2 J. The negative contribution from the j-component reduces the total work because the force component opposes that part of the displacement.
The ball starts and ends at the same point with the same speed, so its kinetic energy does not change over the full journey. By the work-energy theorem, the net work equals the change in kinetic energy, which is zero; gravity's positive work on the way down cancels its negative work on the way up.
Elastic potential energy is U = ½kx². Substituting k = 200 N/m and x = 0.10 m gives U = ½ × 200 × 0.01 = 1 J. The energy depends on the square of the deformation, so small compressions store relatively little energy.
Equal kinetic energy means ½mv_m² = ½(2m)v_2m², so v_m² = 2v_2m². Taking square roots gives v_m = √2 · v_2m, hence v_m : v_2m = √2 : 1. The lighter body must move faster because kinetic energy scales with mass.
The work done against gravity is mgh = 500 × 10 × 10 = 50 000 J. Dividing by the time gives P = 50 000 / 25 = 2000 W. This is the useful power; the actual power drawn by the pump would be larger because of losses.
By energy conservation with friction, the gravitational potential energy mgh is partly dissipated by friction and the rest becomes kinetic energy: ½mv² = mgh − W_f. Solving for v gives v = √(2(gh − W_f)/m). Friction reduces the final speed compared with the frictionless case.
The centripetal force is always perpendicular to the instantaneous velocity, so at every instant F·v = 0 and no work is done. Consequently the kinetic energy stays constant, which is consistent with the constant speed given in the problem.
At the top of the loop the minimum speed satisfies mg = mv²/R, so v² = gR. Energy conservation from the release height H to the loop top (height 2R) gives mgH = mg(2R) + ½mv² = 2mgR + ½mgR = 2.5mgR, hence H = 2.5R. This is the standard minimum-height result for looping the loop.