Exam-Style Problem

9231 P23 - Jun 2018 - Q2 - 9 marks

6189

Two uniform small spheres \(A\) and \(B\) have equal radii and masses \(4 m\) and \(m\) respectively. Sphere \(A\) is moving with speed \(u\) on a smooth horizontal surface when it collides directly with sphere \(B\) which is at rest. The coefficient of restitution between the spheres is \(e\).
(i) Show that after the collision \(A\) moves with speed \(\frac{1}{5} u(4-e)\) and find the speed of \(B\).

Sphere \(B\) continues to move until it collides with a fixed smooth vertical barrier which is perpendicular to the direction of motion of \(B\). The coefficient of restitution between \(B\) and the barrier is \(\frac{3}{4} e\). After this collision, the speeds of \(A\) and \(B\) are equal.
(ii) Find the value of \(e\).

The spheres \(A\) and \(B\) now collide directly again.
(iii) Determine whether sphere \(B\) collides with the barrier for a second time.

Solution Checked by expert

Answer: After the first collision, \(A\) has speed \(\dfrac15u(4-e)\) and \(B\) has speed \(\boxed{\dfrac45(1+e)u}\). Also \(\boxed{e=\dfrac23}\), and \(B\) does collide with the barrier for a second time.

(i) Take the direction of motion of \(A\) before the first collision as positive. Let the velocities of \(A\) and \(B\) after the first collision be \(v_A\) and \(v_B\).

Conservation of momentum gives

\[4mv_A+mv_B=4mu.\]

\[4v_A+v_B=4u.\]

Newton's law of restitution gives

\[v_B-v_A=e(u-0)=eu.\]

Hence \(v_B=v_A+eu\). Substitute this into the momentum equation:

\[4v_A+(v_A+eu)=4u.\]

Therefore

\[5v_A=(4-e)u,\qquad v_A=\frac15u(4-e).\]

Then

\[v_B=v_A+eu=\frac15u(4-e)+eu=\boxed{\frac45(1+e)u}.\]

(ii) When \(B\) hits the fixed barrier, it reverses direction. The coefficient of restitution at the barrier is \(\frac34e\), so the speed of \(B\) after rebounding from the barrier is

\[\frac34e\,v_B=\frac34e\cdot\frac45(1+e)u=\frac35e(1+e)u.\]

After this collision with the barrier, the speeds of \(A\) and \(B\) are equal. Therefore

\[\frac15(4-e)u=\frac35e(1+e)u.\]

Cancel \(u\) and multiply by \(5\):

\[4-e=3e(1+e).\]

\[4-e=3e+3e^2,\]

which gives

\[3e^2+4e-4=0.\]

Factorise:

\[(3e-2)(e+2)=0.\]

Since \(e\) must be between \(0\) and \(1\),

\[\boxed{e=\frac23}.\]

(iii) When \(e=\frac23\), the speed of \(A\) after the first collision is

\[v_A=\frac15u\left(4-\frac23\right)=\frac23u.\]

The speed of \(B\) after rebounding from the barrier is also \(\frac23u\), but it is moving in the opposite direction. So immediately before the second collision,

\[v_A=\frac23u,\qquad v_B=-\frac23u.\]

Let the velocities after the second collision be \(w_A\) and \(w_B\). Conservation of momentum gives

\[4w_A+w_B=4\left(\frac23u\right)-\frac23u=2u.\]

Newton's law of restitution gives

\[w_B-w_A=\frac23\left(\frac23u-\left(-\frac23u\right)\right)=\frac89u.\]

Thus \(w_B=w_A+\frac89u\). Substitute into the momentum equation:

\[4w_A+w_A+\frac89u=2u.\]

\[5w_A=\frac{10}{9}u,\qquad w_A=\frac29u.\]

Hence

\[w_B=\frac29u+\frac89u=\frac{10}{9}u.\]

This is positive, so \(B\) moves towards the barrier again. Therefore \(B\) collides with the barrier for a second time.