FM1-5 Elastic collisions in two dimensions — coverage pack

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

FM1-5.1 · Oblique impact of smooth elastic spheres and a smooth sphere with a fixed surface. Loss of kinetic energy due to impact.

At impact, resolve velocities parallel and perpendicular to the line of centres for spheres, or into tangential and normal components for a fixed surface.
Smoothness makes the impulse normal to the contact surface, so each sphere's component perpendicular to the line of centres is unchanged.
Apply momentum and Newton's law only to the components along the line of centres; then recombine these with the unchanged transverse components.
Only the normal components contribute to kinetic energy loss in a smooth impact. For sphere-sphere questions here, the radii are equal even if the masses need not be.

1. The unit vector $\mathbf i$ points normally towards a fixed smooth wall and $\mathbf j$ is parallel to it. A sphere arrives with velocity $(3\mathbf i-4\mathbf j)\,\text{m s}^{-1}$ and has coefficient of restitution $1/2$ with the wall. Find its velocity after impact.[2 marks]
Answer
- $(-\tfrac32\mathbf i-4\mathbf j)\,\text{m s}^{-1}$
Method: The tangential component $-4\mathbf j$ is unchanged. The normal component $3\mathbf i$ reverses and is multiplied by $e=1/2$ , becoming $-(3/2)\mathbf i$ . Therefore the new velocity is $(-\tfrac32\mathbf i-4\mathbf j)\,\text{m s}^{-1}$ .

1. Two smooth spheres have equal radius and each has mass $2\,\text{kg}$ . At collision their line of centres is parallel to $\mathbf i$ , and their velocities are $(5\mathbf i+4\mathbf j)\,\text{m s}^{-1}$ and $(-\mathbf i+2\mathbf j)\,\text{m s}^{-1}$ . Given $e=1/3$ , determine both velocities after impact and the loss of kinetic energy.[6 marks]
Answer
- Velocities $(\mathbf i+4\mathbf j)\,\text{m s}^{-1}$ and $(3\mathbf i+2\mathbf j)\,\text{m s}^{-1}$
- Kinetic energy lost $16\,\text{J}$
Method: The $\mathbf j$ components remain $4$ and $2$ . Along $\mathbf i$ , momentum for equal masses gives $v_{1i}+v_{2i}=5+(-1)=4$ , while restitution gives $v_{2i}-v_{1i}=(1/3)(5-(-1))=2$ . Hence $v_{1i}=1$ and $v_{2i}=3$ . Since each mass is $2\,\text{kg}$ , the initial kinetic energy is $5^2+4^2+(-1)^2+2^2=46\,\text{J}$ and the final kinetic energy is $1^2+4^2+3^2+2^2=30\,\text{J}$ . The loss is $16\,\text{J}$ .

At each smooth plane, retain the tangential velocity component and reverse the normal component with its magnitude multiplied by that plane's coefficient of restitution.
Use the post-impact vector from one plane as the pre-impact vector for the next; coefficients of restitution may differ between surfaces.
The stated geometry and velocity direction determine the order of impacts. Check that the new vector actually carries the sphere towards the next plane.
Successive normal scalings generally reduce speed and kinetic energy, but a component parallel to a plane is unchanged during that plane's impact.

1. A sphere approaches a vertical smooth wall with velocity $(6\mathbf i+8\mathbf j)\,\text{m s}^{-1}$ , where $\mathbf i$ is normal to the wall. The coefficient of restitution is $1/2$ . Find its velocity and speed immediately after impact.[3 marks]
Answer
- Velocity $(-3\mathbf i+8\mathbf j)\,\text{m s}^{-1}$
- Speed $\sqrt{73}\,\text{m s}^{-1}$
Method: The $\mathbf j$ component is tangential and remains $8$ . The normal component reverses and becomes $-(1/2)(6)=-3$ . Hence the velocity is $-3\mathbf i+8\mathbf j$ , with speed $\sqrt{(-3)^2+8^2}=\sqrt{73}\,\text{m s}^{-1}$ .

1. A sphere moves inside a right-angled corner with velocity $(8\mathbf i+6\mathbf j)\,\text{m s}^{-1}$ . It strikes first the wall normal to $\mathbf i$ and then the wall normal to $\mathbf j$ . Its coefficient of restitution with each wall is $1/2$ . Find its velocity after the second impact and the fraction of its initial kinetic energy that remains.[5 marks]
Answer
- $(-4\mathbf i-3\mathbf j)\,\text{m s}^{-1}$
- Fraction remaining $1/4$
Method: The first impact changes $8\mathbf i$ to $-4\mathbf i$ and leaves $6\mathbf j$ , giving $-4\mathbf i+6\mathbf j$ . The second leaves $-4\mathbf i$ and changes $6\mathbf j$ to $-3\mathbf j$ . The final speed squared is $4^2+3^2=25$ , compared with initial speed squared $8^2+6^2=100$ . For unchanged mass, the kinetic energy fraction is $25/100=1/4$ .

1. A sphere travels inside a right-angled corner with velocity $(10\mathbf i+6\mathbf j)\,\text{m s}^{-1}$ . It hits the wall normal to $\mathbf i$ with coefficient of restitution $e$ , then the wall normal to $\mathbf j$ with coefficient $1/2$ . After both impacts its direction is $30^\circ$ below the negative $\mathbf i$ direction. Find $e$ and the fraction of its initial kinetic energy lost.[6 marks]
Answer
- $e=3\sqrt3/10$
- Fraction lost $25/34$
Method: After the first impact the velocity is $-10e\mathbf i+6\mathbf j$ ; after the second it is $-10e\mathbf i-3\mathbf j$ . The final direction gives $3/(10e)=\tan30^\circ=1/\sqrt3$ , so $e=3\sqrt3/10$ . The final speed squared is $100e^2+9=27+9=36$ , whereas the initial speed squared is $10^2+6^2=136$ . Thus the fraction of kinetic energy lost is $(136-36)/136=25/34$ .