FM1-4 Elastic collisions in one dimension — coverage pack

2 specification leaves · notes, questions, answers and worked methods

FM1-4.1 · Direct impact of elastic spheres. Newton's law of restitution. Loss of kinetic energy due to impact.

For a direct impact, conserve signed momentum along the line of centres when the external impulse during impact is negligible.
Newton's law gives speed of separation $=e\times$ speed of approach, with $0\le e\le1$ for the collisions in this course.
Solve the momentum and restitution equations together; check afterwards that the calculated velocities really describe separation rather than continued approach.
Momentum is conserved in an isolated impact, but kinetic energy is conserved only when $e=1$ ; calculate loss as kinetic energy before minus kinetic energy after.

1. Sphere $A$ approaches stationary sphere $B$ at $7\,\text{m s}^{-1}$ . After their direct impact, $A$ continues in the same direction at $2\,\text{m s}^{-1}$ . The coefficient of restitution is $0.4$ . Find the velocity of $B$ .[2 marks]
Answer
- $4.8\,\text{m s}^{-1}$ in $A$ 's original direction
Method: Newton's law gives $v_B-v_A=e(u_A-u_B)$ . Hence $v_B-2=0.4(7-0)=2.8$ , so $v_B=4.8\,\text{m s}^{-1}$ .

1. A $2\,\text{kg}$ sphere travelling at $6\,\text{m s}^{-1}$ strikes a stationary $3\,\text{kg}$ sphere directly. The coefficient of restitution is $1/2$ . Find both velocities after impact and the kinetic energy lost.[5 marks]
Answer
- Velocities $0.6\,\text{m s}^{-1}$ and $3.6\,\text{m s}^{-1}$ in the original direction
- Kinetic energy lost $16.2\,\text{J}$
Method: Momentum gives $12=2v_A+3v_B$ . Restitution gives $v_B-v_A=(1/2)(6)=3$ . Solving yields $v_A=0.6$ and $v_B=3.6$ . Initial kinetic energy is $\tfrac12(2)(6^2)=36\,\text{J}$ ; final kinetic energy is $\tfrac12(2)(0.6^2)+\tfrac12(3)(3.6^2)=19.8\,\text{J}$ . The loss is $36-19.8=16.2\,\text{J}$ .

1. A $1\,\text{kg}$ sphere moving at $5\,\text{m s}^{-1}$ collides directly with a stationary $2\,\text{kg}$ sphere. The collision loses $16/3\,\text{J}$ of kinetic energy. Determine the coefficient of restitution and both velocities after impact.[6 marks]
Answer
- $e=3/5$
- The $1\,\text{kg}$ sphere has velocity $-1/3\,\text{m s}^{-1}$ and the $2\,\text{kg}$ sphere has velocity $8/3\,\text{m s}^{-1}$
Method: Conservation of momentum gives $5=v_A+2v_B$ , and Newton's law of restitution gives $v_B-v_A=5e$ . Solving simultaneously, $v_A=(5-10e)/3$ and $v_B=(5+5e)/3$ . The kinetic energy loss is $\tfrac12(1)(5^2)-\tfrac12(1)v_A^2-\tfrac12(2)v_B^2$ ; setting this equal to $16/3$ and simplifying gives $1-e^2=16/25$ , so $e^2=9/25$ . Since $0\le e\le1$ , $e=3/5$ . Substituting back gives $v_A=-1/3$ and $v_B=8/3\,\text{m s}^{-1}$ .

Treat each impact separately and use the velocities immediately before that impact as the next restitution equation's approach velocities.
For direct impact with a fixed plane, the normal velocity reverses and its speed is multiplied by $e$ ; the plane's effectively infinite mass means its velocity stays zero.
After every impact, compare positions and velocities to decide which objects can meet next; an algebraic collision result alone does not prove another impact occurs.
Keep one positive direction throughout a sequence. Reversing the axis between impacts is a common source of incorrect restitution signs.

1. A sphere moving normally towards a fixed smooth wall at $5\,\text{m s}^{-1}$ has coefficient of restitution $0.7$ with the wall. State its velocity immediately after impact, taking motion towards the wall as positive.[2 marks]
Answer
- $-3.5\,\text{m s}^{-1}$
Method: The sphere separates from the fixed wall at $e$ times its approach speed. Its speed becomes $0.7(5)=3.5\,\text{m s}^{-1}$ and the direction reverses, so the velocity is $-3.5\,\text{m s}^{-1}$ .

1. Identical spheres $A$ and $B$ lie on a line, with $B$ between $A$ and a wall. Sphere $A$ moves towards stationary $B$ at $6\,\text{m s}^{-1}$ . Their coefficient of restitution is $1/2$ , and $B$ 's coefficient with the wall is $2/3$ . Find the velocities just after the first sphere-sphere impact and explain why $A$ and $B$ collide again after $B$ rebounds from the wall.[5 marks]
Answer
- After the first impact, $v_A=1.5\,\text{m s}^{-1}$ and $v_B=4.5\,\text{m s}^{-1}$ towards the wall
- $B$ rebounds at $3\,\text{m s}^{-1}$ towards $A$ , so they approach one another
Method: For equal masses, momentum gives $v_A+v_B=6$ and restitution gives $v_B-v_A=(1/2)(6)=3$ . Hence $v_A=1.5$ and $v_B=4.5\,\text{m s}^{-1}$ towards the wall. At the wall, $B$ reverses with speed $(2/3)(4.5)=3\,\text{m s}^{-1}$ . It then moves towards $A$ , while $A$ is still moving towards the wall, so their separation decreases and a second collision occurs.

1. Three identical spheres $A$ , $B$ and $C$ are arranged in that order on a straight line. Initially $A$ moves towards the other two at $8\,\text{m s}^{-1}$ while $B$ and $C$ are at rest. Every impact has coefficient of restitution $1/2$ . Find the velocities after all impacts have occurred, and justify that no further collision follows.[7 marks]
Answer
- Final velocities $v_A=13/8\,\text{m s}^{-1}$ , $v_B=15/8\,\text{m s}^{-1}$ and $v_C=9/2\,\text{m s}^{-1}$ , all in $A$ 's original direction
- No further collision occurs because the velocities increase from the rear sphere to the front sphere
Method: For equal masses, each impact conserves the sum of the two velocities and makes their separation speed half their approach speed. The first $A$ - $B$ impact gives $(v_A,v_B)=(2,6)$ . The following $B$ - $C$ impact gives $(v_B,v_C)=(3/2,9/2)$ . Since $A$ at speed $2$ is behind $B$ at speed $3/2$ , they collide again. Solving $v_A+v_B=2+3/2=7/2$ and $v_B-v_A=(1/2)(2-3/2)=1/4$ gives $v_A=13/8$ and $v_B=15/8$ . Now $13/8<15/8<9/2$ , so no rear sphere can catch the one ahead.