Need help with a Physics problem please! Much appreciated?

Question

Hello guys, I really need help with this physics problem. It has been bugging me for a while and I keep getting stuck. If you can't solve it, if you know where I can find a site that has solutions this would be a great help. Thanks!

A l.0 kilogram object is moving horizontally with a velocity of 10 meters per second, as shown above, when it makes a glancing collision with the lower end of a bar that was hanging vertically at rest before the collision. For the system consisting of the object and bar, linear momentum is not conserved in this collision, but kinetic energy is conserved. The bar, which has a length l of 1.2 meters and a mass m of 3.0 kilograms, is pivoted about the upper end. Immediately after the collision the object moves with speed v at an angle  relative to its original direction. The bar swings freely, and after the collision reaches a maximum angle of 90° to the vertical. The moment of inertia of the bar about the pivot is Ibar = ml²/3 Ignore friction.

a. Determine the angular velocity of the bar immediately after the collision.

b. Determine the speed v of the l kilogram object immediately after the collision.

c. Determine the magnitude of the angular momentum of the object about the pivot just before the collision.

d. Determine the angle .

Answer 1

Ok, I have an idea, but if it's not correct, please, let me know ok, so I can check it out later :

The linear momentum is not conserved, but the kinetic energy is conserved, then :

Initial kinetic energy : 1*100 / 2 = 50 Joules

Final kinetic energy : Kinetic energy of the bar plus kinetic energy of the object :

But the kinetic energy of the bar, is the rotational kinetic energy :

I*w^2 / 2

then :

Final kinetic energy : I*w^2/2 + 1/2*1*v^2

I = moment of inertia of the bar

w = angular velocity

v = velocity of the object after the collision :

50 = I*w^2/2 + 1/2*1*v^2

Now, for the bar :

The bar reaches a maximum angle of 90º, then we can use the conservation of energy, to find the angular velocity :

I*w^2/2 = mg*L

L = lenght of the bar

I = 1/3*mL^2

1/6*3*(1.2)^2*w^2 = 3*9.8*1.2

w^2 = 49

w = 7 rad/s >>> angular velocity of the bar

Now replacing that value on :

50 = I*w^2/2 + 1/2*1*v^2

50 = 1/6*3*(1.2)^2*49 + v^2/2

v = 5.42 m/s >>> velocity of the object

The moment of inertia of the bar :

I = 1/6*3*(1.2)^2*49 = 35.28 kg*m^2

The angular momentum of the object, that could be consider as a point object :

L = r*p = r*(m*v)

r = position respect the pivot

m = mass

v = velocity

L = 10*1*1.2 = 12 kg*m^2/s

Hope that helped you