CAIE A-Level Physics 5.2 Gravitational Potential and Kinetic Energy

Practise deriving and applying gravitational and kinetic-energy relations, then combining them in conservation, braking and acceleration problems.

Exam points

derive and use gravitational potential energy changes ΔEP = mgΔh in a uniform field
derive and use kinetic energy EK = 1/2 mv² and connect it to changes in motion
combine gravitational and kinetic energy in conservation-of-energy, braking and acceleration problems

Covered syllabus points

5.2.1—W = Fs, the formula ∆EP = mg∆h for gravitational potential energy changes in 5.2.2—The formula ∆EP = mg∆h for gravitational potential energy changes in a 5.2.3—The equations of motion, the formula for kinetic energy EK = 1/2 mv2 5.2.4—EK = 1/2 mv2

[Maximum number: 2]

A mass m moves a vertical distance $\Delta h$ in a uniform gravitational field and gains gravitational potential energy $\Delta E_{\mathrm{p}}$ . The acceleration of free fall is g.

Use the concept of work done to show that

\Delta E_{\mathrm{P}}=m g \Delta h .

work (done) = force × displacement

M1

( force =m g and distance $=\Delta h)$

(so) work (done) $=m g \Delta h$ and work $=\Delta E_{(\mathrm{P})}$ (so $\Delta \mathrm{E}_{(\mathrm{P})}=m g \Delta h$ )

A1

[Maximum number: 5]

A pendulum consists of a solid sphere suspended by a string from a fixed point P , as shown in Fig. 3.1.

Fig. 3.1 (not to scale)

The sphere swings from side to side. At one instant the sphere is at its lowest position X , where it has kinetic energy 0.86 J and momentum 0.72 Ns in a horizontal direction. A short time later the sphere is at position Y , where it is momentarily stationary at a maximum vertical height h above position X.

The string has a fixed length and negligible weight. Air resistance is also negligible.

(a)

Show that the mass of the sphere is 0.30 kg .

[ 3 ]

p=m v or 0.72=m v

C1

$E=\frac{1}{2} m v^{2}$ or $0.86=1 / 2 m v^{2}$

C1

$(m=) 0.72^{2} /(2 \times 0.86)=0.30(\mathrm{~kg})$

or
$\begin{aligned} & v=2 E_{\mathrm{K}} / p & v=(0.86 \times 2) / 0.72=2.4 \text { (to } 2 \text { s.f.) } & m=0.72 / 2.4=0.30(\mathrm{~kg}) \end{aligned}$

A1

(b)

Calculate height h.

\begin{aligned} & h= \\ & \text { m [2] } \end{aligned}

[ 2 ]

$(\Delta) E=m g(\Delta) h$

C1

$\begin{aligned} h =0.86 /(0.30 \times 9.81) =0.29 \mathrm{~m} \end{aligned}$

A1

[Maximum number: 1]

A student attempts to derive the formula for kinetic energy $E_{\mathrm{K}}$ . She begins by considering an object of mass m which is initially at rest. A constant force F applied to the object causes it to accelerate to final velocity v in displacement s. The kinetic energy gained by the object is equal to the work done on the object by the force F.

Which equation would the student not need in order to derive the formula for $E_{\mathrm{K}}$ ?

F=m a

W=F s

$E=\frac{1}{2} F s$

$v^{2}=u^{2}+2 a s$

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Covered syllabus points

5.2 Gravitational potential energy and kinetic energy question 1

5.2 Gravitational potential energy and kinetic energy question 2

Question (a)

Question (b)

5.2 Gravitational potential energy and kinetic energy question 3