spring is hanging at 20m height from the ground . A ball of mass 0.2kg is shot towards the spring with velocity 49m/s such that extension of spring is 3m . What will be the rebound velocity of ball on the ground ?
So the ball is launched upwards at 49 m/s, then it hits the spring and compresses it, then the spring shoots the ball back down again and it hits the ground with velocity $v$. And we're trying to calculate $v$. Is that correct?
@NehalSamee The ball hits the spring at some velocity $v'$. As it compresses the spring the KE of the ball turns to PE until the ball stops. Then the spring expands again and the PE turns back into KE. If no energy is lost, when the ball leaves the spring it willhave the same speed that it had when it ht the spring.
What I did : $v²=u_o ² -2g.20$... v is the velocity at the moment of touching spring .... Then I found kinetic energy and said that it will turn into elastic potential energy ...
Then I found k and then restoring force ... I divided by mass and found the acceleration for spring ...
Ah, I see what you mean. The KE when the ball touches the spring is 0.5 mV^2 (where we can work out V from the height). Then 3mg + 0.5 k (3)^2 = 0.5 mV^2
You see ... The ball compresses up to the limit ... So that's the question said...
If we see it like this way ...After compression when the ball starts to rebound ..It's not in free air ...The restoring force at the very beginning might give a push ...
My book calculated it that way ...They took the acceleration of restoring force into account ..And added with acceleration due to gravity ...
@Abcd @Tanuj @GaurangTandon Do you know any good Online resource to quickly revise formulas and properties of Conic sections (Parabola, Hyperbola, Circle , Ellipse and Straight lines)
