OCR MEI M1 (Mechanics 1)

A rock of mass 8 kg is acted on by just the two forces \(-80\)k N and \((-\mathbf{i} + 16\mathbf{j} + 72\)k\()\) N, where \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane and k is a unit vector vertically upward.

1. Show that the acceleration of the rock is \(\left(\frac{1}{8}\mathbf{i} + 2\mathbf{j}\right)\) k\()\) ms\(^{-2}\). [2]

The rock passes through the origin of position vectors, O, with velocity \((\mathbf{i} - 4\mathbf{j} + 3\)k\()\) m s\(^{-1}\) and 4 seconds later passes through the point A.

1. Find the position vector of A. [3]
2. Find the distance OA. [1]
3. Find the angle that OA makes with the horizontal. [2]

Fig. 4 shows the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) in the directions of the cartesian axes \(Ox\) and \(Oy\), respectively. O is the origin of the axes and of position vectors. \includegraphics{figure_1} The position vector of a particle is given by \(\mathbf{r} = 3t\mathbf{i} + (18t^2 - 11)\mathbf{j}\) for \(t \geq 0\), where \(t\) is time.

1. Show that the path of the particle cuts the \(x\)-axis just once. [2]
2. Find an expression for the velocity of the particle at time \(t\). Deduce that the particle never travels in the \(\mathbf{j}\) direction. [3]
3. Find the cartesian equation of the path of the particle, simplifying your answer. [3]

In this question, the unit vectors \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\) are in the directions east and north. Distance is measured in metres and time, \(t\), in seconds. A radio-controlled toy car moves on a flat horizontal surface. A child is standing at the origin and controlling the car. When \(t = 0\), the displacement of the car from the origin is \(\begin{pmatrix} 0 \\ -2 \end{pmatrix}\) m, and the car has velocity \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}\) ms\(^{-1}\). The acceleration of the car is constant and is \(\begin{pmatrix} -1 \\ 1 \end{pmatrix}\) ms\(^{-2}\).

1. Find the velocity of the car at time \(t\) and its speed when \(t = 8\). [4]
2. Find the distance of the car from the child when \(t = 8\). [4]

At time \(t\) seconds, a particle has position with respect to an origin O given by the vector $$\mathbf{r} = \begin{pmatrix} 8t \\ 10t^2 - 2t^3 \end{pmatrix},$$ where \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\) are perpendicular unit vectors east and north respectively and distances are in metres.

1. When \(t = 1\), the particle is at P. Find the bearing of P from O. [2]
2. Find the velocity of the particle at time \(t\) and show that it is never zero. [3]
3. Determine the time(s), if any, when the acceleration of the particle is zero. [3]

A particle of mass 5 kg has constant acceleration. Initially, the particle is at \(\begin{pmatrix} -1 \\ 2 \end{pmatrix}\) m with velocity \(\begin{pmatrix} 2 \\ -3 \end{pmatrix}\) ms\(^{-1}\); after 4 seconds the particle has velocity \(\begin{pmatrix} 12 \\ 9 \end{pmatrix}\) ms\(^{-1}\).

1. Calculate the acceleration of the particle. [2]
2. Calculate the position of the particle at the end of the 4 seconds. [3]
3. Calculate the force acting on the particle. [2]

A toy boat moves in a horizontal plane with position vector \(\mathbf{r} = x\mathbf{i} + y\mathbf{j}\), where \(\mathbf{i}\) and \(\mathbf{j}\) are the standard unit vectors east and north respectively. The origin of the position vectors is at O. The displacements \(x\) and \(y\) are in metres. First consider only the motion of the boat parallel to the \(x\)-axis. For this motion $$x = 8t - 2t^2.$$ The velocity of the boat in the \(x\)-direction is \(v_x\) ms\(^{-1}\).

1. Find an expression in terms of \(t\) for \(v_x\) and determine when the boat instantaneously has zero speed in the \(x\)-direction. [3]

Now consider only the motion of the boat parallel to the \(y\)-axis. For this motion $$v_y = (t - 2)(3t - 2),$$ where \(v_y\) ms\(^{-1}\) is the velocity of the boat in the \(y\)-direction at time \(t\) seconds.

1. Given that \(y = 3\) when \(t = 1\), use integration to show that \(y = t^3 - 4t^2 + 4t + 2\). [4]

The position vector of the boat is given in terms of \(t\) by \(\mathbf{r} = (8t - 2t^2)\mathbf{i} + (t^3 - 4t^2 + 4t + 2)\mathbf{j}\).

1. Find the time(s) when the boat is due north of O and also the distance of the boat from O at any such times. [4]
2. Find the time(s) when the boat is instantaneously at rest. Find the distance of the boat from O at any such times. [5]
3. Plot a graph of the path of the boat for \(0 \leq t \leq 2\). [3]