Oblique Impact Example (Engineering Dynamics)

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This tutorial goes over an example of an oblique collision. The coefficient of restitution in this impact is 0.6. We have to use the equation for coefficient of restitution to solve for the component of velocity along the line of impact, and the conservation of linear momentum for the component of velocity that is perpendicular to it.
TYPO: I mislabelled the mass of the left object as 300g, because I do the calculations as if it is 500g. Sorry for any confusion that it caused.
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Thanks for watching, I hope it helps!

[@YazanSY]

I think you used a wrong value for the first mass sir,it should be 0.3 instead of 0.5

[@Engineer4Free]

Thanks for pointing that out, I mislabelled the mass as 300g, when all of the calculations were for 500g. I put a note in the description and stickied your comment so hopefully others don't get confused.

[@Oscar-gx2yf]

Straightforward and clear. Thank you!

[@Engineer4Free]

Glad it was helpful!! More over @ engineer4free.com/dynamics =)

[@alexfish7792]

This was excellent! Thank you.

[@casimirshabnambibijasbeer5966]

hi got -0.5113 for Vbx2
and -0.393 for Vax2

[@Ray-ej3jb]

Hi Do you do requests?
I am working on a boat. (simplistically) I have a hinged metal hatch on the deck (think of a normal door configuration but flat) with a hinge on the long side. I want to raise and lower this by an actuator situated below the hatch (pushing up) either directly or indirectly via perpendicular metalwork
How do I calculate the forces involved around the actuator - in order to spec which unit to buy bearing in mind the following-
1)The mounting point dimension lengths of the actuator to hatch and 'fixed' points are not fixed, (Hinge to hatch mount point=x, hinge to 'fixed' point =y) is not necessarily equal and at 45degs.
2)The length of actuator is not fixed (dependant on the force required dictates which unit to buy and availability),
3)The angle of the actuator to the hatch is not fixed,
4) The weight of the hatch and dimensions are known, (approx 25Kg, hinge side 1100mm and width 700mm)
5) Lastly the mounting point of the actuator to hatch will be offset to allow the hatch to open (rotate) beyond the 90deg point - rotation from horizontal to beyond vertical might be 110deg say with a drop of 75mm to the actuator pivot point on the hatch. This would dictate the location of the 'fixed' mounting point of the actuator.
I have tied using moments, CofG etc but can't pull the values together to be meaningful or come up with a formula enough to iterate the value mentioned above
Thanks in advance

[@kabbashirley2607]

Thanks a lot "engineering made simple" 😉😉😉🤗🤗🤗

[@Engineer4Free]

Most welcome 😊 You can find the full Dynamics playlist here: ua-cam.com/play/PLOAuB8dR35od-BJkNaUWbT8naOb0Y_gC6.html

[@kabbashirley2607]

Thanks

[@lgl_137noname6]

okay,now will you show us an example where both x1 &y1 are different than x2 & y2 ?
thank you !

[@Engineer4Free]

What do you mean by x1,x2, and y1, y2?

[@lgl_137noname6]

@@Engineer4Free If I'm not mistaken in this problem the momentum before and after are equal in the y axis because the plane of impact being coincident with the Y axis. In cases where the impact plane does not coincide with neither axis, would not one assume that momentum Y1 would be different than Y2 ?

[@Engineer4Free]

@@lgl_137noname6 I would recommend setting up a coordinate axis x' and y' that is oriented in line with the line of impact, do all calculations in that frame of reference, and either leave your answer in that frame, or just clearly indicate the directions of velocity etc from a true horizontal line if needed. Similar to solving friction on a slope type problems