Answer:
h = 13.3 m
Explanation:
Given:-
- The mass of ball, mb = 4.80 kg
- The mass of bar, ml = 7.0 kg
- The height from which ball dropped, H = 15.0 m
- The length of bar, L = 6.0 m
- The mass at other end of bar, mo = 5.10 kg
Find:-
The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?
Solution:-
- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:
- The angular momentum for ball dropped before collision ( M1 ):
M1 = mb*vb*(L/2)
Where, vb is the speed of the ball on impact:
- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:
ΔP.E = ΔK.E
mb*g*H = 0.5*mb*vb^2
vb = √2*g*H
vb = ( 2*9.81*15 ) ^0.5
vb = 17.15517 m/s
- The angular momentum of system before collision is:
M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)
M1 = 247.034448 kgm^2 /s
- After collision, the momentum is transferred to the other ball. The momentum after collision is:
M2 = mo*vo*(L/2)
- From principle of conservation of angular momentum the initial and final angular momentum remains the same.
M1 = M2
vo = 247.03448 / (5.10*3)
vo = 16.14604 m/s
- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:
ΔP.E = ΔK.E
mo*g*h = 0.5*mo*vo^2
h = vo^2 / 2*g
h = 16.14604^2 / 2*(9.81)
h = 13.3 m
