A particle $P$ with mass $0.3\ \text{kg}$ is placed on a smooth plane that is inclined at $30^{\circ}$ to the horizontal, and it is let go from rest. $P$ travels $2.5\ \text{m}$ down the slope before arriving on a horizontal plane. Its speed does not change as $P$ reaches the horizontal plane. A particle $Q$ of mass $0.2\ \text{kg}$ is at rest on the horizontal plane, $1.5\ \text{m}$ from the lower end of the inclined plane (see diagram). $P$ then collides directly with $Q$.
(a)[5]
You are told that the horizontal plane is smooth and that, after the collision, $P$ keeps moving in the same direction at a speed of $2\ \text{m s}^{-1}$. Determine the speed of $Q$ after the collision.
(b)[5]
Now suppose the horizontal plane is rough, and when $P$ and $Q$ collide they stick together and move with speed $1.2\ \text{m s}^{-1}$. Determine the coefficient of friction between $P$ and the horizontal plane.
Worked solution & mark scheme
This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Apply Newton’s second law parallel to plane: $0.3g\sin 30 = 0.3a$” …