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Impacts and Collisions

2022 (Sample Paper) Question 5 (a)

A small smooth sphere, \(P\), of mass \(m\), travels along a horizontal surface at a constant speed of \(8\mbox{ m s}^{-1}\). It collides with another small smooth sphere, \(Q\), of mass \(3m\), which is at rest.

The coefficient of restitution between the spheres is \(\dfrac{3}{8}\).

P8 m s-1Q

(i) Calculate the velocity of \(P\) and the velocity of \(Q\) after impact.

(ii) Calculate, in terms of \(m\), the loss in kinetic energy due to the impact. 


(i) \(\vec{v}_P=-\dfrac{1}{4}\vec{i}\mbox{ m/s}\) and \(\vec{v}_Q=\dfrac{11}{4}\vec{i}\mbox{ m/s}\)

(ii) \(20.625m\)

Sphere P

\(m_P = m\)

Initial Velocity

\(\vec{u}_P=8\vec{i}\mbox{ m/s}\)

Final Velocity


Sphere Q

\(m_Q = 3m\)

Initial Velocity

\(\vec{u}_Q=0 \vec{i}\mbox{ m/s}\)

Final Velocity

\(\vec{v}_Q=v_Q\vec{i}\mbox{ m/s}\)


Conservation of Momentum

\begin{align}m_Pu_P+m_Qu_Q = m_Pv_P+m_Qv_Q\end{align}


\begin{align}m(8)+(3m)(0) = mv_P+(3m)v_Q\end{align}


\begin{align}8 = v_P+3v_Q\end{align}

Law of Restitution






We therefore have the following two equations with two unknowns.

\begin{align}8 = v_P+3v_Q\end{align}






Inserting this into \(-3=v_P-v_Q\) gives:


Therefore, the velocities of sphere \(P\) and sphere \(Q\) after the collision are \(-\dfrac{1}{4}\vec{i}\mbox{ m/s}\) and \(\dfrac{11}{4}\vec{i}\mbox{ m/s}\) respectively.


\begin{align}E_i &= K_{Pi}+K_{Qi}\\&=\frac{1}{2}m_Pu_P^2 + \frac{1}{2}m_Qu_Q^2\\&=\frac{1}{2}(m)(8^2)+ \frac{1}{2}(3m)(0)^2\\&=32m\end{align}


\begin{align}E_f &= K_{Pf}+K_{Qf}\\&=\frac{1}{2}m_Pv_P^2 + \frac{1}{2}m_Qv_Q^2\\&=\frac{1}{2}(m)\left(-\frac{1}{4}\right)^2+ \frac{1}{2}(3m)\left(\frac{11}{4}\right)^2\\&=\frac{m}{32}+\frac{363m}{32}\\&=\frac{364m}{32}\end{align}

Therefore, the total energy lost from the system is


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