Question
(a) State the principle of conservation of momentum. (2 marks)
_________________________________
(b) An object A of mass \(4.0\,\mathrm{kg}\) travels at a velocity of \(6.0\,\mathrm{m\,s^{-1}}\) to the right on a horizontal frictionless surface. It moves towards a second object B of mass \(2.0\,\mathrm{kg}\) that is moving at a velocity of \(3.0\,\mathrm{m\,s^{-1}}\) in the same direction as A, as shown in Fig. 4.1.
Object A collides with object B. The two objects join and move off together with velocity \(v\).
Calculate:
(i) velocity \(v\). (2 marks)
\(v=\) ______________________________ \(\mathrm{m\,s^{-1}}\)
(ii) the percentage of the total initial kinetic energy of the two objects that is transferred to other forms of energy during the collision. (2 marks)
percentage = ______________________________ \(\%\)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
▶️ Answer/Explanation
(a) Principle of conservation of momentum [2 marks]
The total momentum of a system remains constant provided that no resultant external force acts on the system.
Equivalently,
\(\text{total momentum before}=\text{total momentum after}\)
Answer: \( \boxed{\text{Total momentum is conserved in an isolated system.}} \)
(b)(i) Velocity after the collision [2 marks]
The two objects join together, so their combined mass is
\(m_{\mathrm{total}}=4.0+2.0=6.0\,\mathrm{kg}\)
Using conservation of momentum,
\(4.0\times6.0+2.0\times3.0=6.0v\)
\(24+6=6v\)
\(v=5.0\,\mathrm{m\,s^{-1}}\)
Answer: \( \boxed{5.0\,\mathrm{m\,s^{-1}}} \)
(b)(ii) Percentage of kinetic energy transferred [2 marks]
The initial kinetic energy of object A is
\(E_{\mathrm{K,A}}=\dfrac{1}{2}\times4.0\times6.0^2\)
\(E_{\mathrm{K,A}}=72\,\mathrm{J}\)
The initial kinetic energy of object B is
\(E_{\mathrm{K,B}}=\dfrac{1}{2}\times2.0\times3.0^2\)
\(E_{\mathrm{K,B}}=9\,\mathrm{J}\)
Therefore,
\(E_{\mathrm{K,before}}=72+9=81\,\mathrm{J}\)
After the collision, the combined mass is \(6.0\,\mathrm{kg}\) and the velocity is \(5.0\,\mathrm{m\,s^{-1}}\).
\(E_{\mathrm{K,after}}=\dfrac{1}{2}\times6.0\times5.0^2\)
\(E_{\mathrm{K,after}}=75\,\mathrm{J}\)
The kinetic energy transferred to other forms is
\(\Delta E_{\mathrm{K}}=81-75=6\,\mathrm{J}\)
Percentage transferred is
\(\text{percentage}=\dfrac{81-75}{81}\times100\)
\(\text{percentage}=7.4\%\)
Answer: \( \boxed{7\%} \)