## 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 \(\dfrac{3}{8}\).

**(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.Â

**(i)Â **\(\vec{v}_P=-\dfrac{1}{4}\vec{i}\mbox{ m/s}\) 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**

\(\vec{v}_P=v_P\vec{i}\)

**Sphere Q**

\(m_Q = 3m\)

**InitialÂ **__Velocity__

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

**Final Velocity**

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

**(i)**

**Conservation of Momentum**

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

\begin{align}\downarrow\end{align}

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

\begin{align}\downarrow\end{align}

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

**Law of Restitution**

\begin{align}e=-\frac{v_P-v_Q}{u_P-u_Q}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{3}{8}=-\frac{v_P-v_Q}{8-0}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-3=v_P-v_Q\end{align}

We therefore have the following two equations with two unknowns.

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

\begin{align}-3=v_P-v_Q\end{align}

\begin{align}\downarrow\end{align}

\begin{align}11=4v_Q\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_Q=\frac{11}{4}\end{align}

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

\begin{align}v_P&=v_Q-3\\&=\frac{11}{4}-3\\&=-\frac{1}{4}\end{align}

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.

**(ii)**

\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}

and

\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

\begin{align}E_i-E_f&=32m-\frac{364m}{32}\\&=\frac{660m}{32}\\&=20.625m\end{align}