A 100 g steel ball and a 200 g steel ball each hang from 4.0-m-long strings. At

Question

A 100 g steel ball and a 200 g steel ball each hang from 4.0-m-long strings. At rest, the balls hang side by side, barely touching. The 100 g ball is pulled to the left until the angle between its string and vertical is 30 . The 200 g ball is pulled to a 30 angle on the right. The balls are released so as to collide at the very bottom of their swings.

Part A

To what angle does 100 g ball rebound?

Express your answers using two significant figures.

Part B

To what angle does 200 g ball rebound?

Express your answers using two significant figures.

Explanation / Answer

The lenght of the string are 4 m and they made angle of 30 degree with the vertical initially therefore
the potential energy stored in the ball will be
=mass*(4 -4Cos30)*g
So potential eneery in the first ball = 0.2*9.81*(4-4Cos30) = 1.0514 J
Potential energy of the second ball =0.5257 J
Now with energy conservation
There velocity before collision
First ball = (1/2)mV12 = mgh
V1  = 3.2425 m/s
Similarly V2 = 3,2425 m/s
Now considering it as perfect elastic coliision
m1V1 - m2V2 = m1v1 +m2v2
0.32425 = 0.2 v1 + 0.1v2
v2 = (0.32425 - 0.2v1)/0.1 --------(1)
And
(1/2)m1V12 +(1/2)m2V22  = (1/2)m1v12 + (1/2)m2v22
Putting the value of v2 in the above equation from equation 1
3.154 = 0.2v12 + 10 (0.32425 - 0.2v1)2
On solving we get
v1 = 3.243 m/s
v2 = 3.2436 m/s
Therefore angle will be 30 degree for both

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.