Maths › Further Mechanics 1 › Oblique impact and impact with a smooth surface
Oblique impact and impact with a smooth surface
Resolve into the direction of the impact and the direction along it. One component is multiplied by −e; the other is left completely alone. Everything else follows from that split.
Builds on Successive impacts and impacts with a wall and Impulse and momentum as vectors.
IN THIS TOPIC
- Resolve a velocity into components along and perpendicular to a smooth surface.
- Apply restitution to the perpendicular component only, and find the outgoing speed and direction.
- Calculate the kinetic energy lost in an oblique impact.
WHAT YOU PROBABLY THINK
In an oblique impact with a smooth wall, the ball leaves at the same angle to the wall as it arrived, like light in a mirror.
Split it in two
A smooth surface can exert no force along itself, so the component of velocity parallel to the surface is completely unchanged. The perpendicular component is reversed and reduced by the factor e, exactly as in a direct impact. That is the whole method.
Because only one of the two components shrinks, the outgoing path is always closer to the surface than the incoming one. The mirror picture in the opening claim would need e = 1, which is the one case where the angles do match.
WORKED EXAMPLE
A ball off a wall
A ball moving at 10 m/s strikes a smooth wall at 60° to the wall. The coefficient of restitution is 0.5. Find the speed and direction afterwards.
Parallel component = 10 cos60° = 5 m/s, unchanged.
Perpendicular component = 10 sin60° = 8.66 m/s, becoming 0.5 × 8.66 = 4.33 m/s the other way.
Speed = √(25 + 18.75) = 6.61 m/s, and the angle to the wall is arctan(4.33/5) = 40.9°.
Angles and energy
The angle to the surface always shrinks, since the parallel component stays and the perpendicular one is cut. A ball skimming in nearly along a wall leaves at almost the same angle, while one arriving nearly perpendicular loses most of its speed.
Kinetic energy is a scalar, so the loss is found from the speeds: compare ½mv² before and after, or take the ratio of the squared speeds. Only the perpendicular component contributes to the loss, so it is e² of that component's energy that survives while the parallel part is untouched.
YOUR TURN
The energy lost
For the impact above, find the percentage of kinetic energy lost.
Show the working
Speed before 10, so the energy is proportional to 100.
Speed after 6.61, so the energy is proportional to 25 + 18.75 = 43.75.
Fraction remaining = 43.75/100, so the loss is 56.25%.
Checking directly: the perpendicular part keeps e² = 0.25 of its 75, that is 18.75, and the parallel 25 is untouched.
THE EXAM BIT
- Draw the surface and mark the two directions before resolving anything.
- Apply e to the perpendicular component only; leaving it on both is the standard error.
- Recombine with Pythagoras for the speed and an inverse tangent for the angle, and say which line the angle is measured from.
- For the energy, work with squared speeds; the mass cancels if a percentage is wanted.
CHECK YOURSELF
A ball hits a smooth floor at 8 m/s at 30° to the floor with e = 0.5. Find the two components after impact.
Show a hint
Resolve along and perpendicular to the floor.
Show the answer
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A smooth surface leaves the parallel component untouched and multiplies the perpendicular component by −e.
So the outgoing path always hugs the surface more closely than the incoming one, and the energy lost comes entirely from the perpendicular component.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Resolve a velocity into components along and perpendicular to a smooth surface.
- Apply restitution to the perpendicular component only, and find the outgoing speed and direction.
- Calculate the kinetic energy lost in an oblique impact.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.