Exam-Style Problem

9231 P21 - Nov 2017 - Q3 - 10 marks

6166

Three uniform small smooth spheres \(A, B\) and \(C\) have equal radii and masses \(m, k m\) and \(m\) respectively, where \(k\) is a constant. The spheres are moving in the same direction along a straight line on a smooth horizontal surface, with \(B\) between \(A\) and \(C\). The speeds of \(A, B\) and \(C\) are \(2 u, u\) and \(\frac{4}{3} u\) respectively. The coefficient of restitution between any pair of the spheres is \(\frac{1}{2}\). After sphere \(A\) has collided with sphere \(B\), sphere \(B\) collides with sphere \(C\).
(i) Find an inequality satisfied by \(k\).

(ii) Given that \(k=2\), show that after \(B\) has collided with \(C\) there are no further collisions between any of the three spheres.

Solution Checked by expert

Answer: \(\boxed{k<\dfrac72}\). When \(k=2\), the final speeds are \(u\), \(\dfrac{17u}{12}\), and \(\dfrac{3u}{2}\), so no further collisions occur.

(i) Take the common direction of motion as positive.

Let the velocities of \(A\) and \(B\) after the first collision be \(v_A\) and \(v_B\).

Conservation of momentum for the collision between \(A\) and \(B\) gives

\[mv_A+kmv_B=m(2u)+km(u).\]

Cancel \(m\):

\[v_A+kv_B=2u+ku.\]

Using Newton's law of restitution with \(e=\frac12\),

\[v_B-v_A=\frac12(2u-u)=\frac{u}{2}.\]

\[v_A=v_B-\frac{u}{2}.\]

Substitute this into the momentum equation:

\[v_B-\frac{u}{2}+kv_B=2u+ku.\]

Hence

\[(k+1)v_B=\left(k+\frac52\right)u,\]

and so

\[v_B=\frac{(2k+5)u}{2(k+1)}.\]

For \(B\) to collide with \(C\), \(B\) must move faster than \(C\), whose speed is \(\frac43u\). Therefore

\[\frac{2k+5}{2(k+1)}>\frac43.\]

\[3(2k+5)>8(k+1).\]

This simplifies to

\[6k+15>8k+8,\]

\[7>2k.\]

Therefore

\[\boxed{k<\frac72}.\]

(ii) Now let \(k=2\). From the formula above,

\[v_B=\frac{(2(2)+5)u}{2(2+1)}=\frac{9u}{6}=\frac32u.\]

Also, since \(v_B-v_A=\frac{u}{2}\),

\[v_A=\frac32u-\frac12u=u.\]

Now consider the collision between \(B\) and \(C\). Let their velocities after this collision be \(w_B\) and \(v_C\).

Conservation of momentum gives

\[2mw_B+mv_C=2m\left(\frac32u\right)+m\left(\frac43u\right).\]

Cancel \(m\):

\[2w_B+v_C=3u+\frac43u=\frac{13u}{3}.\]

Newton's law of restitution gives

\[v_C-w_B=\frac12\left(\frac32u-\frac43u\right)=\frac12\left(\frac{u}{6}\right)=\frac{u}{12}.\]

\[v_C=w_B+\frac{u}{12}.\]

Substitute into the momentum equation:

\[2w_B+w_B+\frac{u}{12}=\frac{13u}{3}.\]

Thus

\[3w_B=\frac{13u}{3}-\frac{u}{12}=\frac{52u-u}{12}=\frac{51u}{12}.\]

Therefore

\[w_B=\frac{17u}{12}.\]

Then

\[v_C=w_B+\frac{u}{12}=\frac{18u}{12}=\frac32u.\]

After the two collisions, the speeds are

\[A:u,\qquad B:\frac{17u}{12},\qquad C:\frac32u.\]

Since \(u<\frac{17u}{12}\), sphere \(A\) cannot catch sphere \(B\). Also, since \(\frac{17u}{12}<\frac32u\), sphere \(B\) cannot catch sphere \(C\). Hence there are no further collisions.