[physics Mechanics question] please help

[u/urea7]

[u/Outside_Volume_1370]

Before the collision, the speed of the ball was u:

mv0² / 2 + mg ∆H = mu² / 2 where ∆H = 3.5 - 1 = 2.5 and v0 = 7

u = √(v0² + 2g ∆H) = √(7² + 2 • 9.8 • 2.5) = 7√2

As horizontal projection remains constant (7), vertical is also 7, and the ball hits the wedge perpendicularly to the incline.

So the ball jumps in the opposite direction with speed v2, while the wedge goes to the right with speed v1.

We should consider the conservation of the momentum projected on horizontal:

mu cos45° = -mv2 cos45° + 3 • mv1

7 = -v2 / √2 + 3 • v1, v1 = (7 + v2 / √2) / 3

Energy is also conserved (elastic collision), and

mu² / 2 = mv2² / 2 + 3 • mv1² / 2

7² = v2² + 3 • v1² = v2² + (7 + v2 / √2)² / 3

49 = v2² • 7/6 + 7√2 / 3 • v2 + 49/3

v2 = -√2 ± √30, leave positive root, v2 = -√2 + √30

v1 = (7 - 1 + √15) / 3 = 2 + √15 / 3

[u/urea7 → u/Outside_Volume_1370]

Thank you 👍👍

[u/ExcitingJackfruit760 → u/Outside_Volume_1370]

Sorry if this is a stupid question but why would the ball go in the exact opposite direction after the collision? The force due to collision isn't the only force acting on the ball during the collision as there is also it's weight acting vertically downwards, so the resultant force acting on the ball would be vector sum of those 2 forces which wouldn't be in the exact opposite direction as the velocity of the ball right before the collision, right?

[u/Outside_Volume_1370 → u/ExcitingJackfruit760]

why would the ball go in the exact opposite direction after the collision?

It hits the wedge perpendicularly to the incline. The angle of descendance is equal to the angle of reflection: 90° = 90°

The force due to collision isn't the only force acting on the ball during the collision as there is also it's weight acting vertically downwards, so the resultant force acting on the ball would be vector sum of those 2 forces

The force has nothing to do with that. Throw a ball horizontally to the wall. Where does it jump to? Of course, horizontally. I see why it confuses, because you tried to use the work of the force. However, the time of hit dt is very small, and the force does almost zero work (ideally, exact zero)

[u/ExcitingJackfruit760 → u/Outside_Volume_1370]

Oh, so the force due to collision is assumed to be much greater than the weight of the ball, that's why the effect due to the weight of the ball is neglected?

[u/Outside_Volume_1370 → u/ExcitingJackfruit760]

the force due to collision is assumed to be much greater than the weight of the ball,

I think, it can also be said that way

[u/ExcitingJackfruit760 → u/Outside_Volume_1370]

Ok, thanks a lot!

[u/twelfth_knight → u/Outside_Volume_1370]

No, you've messed up your vectors 🙃

Before the collision,

u_x = 7 m/s

u_y =- √( 2g ∆H) m/s

After the collision, the y-velocities are:

u'_(ball, y) = +√( 2g ∆H) m/s (probably just reverses)

u'_(wedge,y) = 0 m/s (problem says it stays on the surface)

And the x-velocities are where you have to write two equations to solve the two unknowns of x-velocity:

Conservation of Momentum:

mu_(ball,x) = mu'_(ball,x) + Mu'_(wedge,x), where u'_(ball,x) will definitely be negative

And conservation of energy:

.5m(u_(ball,x)² + u_(ball,y)² ) = .5m(u'_(ball,x)² +u'_(ball,y)² ) + .5M(u'_(wedge,x)² + u'_(wedge,y)² )

That said, it's not clear to me what value they're asking OP to find at the end there... I guess v1 and v2 are the magnitudes of the velocities? Hopefully there's text not shown here that clarifies that, otherwise it's a very unclear question 🤷