Question
A child kicks a ball so that it leaves horizontal ground with a velocity of \(28\,\mathrm{m\,s^{-1}}\) at an angle of \(34^\circ\) to the horizontal, as shown in Fig. 1.1.
Air resistance is negligible. The ball leaves the ground at time \(t=0\).
(a) (i) Calculate the horizontal component \(v_{\mathrm{H}}\) and the vertical component \(v_{\mathrm{V}}\) of the velocity of the ball immediately after it has left the ground. (2 marks)
\(v_{\mathrm{H}}=\) __________________________ \(\mathrm{m\,s^{-1}}\)
\(v_{\mathrm{V}}=\) __________________________ \(\mathrm{m\,s^{-1}}\)
(ii) Show that the ball reaches its maximum height at \(t=1.6\,\mathrm{s}\). (1 mark)
____________________________________________
(iii) On Fig. 1.2, sketch the variation of \(v_{\mathrm{H}}\) with time \(t\) between \(t=0\) and \(t=3.2\,\mathrm{s}\). Label your line \(H\). (1 mark)
(iv) On Fig. 1.2, sketch the variation of \(v_{\mathrm{V}}\) with time \(t\) between \(t=0\) and \(t=3.2\,\mathrm{s}\). Assume that velocity in the upward direction is positive. Label your line \(V\). (3 marks)
(b) The total change in momentum of the ball between leaving the ground at \(t=0\) and landing on the ground at \(t=3.2\,\mathrm{s}\) is \(13\,\mathrm{kg\,m\,s^{-1}}\).
(i) Define momentum. (1 mark)
____________________________________________
(ii) Calculate the force that acts on the ball while it is in the air. (2 marks)
force = __________________________ \(\mathrm{N}\)
(iii) Determine the mass of the ball. (1 mark)
mass = __________________________ \(\mathrm{kg}\)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702, 2025–2027):
• 3.1: Momentum and Newton’s laws of motion — parts (b)(i)–(iii)
▶️ Answer/Explanation
(a)(i) Velocity components [2 marks]
Resolve the initial velocity \(28\,\mathrm{m\,s^{-1}}\) into horizontal and vertical components.
Horizontal component:
\(v_{\mathrm{H}}=28\cos34^\circ\)
\(v_{\mathrm{H}}=23\,\mathrm{m\,s^{-1}}\)
Vertical component:
\(v_{\mathrm{V}}=28\sin34^\circ\)
\(v_{\mathrm{V}}=16\,\mathrm{m\,s^{-1}}\)
Answer: \( \boxed{v_{\mathrm{H}}=23\,\mathrm{m\,s^{-1}},\quad v_{\mathrm{V}}=16\,\mathrm{m\,s^{-1}}} \)
(a)(ii) Maximum height [1 mark]
At maximum height, the vertical velocity is zero.
Using \(v=u+at\):
\(0=16-9.81t\)
\(t=\dfrac{16}{9.81}=1.63\,\mathrm{s}\)
Therefore, \(t\approx1.6\,\mathrm{s}\).
Answer: \( \boxed{t=1.6\,\mathrm{s}} \)
(a)(iii) Horizontal velocity-time graph [1 mark]
There is no horizontal acceleration because air resistance is negligible. Therefore, the horizontal velocity remains constant.
The graph is a horizontal line at \(v_{\mathrm{H}}=23\,\mathrm{m\,s^{-1}}\) from \(t=0\) to \(t=3.2\,\mathrm{s}\), labelled \(H\).
(a)(iv) Vertical velocity-time graph [3 marks]
The vertical acceleration is constant and equal to \(-9.81\,\mathrm{m\,s^{-2}}\).
The ball starts with \(v_{\mathrm{V}}=+16\,\mathrm{m\,s^{-1}}\), reaches \(v_{\mathrm{V}}=0\) at \(t=1.6\,\mathrm{s}\), and has \(v_{\mathrm{V}}=-16\,\mathrm{m\,s^{-1}}\) at \(t=3.2\,\mathrm{s}\).
Therefore, the graph is a straight line from \((0,+16)\) through \((1.6,0)\) to \((3.2,-16)\), labelled \(V\).
(b)(i) Momentum [1 mark]
Momentum is defined as the product of mass and velocity.
\(p=mv\)
Answer: \( \boxed{\text{momentum}=\text{mass}\times\text{velocity}} \)
(b)(ii) Force on the ball [2 marks]
Force is the rate of change of momentum:
\(F=\dfrac{\Delta p}{\Delta t}\)
\(F=\dfrac{13}{3.2}\)
\(F=4.06\,\mathrm{N}\)
The change in momentum is downward, so the force is downward.
Answer: \( \boxed{4.1\,\mathrm{N}\text{ downward}} \)
(b)(iii) Mass of the ball [1 mark]
The vertical velocity changes from \(+16\,\mathrm{m\,s^{-1}}\) to \(-16\,\mathrm{m\,s^{-1}}\).
Therefore, the magnitude of the change in velocity is
\(\Delta v=16+16=32\,\mathrm{m\,s^{-1}}\)
Using \(\Delta p=m\Delta v\):
\(13=m(32)\)
\(m=\dfrac{13}{32}=0.406\,\mathrm{kg}\)
Answer: \( \boxed{0.41\,\mathrm{kg}} \)