Physics 9702 · AS & A Level · Stress and strain

Stress and strain — practice question

A spacecraft travelling in deep space changes its velocity by expelling jets of hot gas from its thrusters. Fig. 2.1 gives a side-on view of the spacecraft together with some of its thrusters. Thruster A is situated a distance of $1.6\,\text{m}$ to the left of the spacecraft’s centre of gravity. Thruster C is positioned a distance of $0.40\,\text{m}$ above the centre of gravity of the spacecraft. Thrusters A and B are able to exert forces on the spacecraft only upwards. Thruster C is able to exert a force on the spacecraft only to the left. The thrusters shown all act wholly in the same plane as the centre of gravity.
(a(i))[2]

Thruster A is turned on and exerts an upward force of $60\,\text{N}$ on the spacecraft. Thruster C is also turned on and exerts a force of $220\,\text{N}$ to the left on the spacecraft. Calculate the resultant moment due to these forces about the centre of gravity.

(a(ii))[1]

State and explain whether the forces from A and C are a couple.

(b)[2]

Thrusters A and C are now turned off and the spacecraft is at rest. Thruster B is switched on at time $t_1$, producing a constant force on the spacecraft until the fuel is exhausted at time $t_2$. As the fuel is consumed, the total mass of the spacecraft falls. On Fig. 2.2, sketch how the speed of the spacecraft varies with time from $t_1$ to $t_2$.

(c(i))[2]

The spacecraft now divides into a carrier and a payload as shown in Fig. 2.3. During the split, an average force of $5500\,\text{N}$ acts on the payload for a time of $0.36\,\text{s}$. The velocity of the payload increases by $8.5\,\text{m s}^{-1}$ in the upward direction. The combined mass of the carrier and payload is $2.5 \times 10^3\,\text{kg}$. State the principle of conservation of momentum.

(c(ii))[2]

Show that the mass of the payload is $230\,\text{kg}$.

(c(iii))[3]

Calculate the magnitude of the change in velocity of the carrier.

Worked solution & mark scheme

This 12-mark question has a full step-by-step worked solution and mark scheme. One marking point: Either $60 \times 1.6 (= 96\,\text{N m})$ or $220 \times 0.40 (= 88\,\text{N m})$.

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