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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) $$20.625m$$

Solution
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}

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