Whoosh
Problem
A ball $A$ of mass $2m$ is at the top of a frictionless chute at height $5h$ above the ground and another ball $B$ of mass $m$ is at rest on the horizontal surface at the end of the chute at height $h$ above the ground. Ball $A$ starts to slide from rest down the chute and hits ball $B$ which is projected, after the impact, off the end of the chute. If the coefficient of restitution between the balls is $0.5$, how far does ball $B$ go horizontally before it hits the ground?
Getting Started
Student Solutions
Thank you Adam from St Alban's School and Andrei from Tudor Vianu National College, Bucharest, Romania for your solutions to this problem.
Ball A moves down a frictionless inclined plane. Its velocity $v_A$ at the base of the plane is determined from the law of conservation of energy:
From the previous equation, $v_A$ is determined as: $v_A=2\sqrt{2gh}$.
After that, ball $A$ collides with ball $B$ at rest. We assume ball $A$ is moving horizontally with velocity $v_A$ when it collides with ball $B$. After the collision the horizontal velocities of balls $A$ and $B$ are $u_A$ and $u_B$ respectively. The collision is characterised by the coefficient of restitution k:
Teachers' Resources
For a falling particle the gravitational acceleration is given by