IBDP Physics- A.2 Forces and momentum- IB Style Questions For SL Paper 2 -FA 2025
Question
A ball is connected to a string that is attached to the ceiling of a truck. The truck accelerates to the right on a horizontal road. The string makes an angle of \(30^\circ\) with the vertical.
(a)
(i) Draw a labelled free-body diagram of the forces on the ball.
(ii) Determine the acceleration of the truck.
\(\boxed{\hspace{10cm}}\)
(b) The truck now moves at constant speed up an incline that makes an angle of \(10^\circ\) to the horizontal. State and explain the angle between the string and the vertical.
\(\boxed{\hspace{10cm}}\)
(c) On another occasion, the truck enters a horizontal semicircular road of radius \(75\,\mathrm{m}\). The coefficient of static friction between the road and tyres is \(0.6\).
Calculate the maximum speed at which the truck can travel on this road.
\(\boxed{\hspace{10cm}}\)
(d) The truck runs out of fuel just as it enters the semicircle with speed \(15\,\mathrm{m\,s^{-1}}\). The resultant force opposing the motion has magnitude \(520\,\mathrm{N}\). The mass of the truck is \(1400\,\mathrm{kg}\). Service station is at the other end of the semicircle.
Determine whether the truck will be able to reach the service station.
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Most-appropriate topic codes (IBDP Physics HL 2025):
▶️ Answer/Explanation
(a)(i) Correct Answer:
The free-body diagram should show two forces acting on the ball:
• Weight \(mg\), vertically downward.
• Tension \(T\), directed along the string towards the ceiling.
The vertical component of the tension balances the weight.
(a)(ii) Correct Answer: \( \boxed{5.7\,\mathrm{m\,s^{-2}}} \)
Resolve the tension vertically:
\(T\cos30^\circ=mg\)
Resolve horizontally in the direction of acceleration:
\(T\sin30^\circ=ma\)
Dividing the two equations gives
\(\tan30^\circ=\dfrac{a}{g}\)
Therefore,
\(a=g\tan30^\circ\)
\(a=(9.8)\tan30^\circ\)
\(a\approx5.7\,\mathrm{m\,s^{-2}}\)
(b) Correct Answer: \( \boxed{0^\circ} \)
The truck is moving at constant speed, so its acceleration is zero.
Therefore, the resultant force on the ball is zero and the forces are balanced.
The vertical component of the tension balances the weight, so there is no horizontal component of tension.
Hence, the string hangs vertically and the angle between the string and the vertical is \( \boxed{0^\circ} \).
(c) Correct Answer: \( \boxed{21\,\mathrm{m\,s^{-1}}} \)
For maximum speed, the frictional force provides the required centripetal force:
\(F_{\mathrm{friction}}=F_{\mathrm{centripetal}}\)
The maximum frictional force is
\(F_{\mathrm{friction}}=\mu mg\)
and
\(F_{\mathrm{centripetal}}=\dfrac{mv^2}{r}\)
Therefore,
\(\mu mg=\dfrac{mv^2}{r}\)
\(v=\sqrt{\mu gr}\)
\(v=\sqrt{(0.6)(9.8)(75)}\)
\(v\approx21\,\mathrm{m\,s^{-1}}\)
(d) Correct Answer: \( \boxed{\text{Yes, the truck reaches the service station.}} \)
The initial kinetic energy of the truck is
\(E_{\mathrm{k}}=\dfrac{1}{2}mv^2\)
\(E_{\mathrm{k}}=\dfrac{1}{2}(1400)(15)^2\)
\(E_{\mathrm{k}}=1.575\times10^5\,\mathrm{J}\)
The distance travelled around the semicircle is
\(s=\pi r=\pi(75)\)
\(s\approx236\,\mathrm{m}\)
The work done against the resistive force is
\(W=Fs\)
\(W=(520)(75\pi)\)
\(W\approx1.23\times10^5\,\mathrm{J}\)
Since
\(1.575\times10^5\,\mathrm{J}>1.23\times10^5\,\mathrm{J}\)
the truck has sufficient kinetic energy to overcome the resistive forces and reach the service station. Hence, \( \boxed{\text{the truck will reach the service station}} \).