Mathematics 9709/43 — October/November 2010

Walkthrough

The particle is released from rest on a smooth plane inclined at $30^\circ$ to the horizontal. Because the plane is smooth, there is no friction, so the only forces acting are the weight $mg$ vertically downwards and the normal reaction force $R$ perpendicular to the plane.

The normal reaction has no component down the plane. The weight does have a component $mg\sin 30^\circ$ down the plane. Applying Newton's second law parallel to the plane:

so the mass cancels, and

This acceleration is constant, so we can use the suvat equations for motion in a straight line. In both parts the initial velocity is $u=0$.

For part (i), we know $u$, $a$ and $s=0.9$ and want $v$; the equation linking these is $v^2=u^2+2as$. Substituting gives $v^2=9$, so $v=3\ \text{m s}^{-1}$.

For part (ii), we know $u$, $a$ and $t=0.8$ and want $v$; the equation linking these is $v=u+at$. Substituting gives $v=4\ \text{m s}^{-1}$.

An equivalent energy approach for part (i) would be to equate the loss in gravitational potential energy $mg(0.9\sin30^\circ)$ to the gain in kinetic energy $\frac12 mv^2$; the mass cancels and gives the same result.

Key Takeaways

• On a smooth inclined plane the acceleration down the plane is $g\sin\theta$, and it does not depend on the mass of the particle.
• The component of weight along a plane inclined at angle $\theta$ is $mg\sin\theta$.
• Choose the suvat equation that contains the three known quantities and the required unknown.
• Smooth contact means no friction, so only the gravitational component causes acceleration along the plane.

Common Mistakes

• Including a frictional force on a smooth plane. A smooth plane means friction is zero.
• Using the whole weight $mg$ instead of the component $mg\sin30^\circ$ along the plane.
• Trying to use $v=u+at$ in part (i), where the distance is given but the time is not directly known.
• Forgetting to take the positive square root when using $v^2 = u^2 + 2as$.
• Forgetting units or writing the answers without $\text{m s}^{-1}$.

Things to Be Careful About

• Use $g = 10\ \text{m s}^{-2}$ unless the question states another value.
• Keep all quantities in metres and seconds; displacement $0.9$ m and time $0.8$ s are already in consistent units.
• The acceleration is down the plane; take this direction as positive, so the speed is positive.
• If using the energy method, remember the vertical height lost is $s\sin30^\circ$, not the distance along the plane.