Questions on forces and Newton's laws

Question 1

State Newton's first law, and what it implies about the forces on a car cruising at a steady 30 m s⁻¹ in a straight line.

Accepted Answer

A body keeps a constant velocity unless a resultant force acts on it. Steady speed in a straight line therefore means zero resultant, so the driving force exactly balances the resistances. The car's speed is irrelevant to that conclusion. B1 for the law, B1 for the zero resultant.

Question 2

A 1000 kg car accelerates at 2 m s⁻² against a total resistance of 500 N. Find the driving force.

Accepted Answer

F = ma applies to the resultant, so D − 500 = 1000 × 2 and D = 2500 N. The engine has two jobs, causing the acceleration and paying off the resistance. Both forces act, and omitting the second gives 2000 N. M1 for the equation of motion, A1 for 2500 N.

Question 3

A horse pulls a cart along a road. State Newton's third law, name the third-law partner of the force the horse exerts on the cart, and explain why that pair of forces does not prevent the cart from accelerating.

Accepted Answer

Newton's third law: when one body exerts a force on a second, the second exerts a force on the first of the same type, equal in magnitude and opposite in direction. The partner of the horse's pull on the cart is the cart's pull on the horse, equal and opposite and acting on the horse. The two forces act on different bodies, so they never appear together in one equation of motion: the cart accelerates because the forces on the cart alone, the horse's pull minus the resistance, have a non-zero resultant. B1 for the law, B1 for the partner named as acting on the horse, B1 for the different-bodies reason.

Question 4

Forces (4i − 2j) N and (2i + 5j) N act on a particle of mass 3 kg. Find the acceleration of the particle as a vector, and its magnitude.

Accepted Answer

The resultant is (4 + 2)i + (−2 + 5)j = (6i + 3j) N, and F = ma gives a = (6i + 3j)/3 = (2i + j) m s⁻². Its magnitude is √(2² + 1²) = √5 = 2.24 m s⁻² (3 s.f.). M1 for adding the forces component by component, A1 for the acceleration vector, A1 for the magnitude.

Question 5

A 6 kg crate is dragged along a floor by a horizontal rope of tension 42 N against friction of 12 N. Find the acceleration, and the normal reaction from the floor.

Accepted Answer

Along the motion the resultant is 42 − 12 = 30 N, so a = 30/6 = 5 m s⁻². Vertically the crate neither rises nor falls, so R = mg = 6 × 9.8 = 58.8 N. The two directions are handled separately, and the vertical equation is an equilibrium even though the crate is accelerating horizontally. M1 for the horizontal resultant, A1 for a = 5, M1 for R = mg, A1 for 58.8 N. Draw the force diagram with all four forces before writing either equation.

Question 6

A person of mass 70 kg stands on scales in a lift. Taking g = 9.8 m s⁻², find the scale reading (the normal reaction) when the lift accelerates upwards at 1.5 m s⁻², and when it accelerates downwards at 1.5 m s⁻².

Accepted Answer

Take upwards as positive. Accelerating up, R − mg = ma gives R = 70(9.8 + 1.5) = 791 N, heavier than the true weight of 686 N. Accelerating down, mg − R = ma gives R = 70(9.8 − 1.5) = 581 N, lighter. The scales read the normal reaction, not the weight, and the reaction is whatever F = ma demands. Both answers must straddle 686 N; if they do not, a sign has gone wrong. M1 for R − mg = ma, A1 for 791 N, M1 for mg − R = ma, A1 for 581 N.

Question 7

A book rests on a table. Identify the Newton's third law partner of the normal reaction on the book, and explain why the reaction and the book's weight are not a third-law pair.

Accepted Answer

The partner is the push of the book down on the table. Same type of force, same magnitude, opposite direction, acting on the other body. The weight's own partner is the gravitational pull of the book on the Earth. Reaction and weight act on the same body and can differ in size, as they do in a lift, so they fail both tests for a third-law pair. B1 for naming the partner, B1 for the weight's own partner, B1 for the reason the pairing fails.

Question 8

A 12 kg sledge is pulled along horizontal ground by a rope inclined at 30° above the horizontal, with tension 40 N, against a horizontal resistance of 8 N. Taking g = 9.8 m s⁻², find the acceleration of the sledge and the normal reaction of the ground.

Accepted Answer

Resolve the tension before writing either equation. Horizontally, 40 cos 30° − 8 = 34.64 − 8 = 26.64 N, so a = 26.64/12 = 2.22 m s⁻². Vertically nothing moves, so R + 40 sin 30° = 12g, giving R = 117.6 − 20 = 97.6 N. M1 for resolving the tension and forming the horizontal equation, A1 for the acceleration, M1 for the vertical equation with the sin component included, A1 for 97.6 N. R has dropped below mg because the rope carries 20 N of the sledge's weight; writing R = mg here would ignore that, and in a friction question the 20 N error would propagate into everything.

Question 9

A ball of mass 0.5 kg is released from rest and falls vertically against a constant air resistance of 1.5 N. Taking g = 9.8 m s⁻², find the acceleration of the ball, and its speed after it has fallen 2 m.

Accepted Answer

Take downwards as positive. The weight is 0.5 × 9.8 = 4.9 N, so the resultant is 4.9 − 1.5 = 3.4 N downwards and a = 3.4/0.5 = 6.8 m s⁻². The resistance is constant, so the acceleration is constant and suvat applies: v² = 0 + 2 × 6.8 × 2 = 27.2, so v = 5.22 m s⁻¹ (3 s.f.). M1 for the equation of motion with both forces, A1 for 6.8, M1 for v² = u² + 2as, A1 for 5.22. The ball does not fall with acceleration g: g is the acceleration when the weight is the only force, and here it is not.

Question 10

Two forces act on a particle of mass 4 kg: a force of 6 N, and a force of 10 N at 60° to it. Find the magnitude of the resultant force and the angle it makes with the 6 N force, and hence find the magnitude of the acceleration of the particle.

Accepted Answer

Take the 6 N force along the x direction and resolve the 10 N force: components 10 cos 60° = 5 and 10 sin 60° = 8.66. The resultant has components (6 + 5, 8.66) = (11, 8.66), so its magnitude is √(11² + 8.66²) = √196 = 14 N, at tan⁻¹(8.66/11) = 38.2° to the 6 N force. Then a = 14/4 = 3.5 m s⁻², in the direction of the resultant. M1 for resolving the 10 N force, A1 for the components of the resultant, M1 for the magnitude, A1 for 14 N and 38.2°, A1 for 3.5. Adding the magnitudes to get 16 N treats the forces as parallel, which they are not; the resultant of two forces at an angle is always shorter than their sum.

Question 11

A cyclist and bicycle have a combined mass of 80 kg. Freewheeling along a level road, the cyclist slows from 12 m s⁻¹ to 8 m s⁻¹ over a distance of 40 m. Modelling the resistance as a constant force, find its magnitude.

Accepted Answer

With the resistance constant the deceleration is constant, so v² = u² + 2as gives 64 = 144 + 80a and a = −1 m s⁻². Then F = ma: the resultant is 80 × (−1) = −80 N, so the resistance has magnitude 80 N, acting against the motion. M1 for the suvat equation, A1 for a = −1, M1 for F = ma with that acceleration, A1 for 80 N.

Question 12

A car of mass 900 kg travelling at 20 m s⁻¹ brakes uniformly to rest in 50 m on a level road. Find the braking force. The same braking force is then applied at 20 m s⁻¹ while the car descends a slope inclined at 5° to the horizontal. Find the new stopping distance.

Accepted Answer

On the level, v² = u² + 2as gives 0 = 400 + 100a, so a = −4 m s⁻² and the braking force is F = 900 × 4 = 3600 N against the motion. On the slope the weight now has a component down the hill of 900 × 9.8 × sin 5° = 769 N, which works with the car. Taking down the slope as positive, 769 − 3600 = 900a gives a = −3.146 m s⁻². Then 0 = 400 + 2(−3.146)s, so s = 63.6 m. M1 A1 for the suvat step and the braking force, M1 for resolving the weight along the slope, A1 for the new acceleration, M1 A1 for the stopping distance.

Question 13

Explain why the statement that a particle is in equilibrium under coplanar forces written in i–j form always yields two scalar equations, and find the force F for which a particle is in equilibrium under (2i − 3j) N, (−5i + j) N and F.

Accepted Answer

Equilibrium says the resultant vector is zero. B1 for that statement. A vector is zero only when both of its components are zero, so the i components must sum to zero and the j components must sum to zero: one vector statement, two scalar equations. B1 for the component argument. Adding the two known forces gives (2 − 5)i + (−3 + 1)j = (−3i − 2j) N, so F must cancel it. M1 for summing the known forces. F = (3i + 2j) N, and both components close: −3 + 3 = 0 and −2 + 2 = 0. A1 for F.

Question 14

A parachutist of mass 80 kg falls vertically. The air resistance R newtons acting on her increases as her speed increases. Explain, using Newton's laws, why her speed eventually becomes constant, and find the value of R when it does. Find also her acceleration at an instant when R = 300 N. Take g = 9.8 m s⁻².

Accepted Answer

Take downwards as positive. Her weight is 80 × 9.8 = 784 N, so the equation of motion is 784 − R = 80a. While R is less than 784 N the resultant is downwards and she speeds up, which by the model increases R and so reduces the resultant. As R approaches 784 N the acceleration approaches zero, and once R = 784 N the resultant is zero: by Newton's first law her velocity then stays constant. So at constant speed R = 784 N. When R = 300 N, 784 − 300 = 80a gives a = 6.05 m s⁻² downwards. B1 for the equation of motion with both forces and the argument that a decreases as R grows, B1 for zero resultant, hence constant velocity by the first law, A1 for R = 784 N, M1 for substituting R = 300 into the equation of motion, A1 for 6.05.

Question 15

A particle of mass m starts from rest and is acted on by a constant resultant force of magnitude F. Show that the distance it travels in time t is Ft²/(2m). A car of mass 1500 kg starts from rest on a level road and covers 100 m in 10 s against a constant resistance of 400 N. Find the driving force.

Accepted Answer

Newton's second law gives a = F/m, and because F is constant the acceleration is constant, so s = ut + ½at² applies with u = 0: s = ½(F/m)t² = Ft²/(2m). For the car, F here is the resultant, so F = 2ms/t² = 2 × 1500 × 100/100 = 3000 N. The driving force D satisfies D − 400 = 3000, so D = 3400 N. M1 for a = F/m with the statement that a constant force gives constant acceleration, M1 for the suvat equation with u = 0, A1 for the result in the required form, M1 for rearranging to find the resultant, A1 for 3000 N, A1 for 3400 N. The derived formula is for the resultant force, and substituting the driving force into it without the resistance gives a car that reaches 100 m late.