Mechanics & materials (projectiles, momentum, work/power, Young modulus) – A-level Physics exam questions with mark scheme

1Core3 marks

A rope pulls a sledge with a force of 25 N at an angle of 35° above the horizontal.

(a) Calculate the horizontal and vertical components of this force. [2]

(b) State the difference between a scalar and a vector quantity. [1]

(a) horizontal = $25\cos35° = 20$ N allow 20.5 N(1)
(a) vertical = $25\sin35° = 14$ N allow 14.3 N(1)
(b) a vector has magnitude and direction; a scalar has magnitude only(1)
Answer: (a) 20 N horizontal, 14 N vertical (b) vector has direction as well as magnitude
Examiner tip: The component adjacent to the given angle uses cos; check the horizontal component is the larger one here since 35° < 45°.

2Core4 marks

A car travelling at 24 m s$^{-1}$ brakes with uniform deceleration and comes to rest in 3.2 s.

(a) Calculate the deceleration of the car. [2]

(b) Calculate the braking distance. [2]

(a) $a = \dfrac{v - u}{t} = \dfrac{0 - 24}{3.2}$(1)
(a) 7.5 m s$^{-2}$(1)
(b) $s = \dfrac{(u+v)}{2}t = \dfrac{24 + 0}{2} \times 3.2$ or any correct suvat route(1)
(b) 38 m allow 38.4 m(1)
Answer: (a) 7.5 m s$^{-2}$ (b) 38 m
Examiner tip: Write the suvat list first (s, u, v, a, t) — it shows which equation to pick and earns method marks even if you slip later.

3Standard6 marks

A ball is kicked from level ground with a velocity of 18 m s$^{-1}$ at 40° above the horizontal. Air resistance can be ignored.

(a) Calculate the time the ball is in the air. [2]

(b) Calculate the horizontal distance travelled before the ball lands. [2]

(c) Air resistance is now taken into account. State and explain its effect on the range and on the maximum height. [2]

(a) vertical component $u_v = 18\sin40° = 11.6$ m s$^{-1}$(1)
(a) $t = \dfrac{2u_v}{g} = \dfrac{2 \times 11.57}{9.81} = 2.4$ s allow 2.36 s(1)
(b) horizontal component $u_h = 18\cos40° = 13.8$ m s$^{-1}$(1)
(b) range $= 13.79 \times 2.36 = 33$ m allow 32.5 m; ecf from (a)(1)
(c) range is reduced because the horizontal velocity decreases (horizontal deceleration due to drag)(1)
(c) maximum height is reduced because drag acts downwards with weight while rising, so greater vertical deceleration(1)
Answer: (a) 2.4 s (b) 33 m (c) both reduced
Examiner tip: Horizontal and vertical motion are independent: only the vertical component goes into the time of flight.

4Standard4 marks

Trolley A, of mass 0.60 kg, moves at 2.5 m s$^{-1}$ towards stationary trolley B, of mass 0.90 kg. The trolleys collide and stick together.

Calculate the velocity of the trolleys after the collision and show whether the collision is elastic or inelastic. [4]

$0.60 \times 2.5 = (0.60 + 0.90)v$(1)
$v = 1.0$ m s$^{-1}$(1)
kinetic energy before = $\frac{1}{2}\times0.60\times2.5^2 = 1.9$ J; after = $\frac{1}{2}\times1.50\times1.0^2 = 0.75$ J(1)
kinetic energy is not conserved (decreases), so the collision is inelastic(1)
Answer: 1.0 m s$^{-1}$; KE falls from 1.9 J to 0.75 J, so inelastic
Examiner tip: Momentum is conserved in every collision; only kinetic energy decides elastic vs inelastic — you must calculate both KE values.

5Standard5 marks

The graph shows how the force on a ball of mass 0.16 kg, initially at rest, varies with time while it is struck by a bat.

(a) Calculate the speed of the ball as it leaves the bat. [3]

(b) A cricket player 'follows through' with the bat. Explain how this increases the speed of the ball. [2]

(a) impulse = area under graph = $\frac{1}{2}\times10\times10^{-3}\times600$(1)
(a) impulse = 3.0 N s (= change in momentum)(1)
(a) $v = \dfrac{3.0}{0.16} = 19$ m s$^{-1}$ allow 18.75; ecf(1)
(b) the bat stays in contact with the ball for longer (increases contact time) allow the force acts on the ball for a longer time(1)
(b) so the impulse $F\Delta t$ / change in momentum of the ball is greater (so greater final speed)(1)
Answer: (a) 19 m s$^{-1}$ (b) longer contact time → greater impulse → greater change in momentum
Examiner tip: Change ms to s before finding the area — the impulse is in N s.

6Hard5 marks

A car of mass 1200 kg travels at a constant speed of 15 m s$^{-1}$ up a straight road inclined at 4.0° to the horizontal. The total resistive force on the car is 450 N.

(a) Calculate the useful power output of the car's engine. [4]

(b) The engine's input power is 64 kW. Calculate its efficiency. [1]

(a) component of weight down the slope = $1200 \times 9.81 \times \sin4.0° = 821$ N(1)
(a) driving force = $821 + 450 = 1270$ N (constant speed so resultant force zero)(1)
(a) $P = Fv = 1271 \times 15$(1)
(a) $1.9\times10^{4}$ W allow 19 kW(1)
(b) efficiency = $\dfrac{19.1}{64} = 0.30$ / 30% ecf(1)
Answer: (a) $1.9\times10^{4}$ W (b) 30%
Examiner tip: At constant speed the driving force equals the TOTAL opposing force — include the component of weight down the slope, not just friction.

7Standard5 marks

A student determines the Young modulus of steel using a long wire. The wire has an original length of 2.00 m and a diameter of 0.50 mm. A load of 60 N produces an extension of 2.9 mm.

(a) Calculate the Young modulus of steel. [4]

(b) Explain why a long, thin wire is used in this experiment. [1]

(a) $A = \pi \left(0.25\times10^{-3}\right)^2 = 1.96\times10^{-7}$ m$^2$(1)
(a) stress = $\dfrac{60}{1.96\times10^{-7}} = 3.1\times10^{8}$ Pa(1)
(a) strain = $\dfrac{2.9\times10^{-3}}{2.00} = 1.45\times10^{-3}$ or correct substitution into $E = \dfrac{FL}{A\Delta L}$ (covers both stress and strain marks)(1)
(a) $E = \dfrac{\text{stress}}{\text{strain}} = 2.1\times10^{11}$ Pa ecf from their area(1)
(b) to give a larger extension (for a given load), so the percentage uncertainty in the extension is smaller allow thin wire gives a larger stress / long wire gives a larger extension for the same strain(1)
Answer: (a) $2.1\times10^{11}$ Pa (b) larger extension → smaller percentage uncertainty
Examiner tip: Use the radius, not the diameter, in πr² — using 0.50 mm makes the answer four times too small.

8Hard5 marks

A ballistic pendulum is used to measure the speed of a pellet. A pellet of mass 12 g is fired horizontally into a wooden block of mass 1.50 kg hanging from light strings. The pellet embeds in the block, which swings upwards so that its centre of mass rises by 0.18 m.

(a) Calculate the speed of the pellet before it hits the block. [4]

(b) Explain why the speed of the pellet cannot be found by equating its initial kinetic energy to the gain in gravitational potential energy of the block and pellet. [1]

(a) after the collision, $\frac{1}{2}mv^2 = mgh$ so $v = \sqrt{2 \times 9.81 \times 0.18}$(1)
(a) $v = 1.88$ m s$^{-1}$(1)
(a) conservation of momentum: $0.012\,u = (0.012 + 1.50) \times 1.88$(1)
(a) $u = 240$ m s$^{-1}$ allow 236.7–237 m s$^{-1}$ (g = 9.8 or 9.81, more s.f.); allow ecf from their speed after the collision(1)
(b) the collision is inelastic: (most of) the pellet's kinetic energy is transferred to internal (thermal) energy / deformation / sound, so kinetic energy is not conserved in the collision allow 'kinetic energy is not conserved in the collision' or 'collision is inelastic so energy is lost to heat/sound'; reject 'energy is not conserved'(1)
Answer: (a) 240 m s$^{-1}$ (b) collision is inelastic, so kinetic energy is not conserved in it
Examiner tip: Two stages, two principles: energy conservation for the swing, momentum conservation for the collision — never energy for the collision.

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