Moderate -0.3 This is a straightforward impulse-momentum question requiring application of Pythagoras' theorem and basic impulse principles. The setup is clear, the mathematics is routine (resolving momentum components and using the given speeds), and it's a standard 4-mark question with no conceptual surprises—slightly easier than average for M3.
\includegraphics{figure_1}
A particle \(P\) of mass \(0.3\) kg is moving in a straight line with speed \(4\) m s\(^{-1}\) when it is deflected through an angle \(\theta\) by an impulse of magnitude \(I\) N s. The impulse acts at right angles to the initial direction of motion of \(P\) (see diagram). The speed of \(P\) immediately after the impulse acts is \(5\) m s\(^{-1}\). Show that \(\cos \theta = 0.8\) and find the value of \(I\). [4]
For using \(v_t - u_c = 0\) or for a triangle sketched with sides \(l/0.3, 4\) and \(5\) with angles \(\theta\) and \(90°\) opposite \(l/m\) and \(5\) respectively. For using \(l = m(\Delta v)\) in 'y' direction or \(l = \sqrt{(0.3 \times 5)^2 - (0.3 \times 4)^2}\)
$[5\cos\theta - 4 = 0]$ $\cos\theta = 0.8$ $l = 0.3(5\sin\theta - 0)$ or $\sin\theta = l \div (0.3 \times 5)$ $l = 0.9$ | M1, A1, M1, A1 | For using $v_t - u_c = 0$ or for a triangle sketched with sides $l/0.3, 4$ and $5$ with angles $\theta$ and $90°$ opposite $l/m$ and $5$ respectively. For using $l = m(\Delta v)$ in 'y' direction or $l = \sqrt{(0.3 \times 5)^2 - (0.3 \times 4)^2}$ | M1 |
\includegraphics{figure_1}
A particle $P$ of mass $0.3$ kg is moving in a straight line with speed $4$ m s$^{-1}$ when it is deflected through an angle $\theta$ by an impulse of magnitude $I$ N s. The impulse acts at right angles to the initial direction of motion of $P$ (see diagram). The speed of $P$ immediately after the impulse acts is $5$ m s$^{-1}$. Show that $\cos \theta = 0.8$ and find the value of $I$. [4]
\hfill \mbox{\textit{OCR M3 2011 Q1 [4]}}