Single particle, Newton's second law – vector (2D forces)

[In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors.] A particle \(A\) of mass 0.5 kg is at rest on a smooth horizontal plane. At time \(t = 0\), two forces, \(\mathbf{F}_1 = (-3\mathbf{i} + 2\mathbf{j})\) N and \(\mathbf{F}_2 = (p\mathbf{i} + q\mathbf{j})\) N, where \(p\) and \(q\) are constants, are applied to \(A\). Given that \(A\) moves in the direction of the vector \((\mathbf{i} - 2\mathbf{j})\),

1. show that \(2p + q - 4 = 0\) [4] Given that \(p = 5\)
2. Find the speed of \(A\) at time \(t = 4\) seconds. [5]

A particle \(P\), of mass 3 kg, moves under the action of two constant forces (6\(\mathbf{i}\) + 2\(\mathbf{j}\)) N and (3\(\mathbf{i}\) - 5\(\mathbf{j}\)) N.

1. Find, in the form (\(a\mathbf{i}\) + \(b\mathbf{j}\)) N, the resultant force \(\mathbf{F}\) acting on \(P\). [1]
2. Find, in degrees to one decimal place, the angle between \(\mathbf{F}\) and \(\mathbf{j}\). [3]
3. Find the acceleration of \(P\), giving your answer as a vector. [2]

The initial velocity of \(P\) is (-2\(\mathbf{i}\) + \(\mathbf{j}\)) m s\(^{-1}\).

1. Find, to 3 significant figures, the speed of \(P\) after 2 s. [5]

A particle \(P\) of mass 0.4 kg is moving under the action of a constant force \(\mathbf{F}\) newtons. Initially the velocity of \(P\) is \((6\mathbf{i} - 2\mathbf{j})\) m s\(^{-1}\) and 4 s later the velocity of \(P\) is \((-14\mathbf{i} + 2\mathbf{j})\) m s\(^{-1}\).

1. Find, in terms of \(\mathbf{i}\) and \(\mathbf{j}\), the acceleration of \(P\). [3]
2. Calculate the magnitude of \(\mathbf{F}\). [3]