Exam-Style Problem

9231 P21 - Jun 2019 - Q11E - 12 marks

6126

Question 11 EITHER alternative.

A particle \(P\), of mass \(m\), is able to move in a vertical circle on the smooth inner surface of a sphere with centre \(O\) and radius \(a\). Points \(A\) and \(B\) are on the inner surface of the sphere and \(AOB\) is a horizontal diameter. Initially, \(P\) is projected vertically downwards with speed \(\sqrt{\frac{21}{2}ag}\) from \(A\) and begins to move in a vertical circle. At the lowest point of its path, vertically below \(O\), the particle \(P\) collides with a stationary particle \(Q\), of mass \(4m\), and rebounds. The speed acquired by \(Q\), as a result of the collision, is just sufficient for it to reach the point \(B\).

(i) Find the speed of \(P\) and the speed of \(Q\) immediately after their collision.

In its subsequent motion, \(P\) loses contact with the inner surface of the sphere at the point \(D\), where the angle between \(OD\) and the upward vertical through \(O\) is \(\theta\).

(ii) Find \(\cos\theta\).

Solution

Answer: Immediately after the collision, \(P\) has speed \(\boxed{\dfrac{3}{\sqrt2}\sqrt{ag}}\), \(Q\) has speed \(\boxed{\sqrt{2ag}}\), and \(\boxed{\cos\theta=\dfrac56}\).

(i) First find the speed of \(P\) just before the collision at the lowest point. Call this speed \(u_P\).

From \(A\) to the lowest point, \(P\) drops a vertical distance \(a\). By conservation of energy,

\[\frac12m u_P^2=\frac12m\left(\frac{21}{2}ag\right)+mga.\]

\[u_P^2=\frac{21}{2}ag+2ag=\frac{25}{2}ag,\]

and hence

\[u_P=\frac5{\sqrt2}\sqrt{ag}.\]

The speed given to \(Q\) is just enough for \(Q\) to rise from the lowest point to \(B\), a vertical height \(a\). If this speed is \(v_Q\), then

\[\frac12(4m)v_Q^2=4mga.\]

Therefore

\[v_Q^2=2ag,\qquad v_Q=\sqrt{2ag}.\]

During the collision, momentum is conserved along the tangent at the lowest point.

Take the initial direction of motion of \(P\) as positive. Since \(P\) rebounds, its velocity after the collision is in the negative direction. Let its speed after the collision be \(v_P\). Then

\[m u_P=-m v_P+4m v_Q.\]

\[v_P=4v_Q-u_P.\]

Substitute the values:

\[v_P=4\sqrt{2ag}-\frac5{\sqrt2}\sqrt{ag}=\left(\frac8{\sqrt2}-\frac5{\sqrt2}\right)\sqrt{ag}.\]

Thus

\[v_P=\frac3{\sqrt2}\sqrt{ag}.\]

(ii) Now consider the later motion of \(P\) after it has rebounded. Let \(V\) be its speed at the point \(D\), where it loses contact.

At \(D\), the point is \(a\cos\theta\) above \(O\). From the lowest point to \(D\), the vertical rise is

\[a+a\cos\theta=a(1+\cos\theta).\]

Using energy from just after the collision to \(D\),

\[\frac12mV^2=\frac12m\left(\frac92ag\right)-mga(1+\cos\theta).\]

\[V^2=\frac92ag-2ag(1+\cos\theta).\]

This simplifies to

\[V^2=\left(\frac52-2\cos\theta\right)ag.\]

At the instant \(P\) loses contact, the normal reaction is zero. Resolving radially towards the centre gives

\[\frac{mV^2}{a}=mg\cos\theta.\]

Hence

\[V^2=ag\cos\theta.\]

Equate the two expressions for \(V^2\):

\[\left(\frac52-2\cos\theta\right)ag=ag\cos\theta.\]

Cancel \(ag\):

\[\frac52=3\cos\theta.\]

Therefore

\[\boxed{\cos\theta=\frac56}.\]