Part A: Find the frequency of the motion after the collision. Part B: Find the m

Question

Part A: Find the frequency of the motion after the collision.

Part B: Find the maximum angular displacement of the motion after the collision.

In the figure (Figure 1) the upper ball is released from rest, collides with the stationary lower ball, and sticks to it. The strings are both 52.5cm long. The upper ball has mass is 1.80kg and it is initially 10.0 cm higher than the lower ball, which has mass 3.00kg . Part A: Find the frequency of the motion after the collision. Part B: Find the maximum angular displacement of the motion after the collision.

Explanation / Answer

velocity of upper ball just before collision is found by Work Energy theorem

v=sqrt(2gh) sqrt means square root of

v=sqrt(20*0.1)=1.414m/s

conserving momentum

1.8*1.414=(1.8+3)*v

v=0.530m/s

w(angular speed)=v/l= 0.530/0.525= 1.009rad/s

frequency= w/2pie =1.009/6.28= 0.16 hz

let us find the max hieght it goes using Work energy theorem

v=sqrt(2gh)

0.530*0.530=2*10*h

h=0.014m

if x is angular displacement

cosx=l-h/l

= 52.5-1.4/52.5

=0.9733

x=cos-(0.9733)

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---