Elastic Collision Homework: Find Max Angle of Swing

[armolinasf]

Homework Statement

There is a ball rolling on a frictionless horizontal surface of mass m and with velocity 5m/s. It collides elastically with a block mass 3m that is initially hanging at rest from a 50 cm wire that is hanging from the ceiling.Find the maximum angle through which the block swings after it is hit

Homework Equations

v=initial velocity=5
Rolling ball=m1=m
Hanging block = m2=3m

Velocity of hanging block after collision: (2m1/(m1+m2))v

Velocity of rolling ball after collision: ((m1-m2)/(m1+m2))v

The Attempt at a Solution

So using the two equations above I get that the velocity of the block is 2.5m/s and velocity of the ball is -2.5m/s. So far this agrees with the fact that this is an elastic collision the knietic energy afterwards is equal to the knietic energy before the collision.

So to find the angle I'm thinking that I need to find how far the block travels in the x direction, that way I can take the arctangent of the length of the string divided by the distance x to find an angle W.

I use the work energy theorem:W=Fx=max=K after. The masses will cancel so I have ax=K after. But the acceleration is just dv/dt which is a change in velocity over a very short time interval. My thinking is the collision is short enough to be called dt and since it goes from rest to 2.5 m/s that a=2.5.

X would then be (.5)(3)(2.5^2)/(2.5)=3.75 meters

angle W would then be arctan(375/50)=82 degrees

This does not seem reasonable since it would mean that the ball would leave the ground and some energy would be converted into gravitation potential energy...help is appreciated

 

[gneill]

The block WILL leave the surface, since it's suspended like a pendulum by the 50cm wire. That means energy WILL be converted into gravitational PE...

 

[armolinasf]

How would I calculate the upward movement if the initial momentum is entirely in the x direction?

 

[gneill]

How would I calculate the upward movement if the initial momentum is entirely in the x direction?

Use a conservation of energy approach. You've already got the block's initial post-collision velocity...

 

[armolinasf]

The initial kinetic energy was equal to 12.5 J and the kinetic energy of the block is 9.4 and the ball is 3.1 unless one of those velocities is wrong there isn't any more energy in the system to be accounted for...

Also I am curious, is the logic behind taking a=2.5 is sound?

 

[gneill]

The initial kinetic energy was equal to 12.5 J and the kinetic energy of the block is 9.4 and the ball is 3.1 unless one of those velocities is wrong there isn't any more energy in the system to be accounted for...

Also I am curious, is the logic behind taking a=2.5 is sound?

Since you don't have the actual masses of the ball or block, you can't really put a numerical KE value to either. (although you could say that the "specific KE" of the block is 12.5 J/kg)

The KE of the block is going to be converted to gravitational PE as it swings upwards on the end of the wire.

Regarding your value of acceleration, no, you really can't put a figure to the acceleration without knowing the actual time duration of the impact event. For nearly perfect undeformable objects this time would be vanishingly small, and the acceleration would approach infinite! Better to think in terms of impulse in those situations.

But you've got the KE of the block, and a way forward with energy conservation. So onward-ho!

 

[armolinasf]

Ok, So I set the KE=12.5=mgH and solved for H since this is where all the KE is converted into gravitational PE. H=.318. Since this is the distance above the ground I subtracted it from the length of the rope. This length then was the adjacent side of a right triangle formed by this side and the hypotenuse equal to the length of the string. arcos(50-32)/50 gave an angle of 67 degrees.

 

[gneill]

Ok, So I set the KE=12.5=mgH and solved for H since this is where all the KE is converted into gravitational PE. H=.318. Since this is the distance above the ground I subtracted it from the length of the rope. This length then was the adjacent side of a right triangle formed by this side and the hypotenuse equal to the length of the string. arcos(50-32)/50 gave an angle of 67 degrees.

Keep in mind that the mass of the block is 3m...

 

[armolinasf]

My mistake KE=12.5 was supposed to be KE=9.4...I should write it out: .5mv^2=mgh ==> H=v^2/2g. v=2.5, So 2.5^2/2(9.8)=.32